Compactness and the norm that survives smoothing

A bounded operator can have large effects on finitely many modes and still be compact. The part that cannot be removed by a compact perturbation is measured by its essential norm. For an ordinary order-zero operator on one canonical graph, this norm is exactly the limiting size of its normalized principal symbol at high frequency. No homogeneous leading term is required.

Original programme exposition, completed proofs and examples: GPT-6 Astra (OpenAI), Ultra, 4 October 2026. This component and its original figure are dedicated under CC0. Earlier linked components retain their own terms.

T0. Precise statements and supplied prerequisites

All symbol classes are ordinary $S^m_{1,0}$. Operators act on complex half-densities, possibly with finite-rank Hermitian bundle coefficients. On a Hilbert space pair define

A compact operator means a bounded linear operator whose image of the unit ball has compact closure. T1 below proves every compact-operator fact used.

For a compact-kernel pseudodifferential operator $P$ of order zero, with principal matrix symbol $p$, we prove

For an order-zero Fourier integral operator $A$ whose kernel is compact and whose relation is one canonical graph $C$, let $\mu_C$ be the symplectic density pulled back from either cotangent space. Write its principal symbol as $\sigma(A)=b\,\mu_C^{1/2}$, where $b$ takes values in the unitary Maslov line and the bundle of coefficient maps. Then

Representatives are compactly based on the relevant support; outside it they may be taken zero. Changes by one lower symbol order do not affect these limits. Over compact base sets, different cotangent norms are comparable, so the choice of radial norm also does not affect the limits.

For a local canonical graph, with possibly several branches, we prove the full local compactness criterion: every compact input/output localization of $A$ is compact if and only if $b\to0$ uniformly at infinity over each compact base subset. Formula (T3), however, is asserted for one graph; Example T4 shows why it cannot be applied to separate branches and then maximized. Proper support gives the usual compact-input and local-space interpretation, detailed in T7.

Use the exact earlier proofs:

• WM W0–W6: graph $L^2$ boundedness, smooth transpose, preliminary weak norm and all real Sobolev orders;
• AC A0–A9: clean analytic composition and its complete principal-symbol formula;
• MC G4–G7 and CC C4–C5: the Maslov and density maps;
• P1 B1–B5, P2 O0–O6, P3 M0–M8 and L0–L3: measure, completeness, density, Fourier estimates and ordinary calculus;
• PS2 and PS5–PS6: geometric symbol estimates, finite phase partitions and the symbol kernel;
• K2–K4: symbol inverses, proper parametrices and conic localization.

Finite-dimensional compactness, basis extension, smooth inverses, roots, cutoffs and integration are the exact U001 proofs already bound to these components. No spectral theorem for compact operators, Rellich theorem, general Hilbert duality theorem or positivity inequality is assumed.

The diagram records T1–T6. The two scalar curves are actual ordinary symbols from Example T1; their different high-frequency limits distinguish compactness from boundedness.

T1. The Hilbert-space facts, with proofs

We use the inner product linear in its first argument. Here are the needed elementary facts for complete inner-product spaces, including the $L^2$ spaces constructed in P3.

Representation and adjoints. A nonzero bounded linear functional $\ell$ has the form $\ell(f)=\langle f,h\rangle$ for one $h$. Indeed minimize the norm on the nonempty closed affine set $\ell(z)=1$. Its infimum $d$ is positive since $1\leq\|\ell\|\|z\|$. For a minimizing sequence the parallelogram identity gives

Completeness supplies a minimizing $z$ with $\ell(z)=1$. For $v\in\ker\ell$, minimize $\|z+tv\|^2$ for real and purely imaginary $t$; its linear terms vanish, giving $\langle v,z\rangle=0$. Now $f-\ell(f)z\in\ker\ell$, so $\ell(f)=\langle f,z\rangle/\|z\|^2$. This proves the claim with $h=z/\|z\|^2$; uniqueness follows by testing the difference of two representatives against itself. The zero functional has representative zero.

Apply this to $f\mapsto\langle Af,g\rangle$ to construct $A^*g$. Uniqueness proves linearity, and Cauchy–Schwarz gives $\|A^*g\|\leq\|A\|\|g\|$. Taking suprema over two unit vectors, and using $\|v\|=\sup_{\|w\|\leq1}|\langle v,w\rangle|$, proves

For the second identity one inequality is submultiplicativity; the other follows from $\|Af\|^2=\langle A^*Af,f\rangle$ on unit vectors. This construction agrees with an already specified adjoint, by uniqueness.

