Section8.4Continuity and the derivative

Note: 1–2 lectures

Subsection8.4.1Bounding the derivative

Let us prove a “mean value theorem” for vector-valued functions.

Proof.

By the mean value theorem on the scalar-valued function $$t \mapsto \bigl(\varphi(b)-\varphi(a) \bigr) \cdot \varphi(t)\text{,}$$ where the dot is the dot product, we obtain that there is a $$t_0 \in (a,b)$$ such that

\begin{equation*} \begin{split} \snorm{\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 $$\varphi'$$ as a vector in $$\R^n$$ by the abuse of notation we mentioned in the previous section. If we think of $$\varphi'(t)$$ as a vector, then by Exercise 8.2.6, $$\snorm{\varphi'(t)}_{L(\R,\R^n)} = \snorm{\varphi'(t)}_{\R^n}\text{.}$$ That is, the euclidean norm of the vector is the same as the operator norm of $$\varphi'(t)\text{.}$$

By the Cauchy–Schwarz inequality

\begin{equation*} \snorm{\varphi(b)-\varphi(a)}^2 = (b-a)\bigl(\varphi(b)-\varphi(a) \bigr) \cdot \varphi'(t_0) \leq (b-a) \snorm{\varphi(b)-\varphi(a)} \, \snorm{\varphi'(t_0)} . \qedhere \end{equation*}

Recall that a set $$U$$ is convex if whenever $$x,y \in U\text{,}$$ the line segment from $$x$$ to $$y$$ lies in $$U\text{.}$$

Proof.

Fix $$x$$ and $$y$$ in $$U$$ and note that $$(1-t)x+ty \in U$$ for all $$t \in [0,1]$$ by convexity. Next

\begin{equation*} \frac{d}{dt} \Bigl[f\bigl((1-t)x+ty\bigr)\Bigr] = f'\bigl((1-t)x+ty\bigr) (y-x) . \end{equation*}

By Lemma 8.4.1 there is some $$t_0 \in (0,1)$$ such that

\begin{equation*} \begin{split} \snorm{f(x)-f(y)} & \leq \norm{\frac{d}{dt} \Big|_{t=t_0} \Bigl[ f\bigl((1-t)x+ty\bigr) \Bigr] } \\ & \leq \norm{f'\bigl((1-t_0)x+t_0y\bigr)} \, \snorm{y-x} \leq M \snorm{y-x} . \qedhere \end{split} \end{equation*}

Example8.4.3.

If $$U$$ is not convex the proposition is not true: Consider the set

\begin{equation*} U := \bigl\{ (x,y) : 0.5 < x^2+y^2 < 2 \bigr\} \setminus \bigl\{ (x,0) : x < 0 \bigr\} . \end{equation*}

For $$(x,y) \in U\text{,}$$ let $$f(x,y)$$ be the angle that the line from the origin to $$(x,y)$$ makes with the positive $$x$$ axis. We even have a formula for $$f\text{:}$$

\begin{equation*} f(x,y) = 2 \operatorname{arctan}\left( \frac{y}{x+\sqrt{x^2+y^2}}\right) . \end{equation*}

Think a spiral staircase with room in the middle. See Figure 8.9.

The function is differentiable, and the derivative is bounded on $$U\text{,}$$ which is not hard to see. Now think of what happens near where the negative $$x$$-axis cuts the annulus in half. As we approach this cut from positive $$y\text{,}$$ $$f(x,y)$$ approaches $$\pi\text{.}$$ From negative $$y\text{,}$$ $$f(x,y)$$ approaches $$-\pi\text{.}$$ So for small $$\epsilon > 0\text{,}$$ $$\sabs{f(-1,\epsilon)-f(-1,-\epsilon)}$$ approaches $$2\pi\text{,}$$ but $$\snorm{(-1,\epsilon)-(-1,-\epsilon)} = 2\epsilon\text{,}$$ which is arbitrarily small. The conclusion of the proposition does not hold for this nonconvex $$U\text{.}$$

Let us solve the differential equation $$f' = 0\text{.}$$

Proof.

