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The assumptions used here · Finite-dimensional norms and exact sequences · Completeness of operator spaces · Quotient norms and completeness · Hahn–Banach and scalar norm tests · Baire's theorem for complete metric spaces · Uniform boundedness for any operator family · Open mapping through summable corrections

Banach estimates, quotient spaces and compact parameter arguments

A Banach-space argument uses completeness, scalar separation and compactness for different reasons. This lesson proves the forms needed later for Fredholm operators, symbol actions and positivity estimates. Its examples show exactly where each hypothesis matters.

The useful distinction is between three mechanisms. Completeness turns summable errors into actual vectors. Hahn–Banach produces scalar tests without a completeness assumption. Compactness permits uniform control of a parameter family, even when that parameter space has no countable neighborhood basis. Keeping these mechanisms separate prevents a Banach-space estimate from being applied silently to a nonnormable space.

1. The assumptions used here

Work over \(\mathbb K=\mathbb R\) or \(\mathbb C\), with complex-linear duals when \(\mathbb K=\mathbb C\). A normed space has the metric \(d(x,y)=\|x-y\|\); it is Banach when every Cauchy sequence converges in that norm. Write \(\mathcal L(X,Y)\) for continuous linear maps. No space below is assumed separable.

Begin with Metric and topological foundations: the complete real field, its Archimedean property, finite linear algebra, scalar Cauchy–Schwarz, compactness in finite dimensions, and Zorn's maximality principle. The measure and differentiation results of that lesson are not used here. Complex scalar completeness follows coordinatewise from real completeness. Zorn's principle is the stated assumption in the Hahn–Banach proof in Section 5.

We use the definitions of open sets, neighborhoods, closure, the subspace topology and the product topology. Closure means that every open neighborhood meets the set. A map is continuous when inverse images of open sets are open; equivalently, each neighborhood of its value contains the image of a sufficiently small neighborhood of the input. The latter equivalence follows by taking inverse images in one direction and the union of these input neighborhoods in the other. Complements give the corresponding inverse-image statement for closed sets. Sections 9–10 prove the needed compactness and net assertions.

The proofs proceed from finite-dimensional norm control through completeness of operator spaces and quotients, Hahn–Banach separation, uniform boundedness, open mapping, nets, compactness and square-summable sequences. The complete-metric Baire theorem is proved in Section 6. Bochner integration, Hilbert-space representation and a closed-graph theorem for Fréchet spaces need separate arguments.

2. Finite-dimensional norms and exact sequences

Let \(E\) have basis \(e_1,\ldots,e_d\) and any norm \(N\). Set \(|a|=(\sum_{j=1}^d|a_j|^2)^{1/2}\) for its coordinate vector. The triangle inequality and the finite scalar Cauchy–Schwarz inequality give

Also \(|N(u)-N(v)|\leq N(u-v)\leq C|u-v|\), so \(N\) is continuous in coordinates. For \(d>0\) there is \(c>0\) such that \(N(a)\geq c\) on \(|a|=1\). Otherwise choose unit vectors with norms tending to zero. The compactness interface Section 4 of Metric and topological foundations gives a coordinate-convergent subsequence whose limit is still a unit vector; continuity makes its norm zero, a contradiction. Scaling yields

Thus every norm is equivalent to the coordinate norm. Identifying \(\mathbb C^d\) with \(\mathbb R^{2d}\) includes complex spaces. Dimension zero has one vector and is treated without choosing positive comparison constants.

A closed bounded subset of \(E\) is coordinate closed and bounded by (B2), hence compact by the same finite-dimensional interface. Every linear map out of \(E\) into a normed space is bounded: apply the estimate in (B1) with \(N(e_j)\) replaced by the norms of the images and then the lower bound in (B2). In particular all coordinate functionals are continuous. Coordinate completeness and (B2) make \(E\) Banach.

If \(E\) is a finite-dimensional subspace of an arbitrary normed space \(X\), a sequence of points of \(E\) converging in \(X\) is Cauchy in the restricted norm. Its coordinates converge, so the sequence converges to a point of \(E\). Norm limits are unique by the triangle inequality. Therefore \(E\) contains the limit. A point in the closure of a set in a metric space is the limit of a sequence in that set, by selecting a point within \(1/n\) at step \(n\). Thus \(E\) is closed in \(X\), without any completeness assumption on \(X\).