Compactness and finite nets. In a complete metric space a set has compact closure exactly when it has a finite $\varepsilon$-net for every $\varepsilon>0$. One direction follows by covering its compact closure by $\varepsilon$-balls. For the other, successive finite $2^{-j}$-covers extract from any sequence a diagonal Cauchy subsequence; completeness gives a limit. To pass from this sequential assertion to open-cover compactness, suppose an open cover has no positive radius such that each ball of that radius about the closure lies in one cover member. Choose offending centers for radii $1/j$; a convergent subsequence has a limit in a cover member, and eventually its small balls lie there, a contradiction. A finite net for such a positive radius then gives a finite subcover. This proves the assertion.

It follows that finite-rank bounded operators are compact: their bounded images lie in a finite-dimensional space, whose closed bounded balls are compact by U001. A norm limit of compact operators is compact: approximate the operator to within $\varepsilon/2$ on the unit ball and use a finite $\varepsilon/2$-net for that approximant. Sums and compositions with bounded operators are compact by the same nets and boundedness estimates.

Every compact operator $K$ is a norm limit of finite-rank operators. Take a finite $\varepsilon$-net in $K$'s unit-ball image and project orthogonally onto its span $E$. Such a projection is obtained by the finite Gram–Schmidt procedure: subtract earlier orthogonal components and normalize each nonzero remainder; discard zero remainders. Pythagoras shows that projection minimizes distance to the span. Hence

Adjoint norm equality now shows that $K^*$ is also compact, since it is the norm limit of the finite-rank $K^*P_E$.

If $u_j$ is bounded and tends weakly to zero, then $Ku_j\to0$ in norm for compact $K$. Otherwise compactness gives a subsequence converging in norm to a nonzero vector. But $\langle Ku_j,v\rangle=\langle u_j,K^*v\rangle\to0$, so that norm limit pairs to zero against itself, a contradiction.

Adjoint products and the essential norm. Put $P=A^*A$. If $A$ is compact then $P$ is compact. Conversely, let $P$ be compact and take a finite $\varepsilon$-net $Pu_i$, with $\|u_i\|\leq1$, in its unit-ball image. For each unit $u$,

for an appropriate $i$. Thus $Au_i$ is a finite $\sqrt{2\varepsilon}$-net and $A$ is compact. Moreover

To prove the lower inequality, subtract any compact $K$ from $A$. The difference between $A^*A$ and $(A-K)^*(A-K)$ is compact; use (T5) and take the infimum. For the reverse inequality choose compact $R$ with $\|P-R\|<\|P\|_{\mathrm{ess}}+\varepsilon$. Replacing $R$ by $(R+R^*)/2$ preserves this bound. By (T6) and self-adjointness there is a finite-dimensional orthogonal projection $Q$ such that $\|R(I-Q)\|<\varepsilon$. For $u=(I-Q)u$, $\|Au\|^2\leq(\|P\|_{\mathrm{ess}}+2\varepsilon)\|u\|^2$. The operator $AQ$ has finite rank, proving the reverse inequality as $\varepsilon\downarrow0$. Finally, (T1) is zero exactly for compact operators, by their proved closure in operator norm.

T2. Square-integrable kernels and negative orders

If a matrix kernel $K(x,y)$ has square-integrable entries, its operator is bounded by its matrix-entry $L^2$ norm. Apply Cauchy–Schwarz in $y$, sum the finitely many entries and integrate in $x$. It is compact. To prove this last statement, first approximate each entry in $L^2$ by compact smooth functions, using P3 M7 in the product dimension. On a containing rectangle, uniform continuity approximates each such function uniformly by a function constant on finitely many small product rectangles. Extend those step functions by zero; their $L^2$ error tends to zero because the containing rectangle has finite volume. Each resulting kernel is a finite sum $c_{ij}\mathbf1_{E_i}(x)\mathbf1_{F_j}(y)$ and has finite rank. The norm estimate and T1 prove compactness. This also covers a kernel whose support is not compact but whose square integral is finite.