For any given $$x \in U\text{,}$$ there is a ball $$B(x,\delta) \subset U\text{.}$$ The ball $$B(x,\delta)$$ is convex. Since $$\snorm{f'(y)} \leq 0$$ for all $$y \in B(x,\delta)\text{,}$$ then by the proposition, $$\snorm{f(x)-f(y)} \leq 0 \snorm{x-y} = 0\text{.}$$ So $$f(x) = f(y)$$ for all $$y \in B(x,\delta)\text{.}$$

This means that $$f^{-1}(c)$$ is open for all $$c \in \R^m\text{.}$$ Suppose $$f^{-1}(c)$$ is nonempty. The two sets

\begin{equation*} U' = f^{-1}(c), \qquad U'' = f^{-1}\bigl(\R^m\setminus\{c\}\bigr) \end{equation*}

are open and disjoint, and further $$U = U' \cup U''\text{.}$$ As $$U'$$ is nonempty and $$U$$ is connected, then $$U'' = \emptyset\text{.}$$ So $$f(x) = c$$ for all $$x \in U\text{.}$$

Subsection8.4.2Continuously differentiable functions

Definition8.4.5.

Let $$U \subset \R^n$$ be open. We say $$f \colon U \to \R^m$$ is continuously differentiable, or $$C^1(U)\text{,}$$ if $$f$$ is differentiable and $$f' \colon U \to L(\R^n,\R^m)$$ is continuous.

Without continuity the theorem does not hold. Just because partial derivatives exist does not mean that $$f$$ is differentiable, in fact, $$f$$ may not even be continuous. See the exercises for the last section and also for this section.

Proof.

We proved that if $$f$$ is differentiable, then the partial derivatives exist. The partial derivatives are the entries of the matrix of $$f'(x)\text{.}$$ If $$f' \colon U \to L(\R^n,\R^m)$$ 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 $$x \in U\text{.}$$ If we show that $$f'(x)$$ exists we are done, because the entries of the matrix $$f'(x)$$ are the partial derivatives and if the entries are continuous functions, the matrix-valued function $$f'$$ is continuous.

We do induction on dimension. First, the conclusion is true when $$n=1\text{.}$$ In this case the derivative is just the regular derivative (exercise, noting that $$f$$ is vector-valued).

Suppose the conclusion is true for $$\R^{n-1}\text{,}$$ that is, if we restrict to the first $$n-1$$ variables, the function is differentiable. It is easy to see that the first $$n-1$$ partial derivatives of $$f$$ restricted to the set where the last coordinate is fixed are the same as those for $$f\text{.}$$ In the following, by a slight abuse of notation, we think of $$\R^{n-1}$$ as a subset of $$\R^n\text{,}$$ that is the set in $$\R^n$$ where $$x_n = 0\text{.}$$ In other words, we identify the vectors $$(x_1,x_2,\ldots,x_{n-1})$$ and $$(x_1,x_2,\ldots,x_{n-1},0)\text{.}$$ Let

\begin{equation*} A := \begin{bmatrix} \frac{\partial f_1}{\partial x_1}(x) & \ldots & \frac{\partial f_1}{\partial x_n}(x) \\ \vdots & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_1}(x) & \ldots & \frac{\partial f_m}{\partial x_n}(x) \end{bmatrix} , \qquad A' := \begin{bmatrix} \frac{\partial f_1}{\partial x_1}(x) & \ldots & \frac{\partial f_1}{\partial x_{n-1}}(x) \\ \vdots & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_1}(x) & \ldots & \frac{\partial f_m}{\partial x_{n-1}}(x) \end{bmatrix} , \qquad v := \begin{bmatrix} \frac{\partial f_1}{\partial x_n}(x) \\ \vdots \\ \frac{\partial f_m}{\partial x_n}(x) \end{bmatrix} . \end{equation*}

Let $$\epsilon > 0$$ be given. By the induction hypothesis, there is a $$\delta > 0$$ such that for every $$k \in \R^{n-1}$$ with $$\snorm{k} < \delta\text{,}$$ we have

\begin{equation*} \frac{\snorm{f(x+k) - f(x) - A' k}}{\snorm{k}} < \epsilon . \end{equation*}