For a finite-dimensional exact sequence

rank-nullity gives \(\dim E_j=\dim\ker A_j+\dim\operatorname{im}A_j\), with the terminal zero map understood. Exactness identifies \(\ker A_j=\operatorname{im}A_{j-1}\), while the initial kernel is zero. Multiply by \((-1)^j\) and sum. Each internal image dimension occurs twice with opposite signs, yielding \(\sum_{j=0}^r(-1)^j\dim E_j=0\). In particular a short exact sequence has middle dimension equal to the sum of its endpoint dimensions. This calculation includes zero spaces and makes no topological exactness assumption.

3. Completeness of operator spaces

First note that a linear map \(T:X\to Y\) is continuous exactly when \(\|Tx\|\leq C\|x\|\) for some finite \(C\). The estimate implies continuity. Conversely, continuity at zero gives \(\|Tx\|<1\) when \(\|x\|<r\), for some \(r>0\). For \(x\ne0\), apply this to \(rx/(2\|x\|)\) to obtain \(\|Tx\|\leq2\|x\|/r\). Define

It is finite, and scaling gives \(\|Tx\|\leq\|T\|\|x\|\). The triangle inequality and homogeneity pass through the supremum. If \(\|T\|=0\), scaling shows that \(T\) vanishes everywhere. Thus this is a norm, also for the zero domain, whose unit ball still contains zero. Composition satisfies \(\|ST\|\leq\|S\|\|T\|\).

A closed linear subspace of a Banach space is Banach: a Cauchy sequence has a limit in the ambient space, and closedness keeps the limit in the subspace. A finite product \(X_1\times\cdots\times X_q\) of Banach spaces is Banach under \(\|(x_1,\ldots,x_q)\|_{\max}=\max_j\|x_j\|\). A Cauchy sequence has a limit in every coordinate; taking the maximum of finitely many coordinate errors proves norm convergence. The sum norm is equivalent, since

The empty product is the zero vector space. A product of the coordinate norms is not a norm and is never used here.

If \(Y\) is Banach and \(X\) merely normed, then \(\mathcal L(X,Y)\) is Banach. For an operator-norm Cauchy sequence \(T_n\), the values \(T_nx\) are Cauchy for every fixed \(x\). Let \(Tx\) be their limit in \(Y\). Passing to the limit in \(T_n(ax+by)=aT_nx+bT_ny\) proves linearity. The operator norms of the sequence are bounded: the tail lies within one of a fixed operator, and the remaining initial segment is finite. Calling a bound \(M\), passage to the limit gives \(\|Tx\|\leq M\|x\|\), so \(T\in\mathcal L(X,Y)\). Given \(\varepsilon>0\), take \(n,m\geq N\) with \(\|T_n-T_m\|\leq\varepsilon\). Holding \(n\) and \(x\) fixed and taking \(m\to\infty\) yields

and taking the unit-ball supremum gives \(\|T_n-T\|\leq\varepsilon\). The completeness of \(X\) was unnecessary. In particular \(X'=\mathcal L(X,\mathbb K)\) is Banach.

We shall repeatedly use the resulting series criterion: if \(X\) is Banach and \(\sum_j\|x_j\|<\infty\), then \(\sum_jx_j\) converges in \(X\), with every tail norm bounded by the corresponding scalar tail sum. Indeed the finite partial sums are Cauchy by the triangle inequality; pass to their limit in that same inequality. No integration theorem is involved.

4. Quotient norms and completeness

Let \(M\) be a linear subspace of a normed space \(X\). The formula

does not depend on the representative: replacing \(x\) by \(x+m_0\) simply translates the set of vectors over which the infimum is taken. Approximate the infima for two cosets within \(\varepsilon\), add the representatives, and let \(\varepsilon\downarrow0\); this proves the triangle inequality. Rescaling the subspace proves homogeneity for nonzero scalars, and the zero scalar case is immediate. Finally \(\|x+M\|_q=0\) exactly when \(x\in\overline M\). Therefore (B3) is a norm precisely when \(M\) is closed. The quotient map \(Q:X\to X/M\) has \(\|Qx\|_q\leq\|x\|\).

Assume now that \(X\) is Banach and \(M\) is closed. Let \(z_n\) be a Cauchy sequence in \(X/M\). Choose an increasing subsequence \(z_{n_j}\) with

Choose a representative \(x_1\) of \(z_{n_1}\) and a representative \(h_j\) of \(z_{n_{j+1}}-z_{n_j}\) with \(\|h_j\|<2^{1-j}\). Such a representative exists by the definition of the infimum; no nearest representative is asserted. Section 3 makes \(x=x_1+\sum_{j\geq1}h_j\) a vector in \(X\). The partial sums represent the successive \(z_{n_j}\), so continuity of \(Q\) shows \(z_{n_j}\to Qx\). A Cauchy sequence with a convergent subsequence converges to that same limit: bound the distance from a late term to a still later subsequence term and then to the limit. Hence \(z_n\to Qx\) and \(X/M\) is Banach.