In a chart let $p\in S^{-\varepsilon}$, $\varepsilon>0$, have compact base support. Choose a fixed smooth radial cutoff $\rho=1$ near zero, supported in a ball, and set $p_R(x,\xi)=\rho(\xi/R)p(x,\xi)$. The kernel of $\operatorname{Op}(p_R)$ is square integrable: Parseval in the difference variable gives

For each fixed finite set of ordinary order-zero symbol seminorms,

When derivatives do not hit the cutoff this is the bound $\langle\xi\rangle^{-\varepsilon}$ on $|\xi|\geq cR$. If $j$ frequency derivatives hit it, their factor is $O(R^{-j})$ where $|\xi|\asymp R$, and the remaining symbol derivative has the compensating factor $\langle\xi\rangle^j$. Base derivatives do not hit the cutoff. The product rule proves (T10) for every finite list. P1 B4's finite-seminorm $L^2$ estimate proves norm convergence of $\operatorname{Op}(p_R)$ to $\operatorname{Op}(p)$. T1 therefore makes the latter compact.

P2 O4–O5 gives the symbol-plus-smooth-kernel representation of a compact proper operator. Its smooth compact kernel is compact by the first paragraph. A finite chart partition thus proves that every compact-kernel ordinary operator of strictly negative order is compact on $L^2$. No compact Sobolev embedding is used in this proof.

T3. Packets that measure a symbol at infinity

Fix $g\in C_c^\infty(\mathbb R^n)$, $\|g\|_2=1$. Let $x_j$ range in a fixed compact set, let $|\xi_j|=R_j\to\infty$, and put $h_j=R_j^{-1/2}$. For unit coefficient vectors $v_j,w_j$, possibly in different finite dimensions, set

Substitution gives unit norms. Both sequences tend weakly to zero. For any $f\in L^2$, Cauchy–Schwarz bounds its pairing with either packet by its $L^2$ norm on a ball of radius $Ch_j$. This bound tends to zero uniformly in the centers: truncate $|f|$ at height $M$, bound the squared truncated part by $M^2$ times the ball volume, and then make the remaining square-integral tail small by monotone convergence.

For an ordinary matrix symbol $p\in S^0$, with uniform estimates on the relevant compact base set,

The constant is independent of the unit vectors and the selected centers. Indeed the Fourier transform of the scalar packet is $h_j^{n/2}e^{-ix_j\cdot\xi} \widehat g(h_j(\xi-\xi_j))$. Substituting $x=x_j+h_j z$, $\xi=\xi_j+\zeta/h_j$ gives the pairing as

This integral is absolutely convergent: $z$ stays compact and $\widehat g$ is Schwartz by the earlier integration-by-parts proof. On $|\zeta|\leq\sqrt{R_j}/2$, the entire frequency segment from $\xi_j$ to $\xi_j+\zeta/h_j$ has size at least $R_j/2$. The fundamental theorem of calculus and the symbol's first derivatives bound its difference from $p(x_j,\xi_j)$ by $C R_j^{-1/2}(|z|+|\zeta|)$. Integrate this against the displayed Schwartz factors. On the complementary region use the uniform bound for $p$; the Schwartz tail of $\widehat g$ is $O(R_j^{-M})$ for any prescribed $M$, by annular summation. Finally, Fourier inversion and $\|g\|_2=1$ show that replacing $p$ by the constant matrix $p(x_j,\xi_j)$ gives exactly the leading term of (T12).

Thus for every compact $K$, T1 gives $Ku_j\to0$, while (T12) retains the high-frequency matrix value. This is the obstruction to compactness used below; it is not inferred merely from a large symbol at one fixed frequency.

T4. The exact ordinary essential norm, including matrices

First work in one Euclidean chart with a compact kernel. P2 supplies a full left symbol $p$, compactly supported in the output base, and a smooth compact remainder. To obtain this full symbol from (O17), absorb its input cutoff into the compact amplitude and apply the exact reduction (O15) once more. T2 makes that remainder irrelevant to (T1). Put $L=\lim_{R\to\infty}\sup_{x,|\xi|\geq R}\|p(x,\xi)\|$. The suprema decrease and are finite, so the limit exists.

Lower bound. Choose $(x_j,\xi_j)$ with $|\xi_j|\to\infty$ and $\|p(x_j,\xi_j)\|\to L$. For each matrix, compactness of the finite-dimensional unit sphere gives unit vectors $v_j,w_j$ with $\langle p(x_j,\xi_j)v_j,w_j\rangle=\|p(x_j,\xi_j)\|$: maximize $\|pv\|$, then choose $w=pv/\|pv\|$, with an arbitrary unit $w$ in the zero case. T3 and T1 show that $\|P-K\|\geq L$ for every compact $K$. When $L=0$, this inequality is immediate without choosing a sequence.

