Exercise 5.2.8

Question

Review the definition of uniform continuity (Definition 4.4.4). Given a differentiable function $f: A ightarrow \mathbf{R}$, let's say that $f$ is \emph{uniformly differentiable} on $A$ if, given $\epsilon>0$ there exists a $\delta>0$ such that $$ \left|\frac{f(x)-f(y)}{x-y}-f^{\prime}(y) ight|<\epsilon \quad ext { whenever } 0<|x-y|<\delta . $$ \begin{enumerate}[label=(\alph*)] \item Is $f(x)=x^{2}$ uniformly differentiable on $\mathbf{R} ?$ How about $g(x)=x^{3} ?$ \item Show that if a function is uniformly differentiable on an interval $A$, then the derivative must be continuous on $A$. \item Is there a theorem analogous to Theorem 4.4.7 for differentiation? Are functions that are differentiable on a closed interval $[a, b]$ necessarily uniformly differentiable? \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item We have
    $$
    \left|\frac{x^2 - y^2}{x - y} - 2y\right| = \left|(x + y) - 2y\right| = |x - y| < \delta
    $$
    Thus $\delta = \epsilon$ suffices to show $x^2$ is uniformly differentiable. Now for $x^3$
    $$
    \left|\frac{x^3 - y^3}{x - y} - 3y^2\right| = \left|\frac{(x-y)(x^2 + xy + y^2)}{x-y} - 3y^2\right| = |x^2 + xy - 2y^2|
    $$
    Let $y = (x+h)$ to get
    $$|x^2 + x(x+h) - 2(x+h)^2| = |x^2 + x^2 xh - 2x^2 - 4xh - 2h^2| = |3xh + 2h^2|$$
    Fix $0 < h < \delta$, since $x$ can be as big as we want we can make $|3xh + 2h^2| > \epsilon$.
    As no fixed $\delta$ works $x^3$ is not uniformly differentiable.

\item We need to show that $\lim_{x \to y} f'(x) = f'(y)$. Now, if we choose $\delta$ small enough such that when $\left|x - y\right| < \delta$, both
    $$\left|f'(x) - \frac{f(x) - f(y)}{x - y}\right| < \frac{\epsilon}{2}$$
    and
    $$\left|\frac{f(x) - f(y)}{x - y} - f'(y)\right| < \frac{\epsilon}{2}$$
    (this is possible because $f$ is uniformly differentiable), then we have
    $$\begin{aligned}
    f'(x) - f'(y) &= f'(x) -\frac{f(x) - f(y)}{x - y} + \frac{f(x) - f(y)}{x - y} - f'(y) \
    & \leq \left|f'(x) - \frac{f(x) - f(y)}{x - y}\right| +\left|\frac{f(x) - f(y)}{x - y} - f'(y)\right|  < \epsilon
    \end{aligned}
    $$
    as desired.

\item
    Consider the counterexample $f(x) = x^2\sin(1/x)$ over $[0,1]$ (where $f(0) = 0$).
    $f$ is differentiable over $[0,1]$ but not uniformly differentiable.

Intuitively this is because I can find $x_n, y_n$ such that the slope between them becomes unbounded, but the derivative $f'$ must stay bounded. To be exact set
    $$
    x_n = \frac{1}{2\pi n + \pi/2}, \quad y_n = \frac{1}{2\pi n}
    $$
    then
    $$
    \begin{aligned}
      \left|\frac{f(x_n) - f(y_n)}{x_n - y_n} - f'(x_n)\right|
      &= \left|\frac{1}{x_n - y_n} - f'(x_n)\right| \
      &= \left|\frac{(2\pi n)(2\pi n + \pi/2)}{\pi/2} - f'(x_n)\right| \
      &= \left|4n(2\pi n + \pi/2) - f'(x_n)\right|
    \end{aligned}
    $$
    Now since $f'(x_n) = 2x\sin(1/x) - \cos(1/x)$ is bounded I can defeat any $\delta$ by picking $n$ large enough so that
    $$
    |x_n - y_n| < \delta \quad\text{and}\quad \left|\frac{f(x_n) - f(y_n)}{x_n - y_n} - f'(x_n)\right| \ge \epsilon
    $$
    Thus $f$ is not uniformly differentiable.

If you try to apply the same proof as for uniform continuity you get stuck at the triangle inequality.
   \end{enumerate}

Exercise 5.2.8

Answers

Comments

Exercise 5.2.8

Answers

Comments

Add answer