This includes \(M=\{0\}\) and \(M=X\). It does not establish completeness of the image of an arbitrary bounded map; that image may fail to be closed.

5. Hahn–Banach and scalar norm tests

Let \(M\subset X\) be a linear subspace of a real normed space and \(f:M\to\mathbb R\) bounded and linear, with \(C=\|f\|\). We first prove an extension by one vector. For \(v\notin M\), consider the real intervals with endpoints

Every lower endpoint is at most every upper endpoint: for \(m,n\in M\),

The endpoints from \(m=0\) show that the lower endpoints are bounded above and the upper endpoints bounded below. Real completeness supplies a real \(c\) between their supremum and infimum. Consequently

Define \(F(m+tv)=f(m)+tc\). The representation is unique because \(v\notin M\), so \(F\) is real-linear and extends \(f\). For \(t\ne0\), apply (B4) to \(-m/t\) and multiply by \(|t|\); it gives \(|F(m+tv)|\leq C\|m+tv\|\). For \(t=0\) use the original bound. Restriction supplies the reverse inequality for the norms, so \(\|F\|=C\). This also handles \(C=0\).

For extension to all of \(X\), order the pairs consisting of a subspace containing \(M\) and an extension of \(f\) bounded by \(C\), by extension. The original pair makes this a nonempty partially ordered set. For a nonempty chain, the union of its subspaces is a subspace: any finite list of vectors belongs to one member of the chain. The functionals agree on overlaps, giving a well-defined linear functional on that union with the same bound. This is an upper bound; the original pair bounds an empty chain. Zorn's explicitly assumed maximality principle gives a maximal pair. The one-vector construction shows that its domain cannot omit any vector of \(X\). Thus the real extension exists on all of \(X\), with norm \(C\).

Now let \(X\) and \(M\) be complex and \(f:M\to\mathbb C\) bounded and complex-linear. Its real part \(h=\operatorname{Re}f\) is a real-linear functional on the underlying real subspace. Its real operator norm equals \(\|f\|\): one inequality is \(|h(m)|\leq|f(m)|\); for the other, when \(f(m)\ne0\), multiply \(m\) by \(\theta=\overline{f(m)}/|f(m)|\), so \(h(\theta m)=|f(m)|\) and \(\|\theta m\|=\|m\|\). Extend \(h\) by the real result to \(a:X\to\mathbb R\), with norm \(\|f\|\), and put

Real-linearity gives additivity and \(F(ix)=iF(x)\); together these prove complex-linearity. On \(M\), \(\operatorname{Re}f(im)=-\operatorname{Im}f(m)\), so \(F(m)=f(m)\). For any \(x\) with \(F(x)\ne0\), choose a unit scalar \(\theta\) making \(\theta F(x)=|F(x)|\). Complex-linearity gives

The zero value is harmless. Thus \(\|F\|\leq\|a\|=\|f\|\), while restriction proves equality. Neither the closedness of \(M\) nor the completeness of \(X\) was required.

For any complex normed space \(E\), this implies the exact norm identity

The right side is at most the left by the operator bound. If \(v\ne0\), define \(f(zv)=z\|v\|\) on the complex line it spans. Its norm is one, and (B5) extends it with the same norm, attaining \(\ell(v)=\|v\|\). If \(v=0\), every value is zero and the dual ball contains the zero functional. Thus the identity holds also for the zero space. It applies in particular to \(E=\mathcal L(B_1,B_2)\); Section 3 separately proves that this operator space is Banach when \(B_2\) is Banach.

The practical consequence is precise. If \(|\ell(v)|\leq K\|\ell\|\) for every continuous complex-linear \(\ell\), with the same \(K\), then \(\|v\|\leq K\). A bound depending arbitrarily on \(\ell\) cannot be substituted for that uniform estimate.

6. Baire's theorem for complete metric spaces

Complete-metric Baire theorem. Every countable intersection of dense open subsets of a complete metric space is dense. Here is its full nested-ball proof. Let \(Z\) be complete, let \(U_j\subset Z\), \(j\geq1\), be dense and open, and let \(V\subset Z\) be nonempty and open. Since \(U_1\) is dense, \(V\cap U_1\) contains a point \(z_1\). Openness supplies \(r_1>0\), chosen also with \(r_1\leq2^{-1}\), such that

