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Can someone check my proof if a function $f:{R\mapsto R}$ is differentiable at all points, it is continuous throughout as well?

Proof:

Let $x$ be a real number, since $f(x)$ is differentiable, or $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ exists and is equal to $f'(x)$. So, for any $\epsilon > 0$ there exists a $\delta > 0$, such that $0<|h - 0| < \delta$

$\implies |\frac{f(x+h)-f(x)}{h} - f'(x)| < \epsilon$

$\implies |\frac{f(x+h)-f(x)-f'(x).h}{h}| < \epsilon$

$\implies |f(x+h)-f(x)-f'(x).h| < |h|\epsilon$

$\implies -|h|\epsilon<f(x+h)-f(x)-f'(x).h < |h|\epsilon$

$\implies f'(x).h-|h|\epsilon<f(x+h)-f(x) < f'(x).h+|h|\epsilon$

Let $\epsilon'$ be min($f'(x).h+|h|\epsilon - f(x+h)+f(x), f'(x).h-|h|\epsilon-f(x+h)+f(x) $), then we can say,

$\implies -\epsilon'<f(x+h)-f(x) < \epsilon'$

$\implies |f(x+h)-f(x)| < \epsilon'$

$\implies \lim_{h \to 0}f(x+h)$ exists and is equal to $f(x)$

$\implies f$ is continuous at $x$

NOTE: This is not a homework question, I am trying to learn basic calculus on my own through the web.

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    $\begingroup$ May I give you a simple proof? $\endgroup$ – md2perpe Jul 28 '18 at 20:26
  • $\begingroup$ @md2perpe, It would be helpful, please post the link. $\endgroup$ – avocado Jul 28 '18 at 20:41
  • $\begingroup$ Should be $h\to 0$, not $h\mapsto 0$. $\endgroup$ – Shalop Jul 28 '18 at 20:53
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    $\begingroup$ Thanks @Shalop, edited $\endgroup$ – avocado Jul 28 '18 at 20:56
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This is not correct. You cannot choose $\varepsilon'$. What you are supposed to prove is that for every $\varepsilon'>0$, there is a $\delta>0$ such that$$|h|<\delta\implies\bigl|f(x+h)-f(x)\bigr|<\varepsilon'.$$

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  • $\begingroup$ Thanks for the edit as well as letting me realise my mistake. I'll think about it more. $\endgroup$ – avocado Jul 28 '18 at 20:39
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Jul 28 '18 at 20:49
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Simple proof without explicit use of $\epsilon$-$\delta$:

For any $x$ we have $$ \left| f(x+h) - f(x) \right| = \left| \frac{f(x+h) - f(x)}{h} h \right| = \left| \frac{f(x+h) - f(x)}{h} \right| |h| \to |f'(x)| \cdot 0 = 0 $$ as $h \to 0.$ This implies that $f(x+h) \to f(x)$ as $h \to 0.$

Rewritten using $\epsilon$-$\delta$

Fix $x \in \mathbb R.$

Let $\delta_0>0$ be such that $$\left| \frac{f(x+h) - f(x)}{h} - f'(x) \right| < 1$$ whenever $|h| < \delta_0.$ Such $\delta_0$ exists by definition of $f'(x)$ as a limit of $(f(x+h)-f(x))/h.$

Given $\epsilon>0,$ let $\delta_1 = \epsilon/(1+|f'(x)|)$ and let $\delta = \min(\delta_0, \delta_1).$ Then $$ \left| f(x+h) - f(x) \right| = \left| \frac{f(x+h) - f(x)}{h} h \right| = \left| \frac{f(x+h) - f(x)}{h} \right| |h| = \left| \frac{f(x+h) - f(x)}{h} - f'(x) + f'(x) \right| |h| \\ = \{ \text{ triangle inequality } \} \\ \leq \left( \left| \frac{f(x+h) - f(x)}{h} - f'(x) \right| + \left| f'(x) \right| \right) |h| \\ < \left( 1 + \left| f'(x) \right| \right) \frac{\epsilon}{1 + |f'(x)|} = \epsilon $$ whenever $|h| < \delta.$ Thus $f(x+h) \to f(x)$ as $h \to 0,$ i.e. $f$ is continuous in $x.$ Since $x$ was taken arbitrary, $f$ is continuous on all of $\mathbb R.$

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  • $\begingroup$ Thanks a lot @md2perpe for taking out time to write this answer. I really like the simplicity of both the proofs. $\endgroup$ – avocado Jul 28 '18 at 21:29
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    $\begingroup$ The $\epsilon$-$\delta$ proof is certainly a bit more complicated that using limit laws. Even so, it could have been simpler if I hadn't cared about ending with $\epsilon$ but accepted $(\epsilon + |f'(x)|) \epsilon$ which obviously can be made arbitrarily small. I don't know which one is least confusing; choosing $\delta$ in a complicated way or not ending with $\epsilon.$ $\endgroup$ – md2perpe Jul 28 '18 at 21:51
  • $\begingroup$ I feel choosing $\delta$ in the proof may look a little strange at the beginning, however since it leads to a form which is seen throughout the text books. I personally feel this is slightly better than ending with $(\epsilon+|f'(x)|)\epsilon$. $\endgroup$ – avocado Jul 29 '18 at 5:28
  • $\begingroup$ I've debated with myself the same question. I feel like maybe the best way to go actually shows the process of learning and doing Math: initially just arrive at something that can be made arbitrarily small and declare "This is pretty good--it is an adequate proof. But if you really need to see it with a simple $\varepsilon$ we can use the work we have already done and just re-jigger the choice of $\delta$..." Basically show the process of making an initial proof that is correct in spirit and then ironing out the details. $\endgroup$ – Addem Jul 30 '18 at 16:49
  • $\begingroup$ @Addem. I agree. $\endgroup$ – md2perpe Jul 30 '18 at 17:42
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Your proof does not work. It seems you have a misunderstanding of the definition of continuity. Your mistake is that your $\epsilon'$ depends on $h$. But this makes no sense (why? hint: continuity means that for every arbitrary margin of error $\epsilon$ you can find a neighborhood of $h$ where your function will be within that margin of error)

Try to fix it now that you understand and check the accepted answer here:

How to prove differentiability implies continuity with $\epsilon-\delta$ definition?

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  • $\begingroup$ Thanks John11, I'll surely check out the other answers. $\endgroup$ – avocado Jul 28 '18 at 20:40

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