A positive matrix factor. Fix $M>L$ and choose $L<M'<M$. At sufficiently high frequency $\|p\|\leq M'$. Multiply $p$ by a cutoff $0\leq\chi\leq1$ supported in that high-frequency region and equal to one above a larger radius, obtaining $p_0$. Then $p-p_0$ has bounded frequency support, and

Here the matrix size of $c$ is the input rank, even if $p$ is rectangular. We construct an upper triangular $b$ with positive diagonal and $b^*b=c$. Its entries, in increasing row order, are

These formulas and multiplication verify $b^*b=c$, once the positive pivots are justified. After completing the first squares in the quadratic form $v^*cv$, the next pivot is the minimum of that form over vectors whose next coordinate is one and later coordinates are zero. Earlier coordinates vary freely. Bound (T14) therefore makes each pivot at least $M^2-(M')^2$, because every such vector has norm at least one. Completing a square subtracts precisely the terms in (T15), proving this minimum assertion inductively. Upper bounds follow from bounded entries of $c$ and the same formulas.

Consequently every entry of $b$ is an ordinary symbol of order zero. For products this follows from the product rule. For reciprocals it is the differentiated inverse identity in K2. Positive square roots are smooth on the fixed compact interval of possible pivots, bounded away from zero; repeatedly differentiating $q^2=t$ proves bounded derivatives of every order there. The chain rule assigns to each frequency derivative the sum of the orders of derivatives of its input, hence one order of decay per frequency differentiation. Base derivatives have no such cost. Induction in (T15) proves every required seminorm. Outside the compact base support of $p_0$, the formula gives $b=MI$.

Upper bound. Quantize $b-MI$ with compact kernel cutoffs equal to one near its compact diagonal, as in P2 O4–O5, and call the result $D$. Set $B=MI+D$. P1 B4 makes $B$ bounded. The ordinary adjoint and product formulas show that

with compact kernel: the identity terms cancel, and all remaining operators have compact kernels. Its principal order-zero term vanishes by (T14)–(T15); changing $p$ to $p_0$ contributes a smoothing symbol. The actual $R$ is self-adjoint. T2 makes it compact. For any $u$,

By T1 take a finite-dimensional orthogonal projection $Q$ with $\|R(I-Q)\|<\varepsilon$. On its orthogonal complement (T17) gives $\|P(I-Q)\|\leq\sqrt{M^2+\varepsilon}$. Since $PQ$ has finite rank, $\|P\|_{\mathrm{ess}}\leq M$. Let $M\downarrow L$. Together with the lower bound this proves (T2).

The formula holds also on manifolds with Hermitian bundles and compact kernel. Here are the gluing details for the upper bound. Choose finitely many relatively compact charts covering the relevant diagonal support, with local orthonormal frames. Smooth Gram–Schmidt constructs those frames by the same positive-root operations used above. Take smooth real $\chi_j$ supported in these charts and $\chi_0$ equal to one outside a compact set, with $\sum_{j\geq0}\chi_j^2=1$ and $\chi_0=0$ near the diagonal support of $P$. To construct them, take a cutoff $t$, equal to $\pi/2$ there and zero outside a larger compact set; use $\chi_0=\cos t$, and multiply $\sin t$ by a finite smooth partition normalized by the square root of the sum of its squares. PS5 supplies the finite partition. All denominators stay positive on the support of $\sin t$.

In each chart construct the preceding factor with principal matrix $\chi_j b_j$, and properly quantize it into the trivial output coefficient space of that chart. Include $B_0=M\chi_0 I$. The sum $P^*P+\sum_{j\geq0}B_j^*B_j-M^2I$ has vanishing principal symbol, since the local factors satisfy (T14) modulo bounded frequencies and $\chi_0p=0$. It has compact kernel after the constant identity cancels, and is of order minus one. Use this compact error in (T17), with the nonnegative sum of squared norms in place of $\|Bu\|^2$. For the lower bound, a sequence approaching the limiting symbol supremum has a subsequence in one of the finitely many smaller charts; apply T3 there in an orthonormal frame. Half-density substitution preserves its $L^2$ norms. This proves the manifold formula. Overlaps change principal representatives by one lower order and unitary conjugation, which preserve the high-frequency limiting matrix norm.

T5. The normalization on a canonical graph

Let $C$ be the graph of a homogeneous canonical diffeomorphism $\kappa:T^*Y\supset V\to T^*X$, so $n_X=n_Y=n$. The two pullbacks of the symplectic density agree. At the tangent level, $(d\kappa)^TJ_X(d\kappa)=J_Y$ implies $|\det d\kappa|=1$ in symplectic volume bases. Thus $\mu_C$ is well defined. The Maslov phase transitions proved in MC and PS have modulus one, so its phase frames define a Hermitian line metric. Accordingly $\|b\|$ in (T3) is intrinsic.