Indeed choose a ball contained in the open set and take a smaller radius, so its closed ball is contained in that original ball. Suppose \(z_j,r_j>0\) have been chosen. The nonempty open ball \(B(z_j,r_j)\) meets dense \(U_{j+1}\). Choose \(z_{j+1}\) in that intersection and a positive radius \(r_{j+1}\leq2^{-j-1}\) whose closed ball lies in the intersection. Thus

The construction works also at isolated points: a sufficiently small ball then consists of that point. For \(k,l\geq j\), both centers lie in \(\overline B(z_j,r_j)\), so \(d(z_k,z_l)\leq2r_j\to0\). Completeness gives a limit \(z\). For each \(j\), the entire tail lies in that closed ball; continuity of distance gives \(z\in\overline B(z_j,r_j)\). The first inclusion puts \(z\) in \(V\), and each successive inclusion puts it in \(U_j\). Hence \(V\cap\bigcap_jU_j\ne\varnothing\). This proves density. If \(Z\) is empty, the assertion is true by the definition of density, and no ball is chosen. Neither separability nor local compactness is required. the complete nested-ball argument above supplies this input here.

The form needed below follows exactly. If a nonempty complete metric space is the union of closed sets \(F_n\), then some \(F_n\) has nonempty interior. If every interior were empty, each open complement would be dense, while their intersection would be empty, contradicting the theorem just proved. Completeness will be checked for each space to which this form is applied.

This is the general complete-metric result. The Banach consequences in the next two sections apply in normed spaces; a later argument in the Schwartz topology must verify completeness for its own metric.

7. Uniform boundedness for any operator family

Let \(X\) be Banach, let \(Y\) be normed, and let \(\mathcal A\subset\mathcal L(X,Y)\) be any family. Assume

Then the operator norms of the family have a common finite bound. The empty family is covered by the bound zero, so assume it is nonempty. Define

Every set in the intersection is closed by continuity. Arbitrary intersections of closed sets are closed, including uncountable intersections. Assumption (B7) says \(X=\bigcup_{n\geq1}F_n\). Baire applies to the complete norm metric on \(X\), a nonempty space even when \(X=\{0\}\). Hence \(B(x_0,r)\subset F_n\) for some \(r>0\) and \(n\). If \(\|h\|<r\), both \(x_0+h\) and \(x_0\) are in \(F_n\), and

For every \(\|x\|\leq1\), substitute \(h=rx/2\). It follows that \(\|Ax\|\leq4n/r\), so \(\sup_{A\in\mathcal A}\|A\|\leq4n/r<\infty\). No countability of \(\mathcal A\) and no completeness of \(Y\) entered the argument. The countable family used by Baire is the family of level sets \(F_n\), not the operator family.

8. Open mapping through summable corrections

Let \(T:X\to Y\) be a bounded surjective linear map between Banach spaces. We show that a fixed ball about zero in \(Y\) lies in the image of the closed unit ball of \(X\). If \(Y=\{0\}\), this holds for every positive radius, so suppose otherwise. Write \(B_X\) for the open unit ball. Surjectivity gives

By Section 6 applied to \(Y\), some closed set on the right contains \(B_Y(y_0,r)\), where \(r>0\). If \(\|y\|<r\), approximate both \(y_0+y\) and \(y_0\) by images of vectors in \(nB_X\). Their differences approximate \(y\) by images of vectors in \(2nB_X\). Consequently

The second inclusion follows by scaling. Scaling again gives \(B_Y(0,t\delta)\subset\overline{T(tB_X)}\) for every \(t>0\).

Take \(\|y\|<\delta/2\). From (B8) with \(t=1/2\), choose \(x_1\) with \(\|x_1\|<1/2\) and \(\|y-Tx_1\|<\delta/4\). If \(x_1,\ldots,x_j\) have been chosen with residual norm less than \(\delta 2^{-j-1}\), use (B8) at \(t=2^{-j-1}\) to choose \(x_{j+1}\) with

Section 3 and completeness of \(X\) give \(x=\sum_{j\geq1}x_j\) with \(\|x\|\leq1\). Continuity of \(T\) and the residual bounds give \(Tx=y\). This is actual surjectivity of a ball, not merely density of its image.

For any open \(O\subset X\) and \(x_0\in O\), choose \(s>0\) with \(x_0+s\overline B_X\subset O\), for example half a radius of an open ball contained in \(O\). The preceding conclusion gives

Therefore \(T\) is open. If \(T\) is bijective, its inverse is linear by uniqueness of preimages. For \(y\ne0\), apply the unit-ball conclusion to \(\delta y/(4\|y\|)\); rescaling and uniqueness yield

The zero vector satisfies the same estimate. Thus the inverse is bounded. The two completeness assumptions have distinct uses: Baire on \(Y\), and convergence of the correction series in \(X\).