By continuity of the partial derivatives, suppose $$\delta$$ is small enough so that

\begin{equation*} \abs{\frac{\partial f_j}{\partial x_n}(x+h) - \frac{\partial f_j}{\partial x_n}(x)} < \epsilon \end{equation*}

for all $$j$$ and all $$h \in \R^n$$ with $$\snorm{h} < \delta\text{.}$$

Suppose $$h = k + t e_n$$ is a vector in $$\R^n\text{,}$$ where $$k \in \R^{n-1}\text{,}$$ $$t \in \R\text{,}$$ such that $$\snorm{h} < \delta\text{.}$$ Then $$\snorm{k} \leq \snorm{h} < \delta\text{.}$$ Note that $$Ah = A' k + tv\text{.}$$

\begin{equation*} \begin{split} \snorm{f(x+h) - f(x) - Ah} & = \snorm{f(x+k + t e_n) - f(x+k) - tv + f(x+k) - f(x) - A' k} \\ & \leq \snorm{f(x+k + t e_n) - f(x+k) -tv} + \snorm{f(x+k) - f(x) - A' k} \\ & \leq \snorm{f(x+k + t e_n) - f(x+k) -tv} + \epsilon \snorm{k} . \end{split} \end{equation*}

As all the partial derivatives exist, by the mean value theorem, for each $$j$$ there is some $$\theta_j \in [0,t]$$ (or $$[t,0]$$ if $$t < 0$$), such that

\begin{equation*} f_j(x+k + t e_n) - f_j(x+k) = t \frac{\partial f_j}{\partial x_n}(x+k+\theta_j e_n). \end{equation*}

Note that if $$\snorm{h} < \delta\text{,}$$ then $$\snorm{k+\theta_j e_n} \leq \snorm{h} < \delta\text{.}$$ We finish the estimate

\begin{equation*} \begin{split} \snorm{f(x+h) - f(x) - Ah} & \leq \snorm{f(x+k + t e_n) - f(x+k) -tv} + \epsilon \snorm{k} \\ & \leq \sqrt{\sum_{j=1}^m {\left(t\frac{\partial f_j}{\partial x_n}(x+k+\theta_j e_n) - t \frac{\partial f_j}{\partial x_n}(x)\right)}^2} + \epsilon \snorm{k} \\ & \leq \sqrt{m}\, \epsilon \sabs{t} + \epsilon \snorm{k} \\ & \leq (\sqrt{m}+1)\epsilon \snorm{h} . \qedhere \end{split} \end{equation*}

A common application is to prove that a certain function is differentiable. For example, let us show that all polynomials are differentiable, and in fact continuously differentiable by computing the partial derivatives.

Proof.

Consider the partial derivative of $$p$$ in the $$x_n$$ variable. Write $$p$$ as

\begin{equation*} p(x) = \sum_{j=0}^d p_j(x_1,\ldots,x_{n-1}) \, x_n^j , \end{equation*}

where $$p_j$$ are polynomials in one less variable. Then

\begin{equation*} \frac{\partial p}{\partial x_n}(x) = \sum_{j=1}^d p_j(x_1,\ldots,x_{n-1}) \, j x_n^{j-1} , \end{equation*}

which is again a polynomial. So the partial derivatives of polynomials exist and are again polynomials. By the continuity of algebraic operations, polynomials are continuous functions. Therefore $$p$$ is continuously differentiable.

Subsection8.4.3Exercises

Exercise8.4.1.

Define $$f \colon \R^2 \to \R$$ as

\begin{equation*} f(x,y) := \begin{cases} (x^2+y^2)\sin\bigl({(x^2+y^2)}^{-1}\bigr) & \text{if } (x,y) \not= (0,0), \\ 0 & \text{if } (x,y) = (0,0). \end{cases} \end{equation*}

Show that $$f$$ is differentiable at the origin, but that it is not continuously differentiable.
Note: Feel free to use what you know about sine and cosine from calculus.

Exercise8.4.2.

Let $$f \colon \R^2 \to \R$$ be the function from Exercise 8.3.5, that is,

\begin{equation*} f(x,y) := \begin{cases} \frac{xy}{x^2+y^2} & \text{if } (x,y) \not= (0,0), \\ 0 & \text{if } (x,y) = (0,0). \end{cases} \end{equation*}

Compute the partial derivatives $$\frac{\partial f}{\partial x}$$ and $$\frac{\partial f}{\partial y}$$ at all points and show that these are not continuous functions.