For precision, $b$ is an ordinary order-zero symbol in graph coordinates $(y,\eta)$. PS2 initially describes symbols in homogeneous frequency coordinates $\lambda$ on the $2n$-dimensional relation. Its coefficient for an order-zero kernel is of order $-n/2$, multiplying $|d\lambda|^{1/2}$. The change $(y,\eta)\mapsto\lambda$ is degree one; its $n$ base columns have degree one and its $n$ frequency columns degree zero. Hence its absolute determinant has degree $n$, and its square root has degree $n/2$, with every derivative bound on compact normalized charts. Multiplying gives order zero. The chain rule gives no loss for a base derivative and one loss for a frequency derivative. Conversely the inverse coordinate change gives the original estimates. This also shows that lowering the kernel order by one lowers the normalized graph coefficient by one.

Let $P=A^*A$. WM W2 and AC prove that it is an ordinary order-zero operator. Its principal coefficient satisfies

uniformly over compact base sets. We verify the normalization, including the density factor. At a matching point choose a symplectic volume basis $e$ on $T(T^*Y)$ and write $T=d\kappa$. The tangent graph bases for $C^{-1}$ and $C$ are $(e,Te)$ and $(Te,e)$. The matching map is $(u,v)\mapsto T(u-v)$. Its kernel basis is $(e,e)$; choose $(e,0)$ as lifts of a middle basis. The combined domain change of basis has determinant of absolute value one, and $Te$ has symplectic density one. CC C4 therefore sends the product of the two graph half-density units to the identity graph half-density unit, with factor one. Excess is zero, so AC supplies no additional $2\pi$ factor.

Reversal and adjoint conjugate the Maslov line, by MC G5, and compose the coefficient maps in the order $b^*b$. MC G4 maps unit phase frames to unit phase frames. Thus, in any of these frames, the resulting coefficient has the norm of $b^*b$; a possible unit phase does not affect that norm. For a finite matrix, $\|b^*b\|=\|b\|^2$, by the proof of (T5). The AC principal-symbol remainder is one lower order, proving (T18). This norm statement suffices here without choosing a particular phase trivialization of the identity symbol.

T6. One graph: compactness and the sharp essential norm

For a compact-kernel order-zero $A$ on one graph, its adjoint is bounded by WM W2. Its actual $L^2$ adjoint agrees with that phase adjoint by density. Combining (T8), (T2) and (T18) gives

Suprema are over the compact base support; all omitted symbols have uniform order-minus-one bounds there. Taking the nonnegative square root proves (T3), and T1 proves compactness exactly when $b\to0$.

In particular a strictly negative-order operator on one graph is compact: its normalized order-zero symbol tends to zero. Equivalently its adjoint product has strictly negative ordinary order, so T2 and (T7) apply. The criterion also covers symbols that tend to zero more slowly than every negative power, as Example T1 demonstrates.

T7. Local graphs, proper support and Sobolev compactness

For a compact input/output localization of a local canonical graph, PS5 partitions its kernel into finitely many graph-phase pieces plus a smooth compact remainder. Each piece has principal coefficient equal to the corresponding smooth partition factor times $b$. If $b\to0$, T6 makes every piece compact, and T2 handles the remainder. T1 makes their finite sum compact. This proves sufficiency without assuming that the adjoint product of a multi-sheeted relation is pseudodifferential.

A relative-frequency filter. Separate homogeneous input/output cutoffs do not by themselves distinguish branches with the same base points and covector directions but different ratios $|\xi|/|\eta|$. We supply the missing filter, including its operator-norm justification. After compact localization to Euclidean charts, let $F\in C_c^\infty(\mathbb R)$ and use the unitary Fourier multipliers $J_X^{it}=\langle D_X\rangle^{it}$, $J_Y^{it}=\langle D_Y\rangle^{it}$. Their unitarity is P3 L3. They are strongly continuous in $t$, since Parseval and dominated convergence apply to their bounded multiplier coefficients. Define, for compact $A$,

This is a norm-convergent integral of compact operators. Indeed a finite-rank operator is a finite sum $f\mapsto\langle f,v_l\rangle w_l$, by T1's representation theorem. Strong continuity on the finitely many $v_l,w_l$ makes its conjugated family norm continuous. Approximation of a compact $A$ by finite-rank operators, uniformly under unitary multiplication, gives the same norm continuity for $A$. On bounded $t$ intervals, uniform continuity makes Riemann sums converge in operator norm: upper bounds for the error are the interval length times the uniform oscillation on each partition. The tails are bounded by $(2\pi)^{-1}\|A\|\int_{|t|>T}|\widehat F(t)|\,dt$, which tends to zero because $\widehat F$ is Schwartz. Norm closure in T1 proves that $\mathcal T_F(A)$ is compact.

