1
$\begingroup$

How would I go about showing that if $f$ is uniformly differentiable then $f'$ is continuous.

my attempt: A differentiable function $f:[a,b]\to \Bbb R$ is said to be uniformly differentiable on $[a,b]$ if $\forall \epsilon>0 \exists \delta>0:\forall x,y\in[a,b]$ we have $$0<|x-y|<\delta\implies |\frac{f(x)-f(y)}{x-y}-f'(y)|<\epsilon$$

I don't know where to go from here, any help would be highly appreciated.

$\endgroup$
  • $\begingroup$ If $f$ was just differentiable (not uniformly), then would the result hold ?? $\endgroup$ – Anik Bhowmick Aug 12 '18 at 4:07
2
$\begingroup$

HINT: $$\vert f'(x) - f'(y)\vert \leq \left\vert f'(x) - \frac{f(x) - f(y)}{x - y} \right\vert + \left\vert \frac{f(x) - f(y)}{x - y} - f'(y)\right\vert$$

$\endgroup$
2
$\begingroup$

Let $\varepsilon > 0$. Since $f$ is uniformly differentiable there exists a $\delta > 0$ such that $\forall x,y\in[a,b]$ if $|x-y| < \delta$ then: $$ \bigg{|}\frac{f(x)-f(y)}{x-y} - f'(x)\bigg{|} < \frac{\varepsilon}{2} $$ interchanging $x$ and $y$ (why can you do this?) you get: $$ \bigg{|}\frac{f(x)-f(y)}{x-y} - f'(y)\bigg{|} < \frac{\varepsilon}{2} $$ Now let $|x-y|< \delta$ then (adding and substracting the same term): $$ |f'(x)-f'(y)| = \bigg{|} f'(x) + \frac{f(x)-f(y)}{x-y} - \frac{f(x)-f(y)}{x-y} - f'(y) \bigg{|} \\ \leq\bigg{|} \frac{f(x)-f(y)}{x-y} - f'(x)\bigg{|} + \bigg{|} \frac{f(x)-f(y)}{x-y} - f'(x) \bigg{|} \\ < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon $$

$\endgroup$
  • $\begingroup$ May i ask why it's less than $\frac{\epsilon}{2}$ and can you interchange $x$ and $y$ because of something to do with the definition of $f'(x)$ and $f'(y)$? $\endgroup$ – user395952 Apr 13 '17 at 0:55
  • $\begingroup$ Choose $\epsilon' = \epsilon/2$ then since $\epsilon>0$ also $\epsilon/2 >0$. Now, pay attention to your definition it says $\forall\epsilon>0$ in particular use that for $\epsilon'$. This is an important trick of Analysis and you should get used to it. $\endgroup$ – user263732 Apr 13 '17 at 1:02
  • $\begingroup$ ahh yes okay, that makes sense, the question just states to prove that it is continuous, does this mean that it is also uniformly continuous? $\endgroup$ – user395952 Apr 13 '17 at 2:06
  • $\begingroup$ yes, it is uniformly continuous. $\endgroup$ – user263732 Apr 13 '17 at 17:03
  • $\begingroup$ Yes, continuity on closed bounded interval implies uniform continuity $\endgroup$ – CHOUDHARY bhim sen Oct 23 '18 at 12:19
1
$\begingroup$

The answers from Ethan and Juan work, but here is a more conceptual approach.

Write $F_n(y)=\frac{f(y+\frac{1}{n})-f(y)}{\frac{1}{n}}$, then the assumption of uniform differentiability implies that $$F_n(y) \text{ converges to }f'(y)\text{ uniformly on }[a,b].$$ Note that the the $F_n$'s are all continuous functions. Then by the fact that uniform limit of continuous functions is also continuous, $f'(y)$ is continuous.

$\endgroup$
  • 1
    $\begingroup$ This one is really neat! $\endgroup$ – user263732 Apr 13 '17 at 0:58
  • $\begingroup$ Why are the $F_{n}$'s continuous? $\endgroup$ – Alex D May 12 '18 at 1:43
  • $\begingroup$ At the end points you need to define separately otherwise your function is not defined at $b$ $\endgroup$ – CHOUDHARY bhim sen Oct 23 '18 at 12:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy