Reading guide · Proof index

Continuity and the derivative

Jiří Lebl, Basic Analysis I–II, version 6.3. Free author edition of this section. Selection and attribution · Notation.

L8.4.1: Full vector mean-value inequality with zero case supplied.

Proof.

By the mean value theorem on the scalar-valued function t↦(φ(b)−φ(a))⋅φ(t),t \mapsto \bigl(\varphi(b)-\varphi(a) \bigr) \cdot \varphi(t)\text{,} where the dot is the dot product, we obtain a t0∈(a,b)t_0 \in (a,b) such that
∥φ(b)−φ(a)∥2=(φ(b)−φ(a))⋅(φ(b)−φ(a))=(φ(b)−φ(a))⋅φ(b)−(φ(b)−φ(a))⋅φ(a)=(b−a)(φ(b)−φ(a))⋅φ′(t0),\begin{equation*} \begin{split} \bnorm{\varphi(b)-\varphi(a)}^2 & = \bigl( \varphi(b)-\varphi(a) \bigr) \cdot \bigl( \varphi(b)-\varphi(a) \bigr) \\ & = \bigl(\varphi(b)-\varphi(a) \bigr) \cdot \varphi(b) - \bigl(\varphi(b)-\varphi(a) \bigr) \cdot \varphi(a) \\ & = (b-a) \bigl(\varphi(b)-\varphi(a) \bigr) \cdot \varphi'(t_0) , \end{split} \end{equation*}
where we treat φ′\varphi' as a vector in Rn\R^n by the abuse of notation we mentioned in the previous section. If we think of φ′(t)\varphi'(t) as a vector, then by Exercise 8.2.6, ∥φ′(t)∥L(R,Rn)=∥φ′(t)∥Rn.\bnorm{\varphi'(t)}_{L(\R,\R^n)} = \bnorm{\varphi'(t)}_{\R^n}\text{.} That is, the euclidean norm of the vector is the same as the operator norm of φ′(t).\varphi'(t)\text{.}
By the Cauchy–Schwarz inequality
∥φ(b)−φ(a)∥2=(b−a)(φ(b)−φ(a))⋅φ′(t0)≤(b−a)∥φ(b)−φ(a)∥ ∥φ′(t0)∥.\begin{equation*} \bnorm{\varphi(b)-\varphi(a)}^2 = (b-a)\bigl(\varphi(b)-\varphi(a) \bigr) \cdot \varphi'(t_0) \leq (b-a) \bnorm{\varphi(b)-\varphi(a)} \, \bnorm{\varphi'(t_0)} . \qedhere \end{equation*}

L8.4.2: Derivative bound implies Lipschitz bound on a convex open domain.

Proof.

Fix pp and qq in UU and note that (1−t)p+tq∈U(1-t)p+tq \in U for all t∈[0,1]t \in [0,1] by convexity. Next
ddt[f((1−t)p+tq)]=f′((1−t)p+tq)(q−p).\begin{equation*} \frac{d}{dt} \Bigl[f\bigl((1-t)p+tq\bigr)\Bigr] = f'\bigl((1-t)p+tq\bigr) (q-p) . \end{equation*}
By Lemma 8.4.1, there is some t0∈(0,1)t_0 \in (0,1) such that
∥f(p)−f(q)∥≤∥ddt∣t=t0[f((1−t)p+tq)]∥≤∥f′((1−t0)p+t0q)∥ ∥q−p∥≤M∥q−p∥.\begin{equation*} \begin{split} \bnorm{f(p)-f(q)} & \leq \norm{\frac{d}{dt} \Big|_{t=t_0} \Bigl[ f\bigl((1-t)p+tq\bigr) \Bigr] } \\ & \leq \norm{f'\bigl((1-t_0)p+t_0q\bigr)} \, \snorm{q-p} \leq M \snorm{q-p} . \qedhere \end{split} \end{equation*}

L8.4.6: C1 iff all partial derivatives exist and are continuous.

Proof.

We proved that if ff is differentiable, then the partial derivatives exist. The partial derivatives are the entries of the matrix representing f′(x).f'(x)\text{.} If f′ ⁣:U→L(Rn,Rm)f' \colon U \to L(\R^n,\R^m) is continuous, then the entries are continuous, and hence the partial derivatives are continuous.
To prove the opposite direction, suppose the partial derivatives exist and are continuous. Fix x∈U.x \in U\text{.} If we show that f′(x)f'(x) exists, we are done, because the entries of the matrix representing f′(x)f'(x) are the partial derivatives and if the entries are continuous functions, the matrix-valued function f′f' is continuous.
We do induction on dimension. First, the conclusion is true when n=1n=1 (exercise, note that ff is vector-valued). In this case, f′(x)f'(x) is essentially the derivative of Chapter 4. Suppose the conclusion is true for Rn−1.\R^{n-1}\text{.} That is, if we restrict to the first n−1n-1 variables, the function is differentiable. When taking the partial derivatives in x1x_1 through xn−1,x_{n-1}\text{,} it does not matter if we consider ff or ff restricted to the set where xnx_n is fixed. In the following, by a slight abuse of notation, we think of Rn−1\R^{n-1} as a subset of Rn,\R^n\text{,} that is, the set in Rn\R^n where xn=0.x_n = 0\text{.} In other words, we identify the vectors (x1,x2,…,xn−1)(x_1,x_2,\ldots,x_{n-1}) and (x1,x2,…,xn−1,0).(x_1,x_2,\ldots,x_{n-1},0)\text{.}
Fix p∈Up \in U and let
A≔[∂f1∂x1(p)…∂f1∂xn(p)⋮⋱⋮∂fm∂x1(p)…∂fm∂xn(p)],A′≔[∂f1∂x1(p)…∂f1∂xn−1(p)⋮⋱⋮∂fm∂x1(p)…∂fm∂xn−1(p)],v≔[∂f1∂xn(p)⋮∂fm∂xn(p)].\begin{equation*} A \coloneqq \begin{bmatrix} \frac{\partial f_1}{\partial x_1}(p) & \ldots & \frac{\partial f_1}{\partial x_n}(p) \\ \vdots & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_1}(p) & \ldots & \frac{\partial f_m}{\partial x_n}(p) \end{bmatrix} , \quad A' \coloneqq \begin{bmatrix} \frac{\partial f_1}{\partial x_1}(p) & \ldots & \frac{\partial f_1}{\partial x_{n-1}}(p) \\ \vdots & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_1}(p) & \ldots & \frac{\partial f_m}{\partial x_{n-1}}(p) \end{bmatrix} , \quad v \coloneqq \begin{bmatrix} \frac{\partial f_1}{\partial x_n}(p) \\ \vdots \\ \frac{\partial f_m}{\partial x_n}(p) \end{bmatrix} . \end{equation*}
Let ϵ>0\epsilon > 0 be given. By the induction hypothesis, there is a δ>0\delta > 0 such that for every h′∈Rn−1h' \in \R^{n-1} with ∥h′∥<δ,\snorm{h'} < \delta\text{,} we have
∥f(p+h′)−f(p)−A′h′∥∥h′∥<ϵ.\begin{equation*} \frac{\bnorm{f(p+h') - f(p) - A' h'}}{\snorm{h'}} < \epsilon . \end{equation*}
By continuity of the partial derivatives, suppose δ\delta is small enough so that
∣∂fk∂xn(p+h)−∂fk∂xn(p)∣<ϵ\begin{equation*} \abs{\frac{\partial f_k}{\partial x_n}(p+h) - \frac{\partial f_k}{\partial x_n}(p)} < \epsilon \end{equation*}
for all kk and all h∈Rnh \in \R^n with ∥h∥<δ.\snorm{h} < \delta\text{.}
Suppose h=h′+tenh = h' + t e_n is a vector in Rn,\R^n\text{,} where h′∈Rn−1,h' \in \R^{n-1}\text{,} t∈R,t \in \R\text{,} such that ∥h∥<δ.\snorm{h} < \delta\text{.} Then ∥h′∥≤∥h∥<δ.\snorm{h'} \leq \snorm{h} < \delta\text{.} Note that Ah=A′h′+tv.Ah = A' h' + tv\text{.}
∥f(p+h)−f(p)−Ah∥=∥f(p+h′+ten)−f(p+h′)−tv+f(p+h′)−f(p)−A′h′∥≤∥f(p+h′+ten)−f(p+h′)−tv∥+∥f(p+h′)−f(p)−A′h′∥≤∥f(p+h′+ten)−f(p+h′)−tv∥+ϵ∥h′∥.\begin{equation*} \begin{split} & \bnorm{f(p+h) - f(p) - Ah} \\ & \qquad \qquad = \bnorm{f(p+h' + t e_n) - f(p+h') - tv + f(p+h') - f(p) - A' h'} \\ & \qquad \qquad \leq \bnorm{f(p+h' + t e_n) - f(p+h') -tv} + \bnorm{f(p+h') - f(p) - A' h'} \\ & \qquad \qquad \leq \bnorm{f(p+h' + t e_n) - f(p+h') -tv} + \epsilon \snorm{h'} . \end{split} \end{equation*}
As all the partial derivatives exist, by the mean value theorem, for each kk there is some θk∈[0,t]\theta_k \in [0,t] (or [t,0][t,0] if t<0t < 0), such that
fk(p+h′+ten)−fk(p+h′)=t∂fk∂xn(p+h′+θken).\begin{equation*} f_k(p+h' + t e_n) - f_k(p+h') = t \frac{\partial f_k}{\partial x_n}(p+h'+\theta_k e_n). \end{equation*}
We have ∥h′+θken∥≤∥h∥<δ,\snorm{h'+\theta_k e_n} \leq \snorm{h} < \delta\text{,} and so we can finish the estimate
∥f(p+h)−f(p)−Ah∥≤∥f(p+h′+ten)−f(p+h′)−tv∥+ϵ∥h′∥≤∑k=1m(t∂fk∂xn(p+h′+θken)−t∂fk∂xn(p))2+ϵ∥h′∥≤m ϵ∣t∣+ϵ∥h′∥≤(m+1)ϵ∥h∥.\begin{equation*} \begin{split} \bnorm{f(p+h) - f(p) - Ah} & \leq \bnorm{f(p+h' + t e_n) - f(p+h') -tv} + \epsilon \snorm{h'} \\ & \leq \sqrt{\sum_{k=1}^m {\left(t\frac{\partial f_k}{\partial x_n}(p+h'+\theta_k e_n) - t \frac{\partial f_k}{\partial x_n}(p)\right)}^2} + \epsilon \snorm{h'} \\ & \leq \sqrt{m}\, \epsilon \sabs{t} + \epsilon \snorm{h'} \\ & \leq (\sqrt{m}+1)\epsilon \snorm{h} . \qedhere \end{split} \end{equation*}