In the joint Fourier variables of the distributional kernel this operation is multiplication by

To verify it, pair first against product Schwartz functions, use the bounded $L^2$ operator pairing to justify integration in $t$, and apply one-dimensional Fourier inversion for $F$. Equivalently the Fourier multiplier identity holds on each Schwartz product before pairing with the kernel; convergence in Schwartz seminorms follows from the rapid decay of $\widehat F$ and polynomial growth in $|t|$ of each multiplier derivative. Product-test uniqueness P2 O0.1 proves the kernel identity. The sign change for the input Fourier variable has no effect on its Euclidean length.

The function $f$ is an ordinary joint order-zero symbol. On the support of it or any of its derivatives the two brackets are comparable, with constants fixed by $\operatorname{supp}F$. Thus each frequency derivative of either logarithm costs the reciprocal joint frequency, and the chain rule gives all joint estimates. At bounded joint frequency smoothness supplies the remaining bounds. Its degree-zero principal part on the comparable cone is $F(\log|\xi|-\log|\eta|)$: the difference between $\log\langle\xi\rangle$ and $\log|\xi|$, with all its large-frequency derivatives, has order minus two there. This follows from $\tfrac12\log(1+|\xi|^{-2})$ and its differentiated integral formula $\log(1+z)=\int_0^z(1+t)^{-1}dt$.

We spell out the symbol action used here. An ordinary order-zero operator $Q$ on the product base acts on a phase distribution $I_\phi(a)$ by multiplying its principal coefficient by $q(x,\phi_x)$. In local left quantization its combined phase is $(x-z)\cdot\zeta+\phi(z,\theta)$, with the ordinary factor $(2\pi)^{-d}$, where $d$ is the product-base dimension. First remove the part of the input phase amplitude away from $\phi_\theta=0$; PH F2 makes it smooth, and P2 O1 preserves that smoothness after compact localization. Near the remaining closed normalized critical support, $|\phi_z|$ is bounded above and below by positive multiples of $|\theta|$, since the represented covectors are nonzero. In the critical region $|\zeta|\asymp|\theta|$, rescale $\zeta=|\theta|\omega$. The stationary variables are $(z,\omega)$, the critical point is $z=x,\ |\theta|\omega=\phi_z(x,\theta)$, and their Hessian is $\begin{pmatrix}\phi_{zz}/|\theta|&-I\\-I&0\end{pmatrix}$. The triangular shift $\omega\mapsto\omega-\phi_{zz}z/(2|\theta|)$ at the tangent level makes it hyperbolic, with determinant magnitude one and signature zero. The full stationary expansion proved in PH F3 therefore gives leading amplitude $q(x,\phi_x)a$: the stationary factor $(2\pi/|\theta|)^d$ cancels both $(2\pi)^{-d}$ and the Jacobian $|\theta|^d$. Every further term and remainder has one additional ordinary order of decay, with all derivatives, by that same expansion. Outside comparable frequency, the derivative in $z$ is bounded below by a positive multiple of $|\zeta|+|\theta|$; repeated integration by parts in $z$ gives arbitrary decay. Away from the critical point in the comparable region use the normalized stationary noncritical operator of PH F2–F3. Compact base cutoffs may be inserted first; separated base portions are smooth by the same argument. This proves the symbol action and its essential-support inclusion, including for (T21). It does not assume an unproved pullback of kernels.

Necessity for local graphs. Suppose $b$ does not tend to zero uniformly over one compact base set. Choose covectors tending to infinity with $\|b\|\geq\varepsilon>0$. Their joint normalization has a convergent subsequence: the primed relation is closed in the ambient nonzero cotangent bundle and avoids both zero axes, as in WM W0–W5. Near its limit, separate input/output base and angular neighborhoods, together with an interval for $\log(|\xi|/|\eta|)$, form a neighborhood basis of the joint normalized cotangent space. Shrink these choices until their intersection with the embedded relation lies in one injective graph patch. Apply compactly based ordinary cutoffs for the separate neighborhoods, equal to one near the limit, and then apply $\mathcal T_F$ with $F=1$ near the limiting logarithmic ratio and supported in the selected interval. Finally apply compact base cutoffs again. If the original localized $A$ is compact, this resulting operator is compact, by bounded composition and (T20).