Exercise8.4.3.

Let $$B(0,1) \subset \R^2$$ be the unit ball, that is, the set given by $$x^2 + y^2 < 1\text{.}$$ Suppose $$f \colon B(0,1) \to \R$$ is a differentiable function such that $$\sabs{f(0,0)} \leq 1\text{,}$$ and $$\babs{\frac{\partial f}{\partial x}} \leq 1$$ and $$\babs{\frac{\partial f}{\partial y}} \leq 1$$ for all points in $$B(0,1)\text{.}$$

1. Find an $$M \in \R$$ such that $$\snorm{f'(x,y)} \leq M$$ for all $$(x,y) \in B(0,1)\text{.}$$

2. Find a $$B \in \R$$ such that $$\sabs{f(x,y)} \leq B$$ for all $$(x,y) \in B(0,1)\text{.}$$

Exercise8.4.4.

Define $$\varphi \colon [0,2\pi] \to \R^2$$ by $$\varphi(t) = \bigl(\sin(t),\cos(t)\bigr)\text{.}$$ Compute $$\varphi'(t)$$ for all $$t\text{.}$$ Compute $$\snorm{\varphi'(t)}$$ for all $$t\text{.}$$ Notice that $$\varphi'(t)$$ is never zero, yet $$\varphi(0) = \varphi(2\pi)\text{,}$$ therefore, Rolle's theorem is not true in more than one dimension.

Exercise8.4.5.

Let $$f \colon \R^2 \to \R$$ be a function such that $$\frac{\partial f}{\partial x}$$ and $$\frac{\partial f}{\partial y}$$ exist at all points and there exists an $$M \in \R$$ such that $$\babs{\frac{\partial f}{\partial x}} \leq M$$ and $$\babs{\frac{\partial f}{\partial y}} \leq M$$ at all points. Show that $$f$$ is continuous.

Exercise8.4.6.

Let $$f \colon \R^2 \to \R$$ be a function and $$M \in R\text{,}$$ such that for every $$(x,y) \in \R^2\text{,}$$ the function $$g(t) := f(xt,yt)$$ is differentiable and $$\sabs{g'(t)} \leq M$$ for all $$t\text{.}$$

1. Show that $$f$$ is continuous at $$(0,0)\text{.}$$

2. Find an example of such an $$f$$ that is discontinuous at every other point of $$\R^2\text{.}$$
Hint: Think back to how we constructed a nowhere continuous function on $$[0,1]\text{.}$$

Exercise8.4.7.

Suppose $$r \colon \R^n \setminus X \to \R$$ is a rational function, that is, let $$p \colon \R^n \to \R$$ and $$q \colon \R^n \to \R$$ be polynomials, $$q$$ not identically zero, where $$X = q^{-1}(0)\text{,}$$ and $$r = \frac{p}{q}\text{.}$$ Show that $$r$$ is continuously differentiable.

Exercise8.4.8.

Suppose $$f \colon \R^n \to \R$$ and $$h \colon \R^n \to \R$$ are two differentiable functions such that $$f'(x) = h'(x)$$ for all $$x \in \R^n\text{.}$$ Prove that if $$f(0) = h(0)\text{,}$$ then $$f(x) = h(x)$$ for all $$x \in \R^n\text{.}$$

Exercise8.4.9.

Prove the base case in Proposition 8.4.6. That is, prove that if $$n=1$$ and “the partials exist and are continuous,” then the function is continuously differentiable. Note that $$f$$ is vector-valued.

Exercise8.4.10.

Suppose $$g \colon \R \to \R$$ is continuously differentiable and $$h \colon \R^2 \to \R$$ is continuous. Show that

\begin{equation*} F(x,y) := g(x) + \int_0^y h(x,s) \,ds \end{equation*}

is continuously differentiable, and that it is the solution of the partial differential equation $$\frac{\partial F}{\partial y} = h\text{,}$$ with the initial condition $$F(x,0) = g(x)$$ for all $$x \in \R\text{.}$$

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