The proved symbol action and AC give it principal coefficient equal to these cutoff factors times $b$, with one lower-order error. Its essential support lies in the selected single graph patch; the rest is a smooth compact error by the nonstationary estimates just proved. All cutoff factors are one along the selected sequence eventually, so its normalized symbol does not tend to zero. T6 contradicts compactness. This proves necessity and the complete local-graph criterion.

For a proper operator, a fixed compact input set has compact output support and a fixed compact output set sees only compact input support, by WM W6. Consequently the preceding local estimates say precisely that bounded sets with fixed compact input support have relatively compact output in each local $L^2$ seminorm. For a bounded set in $L^2_{\mathrm{loc}}$, the proper input cutoff gives the same conclusion for each output compact set. To get one subsequence converging in every local seminorm, choose a countable compact exhaustion with interior coverage, supplied by the second-countable chart construction in PS5, and extract subsequences successively. The diagonal subsequence converges in all these seminorms. The limits agree on overlaps by their distributional pairings, hence define one local $L^2$ output. For the topological assertion, choose cutoffs $\rho_j=1$ on the smaller members of that exhaustion and use the metric $d(u,v)=\sum_{j\geq1}2^{-j}\min(1,\|\rho_j(u-v)\|_2)$. The triangle inequality follows termwise from that for the seminorm and $\min(1,a+b)\leq\min(1,a)+\min(1,b)$. Finite initial sums and the geometric tail show that metric convergence is exactly convergence in all these seminorms. Their compatible $L^2$ limits show completeness. The metric compactness argument in T1 therefore applies to the obtained subsequence property. Conversely compactness in this local topology implies compactness after each fixed output cutoff.

There is an all-real Sobolev version. Let $A\in I^m$ be on a local canonical graph, and interpret compactness after fixed compact input and output localization. Then, for every real $s$,

uniformly over compact base sets. Here $b_m=\sigma_m(A)/\mu_C^{1/2}$ has ordinary order $m$, and the condition is only at high frequency. To prove it, use the properly supported weights and parametrices of WM W6. Conjugating by the output weight of order $s-m$ and input parametrix of order $-s$ gives an order-zero graph operator. Its coefficient is

The principal symbols of the weights are the indicated homogeneous powers; replacing $|\xi|$ by $\langle\xi\rangle$ changes only lower orders. On the compact normalized relation, $|\xi|/|\eta|$ has positive lower and finite upper bounds. Thus (T23) tends to zero exactly when the right-hand side of (T22) does.

The conjugation equivalence respects compactness, not only boundedness. Use both directions of the parametrix identities from WM W6. Every ordinary weight or inverse maps the relevant Sobolev spaces boundedly by P1 B4. Every error has a smooth compact kernel and is compact between any two fixed Sobolev spaces: compose it with the weights and their parametrices to obtain a smooth compact kernel on $L^2$, which is compact by T2; the remaining smooth errors obey the same conclusion. Alternatively truncate the input and output Fourier variables of a smooth compact kernel. After any fixed Sobolev weights its Fourier kernel is Schwartz, so these truncations converge in the square-integrable kernel norm; T2 gives finite-rank approximations in the weighted spaces as well. The Schwartz assertion is repeated integration by parts in both compact base variables. This supplies the direct error argument without circular use of (T22). Bounded compositions and finite sums in the parametrix identities now prove both implications. Proper support and the countable extraction just given yield the corresponding local Sobolev-space statement.

T8. Four worked tests

Exercise T1 — Compact with no negative power. In one dimension take real $\chi\in C_c^\infty$, $0\leq\chi\leq1$, with maximum one, and

Prove compactness, although $q\notin S^{-\varepsilon}$ for any $\varepsilon>0$.

Solution. Repeated differentiation of $1/\log(e+t)$ gives finite sums of constants times $(e+t)^{-j}(\log(e+t))^{-l}$ at derivative order $j$, with $l\geq1$. This follows inductively by differentiating either factor. Derivatives of $\langle\xi\rangle$ have bounds $C_j\langle\xi\rangle^{1-j}$, by differentiating its square-root formula; the chain rule gives $|q^{(j)}(\xi)|\leq C_j\langle\xi\rangle^{-j}$. Thus $q\in S^0$, and $q\to0$. The principal symbol of $P$ is $\chi^2q$, so (T2) gives compactness. On the other hand $t^\varepsilon/\log(e+t)\to\infty$: putting $v=\varepsilon\log t$, the exponential series bound $e^v\geq v^2/2$ proves this after division by $\log t$. Therefore even the zeroth derivative fails the order-minus-$\varepsilon$ estimate. T2's negative-order argument alone would not have settled this example; the complete criterion does.

Exercise T2 — Bounded but oscillating at infinity. Replace $q$ in (T24) by $q(\xi)=\sin(\log\langle\xi\rangle)$. Find the essential norm.

Solution. Differentiation gives finite sums of bounded sine or cosine factors times derivatives of $\log\langle\xi\rangle$, whose derivative of order $j\geq1$ is $O(\langle\xi\rangle^{-j})$, by its explicit logarithmic derivative and induction. The product rule proves $q\in S^0$. Its absolute value is at most one and equals one along sequences with $\langle\xi_j\rangle=\exp(\pi/2+2\pi j)$. Since $\chi$ attains one, (T2) gives $\|P\|_{\mathrm{ess}}=1$. In particular $P$ is not compact. No classical homogeneous leading coefficient is present or needed.

Exercise T3 — A dilation, half-densities and a rectangular matrix. On $\mathbb R^2$ set $M=\operatorname{diag}(2,3)$, $Uf(x)=\sqrt6\,f(Mx)$, and $B=\begin{pmatrix}2&0\\0&1\\0&0\end{pmatrix}$. Let $A=\chi_X B U\chi_Y$, where both real compact smooth cutoffs lie between zero and one and equal one near a matched pair $y=Mx$. Find the graph and essential norm.

Solution. Substitution $y=Mx$ proves $\|Uf\|_2=\|f\|_2$. The phase is $(Mx-y)\cdot\eta$, and its canonical graph is $(y,\eta)\mapsto(M^{-1}y,M^T\eta)$. Its mixed graph matrix is $\begin{pmatrix}0&M^T\\-I&0\end{pmatrix}$, of determinant magnitude six. The normalized amplitude contains $\sqrt6$, so division by the square root of this critical Jacobian leaves the graph coefficient $\chi_X(x)B\chi_Y(y)$, of maximum norm two. Formula (T3) gives $\|A\|_{\mathrm{ess}}=2$. The rank-three output and rank-two input cause no change in the argument: $B^*B=\operatorname{diag}(4,1)$.

Exercise T4 — Why separate branches cannot be maximized. On the circle of length $2\pi$, let $Tf(x)=f(x+\pi)$ and $A=I+T$. Each of its two disjoint canonical graphs has scalar principal coefficient one. Compute $\|A\|_{\mathrm{ess}}$.

Solution. Translation is unitary by substitution, so $\|A\|\leq2$. The unit functions $u_j(x)=(2\pi)^{-1/2}e^{2ijx}$ are orthonormal, by direct integration, and $Au_j=2u_j$. Any orthonormal sequence is weakly zero: for a fixed $f$, Pythagoras applied to its projection on the first $N$ vectors gives $\sum_{j\leq N}|\langle f,u_j\rangle|^2\leq\|f\|^2$, so its individual terms tend to zero. T1 therefore gives $Ku_j\to0$ for every compact $K$, proving $\|A-K\|\geq2$. Thus $\|A\|_{\mathrm{ess}}=2$, although the maximum of the two individual symbol norms is one. The two-branch relation is a local canonical graph, so the compactness criterion in T7 applies and correctly says that $A$ is not compact. The single-graph numerical formula (T3) does not apply.

Free source and the remaining course

The freely readable human source is Lars Hörmander's Fourier integral operators. I, Acta Mathematica 127 (1971), Section 2.2 and Section 4.3. These give the positivity-factor route, compactness direction and graph compactness statement. The converse referred elsewhere in that paper is proved here by T3; its referenced work is not adopted. T1 supplies the functional analysis, T4 the finite matrix factor and manifold assembly, and T5–T7 the graph normalization, exact essential norm and full local/Sobolev interpretation. Citations replace none of these proofs.

The ordinary graph compactness and essential-norm obligations are now proved in this component. Symbol classes with derivative losses, the remaining intrinsic generality, propagation, hyperbolic and boundary problems, glancing and complex phases remain part of the active full course.