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Let $ f: \mathbb{R} \rightarrow \mathbb{R}$ . Show that $f$ is continuously differentiable if and only if, for every $x \in \mathbb{R}$ there exists a $l \in \mathbb{R}$ with the property that for all $\epsilon >0$ there exists a $\delta>0$, such that for all $h,t, \in B(0,\delta)$ the following holds:

$$ |f(x+h) -f(x+t) - l(h-t)| \leq \epsilon |h-t| $$

I've been staring at this exercise for way too long and still have no clue on how to proof this. The statement looks very much like the definition of the differential but I don't see how to get both a $h$ and a $t$ in my expression.
Furthermore I don't see how the continuously differentiable should be used to proof this property. I tried to use the mean value theorem but this makes it all to confusing for me as I suppose I would need the limit definition but then I don't see how to get rid of it.

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Let's call the condition (1) (so that we have to show that (1) is equivalent to continuous differentiability). If $f$ is continuously differentiable, you let $l=f'(x)$. Sice $f'$ is continuous, then for $\epsilon>0$ there exists $\delta>0$ such that $|f'(y)-l|<\epsilon$ for all $y\in\mathbb{R}$ with $|y-x|<\delta$. Take $h,t\in B(0,\delta)$. If $h=t$, then we are obviously done, so assume that $h\neq t$. By the mean value theorem there exists $c$ between $t$ and $h$ such that $${f(x+h)-f(x+t)}=f'(x+c)(h-t).$$ Thus \begin{align*}|f(x+h)-f(x+t)-l(h-t)|&=|f'(x+c)(h-t)-l(h-t)|=|f'(x+c)-l||h-t|.\end{align*} Since $|c|\leq\max\{|t|,|h|\}<\delta$, we have $|f'(x+c)-l|<\epsilon$, proving one direction.

Conversely, if (1) holds, then if we first fix $x\in\mathbb{R}$ and let $l\in\mathbb{R}$ such that property in (1) is satisfied, then let $\epsilon>0$. Take $\delta>0$ such that the inequality in (1) is satisfied for all $h,t\in B(0,\delta)$. By letting $t=0$ we have $$|f(x+h)-f(x)-lh|\leq\epsilon|h|.$$ Dividing by $|h|$, we see that $f$ is differentiable in $x$ with $f'(x)=l$. Now, to see that $f'$ is continuous at any $x\in\mathbb{R}$, let $\epsilon>0$ and take $\delta>0$ such that the inequality in (1) is satisfied again for all $h,t\in B(0,\delta)$. Fix $h\in B(0,\delta)$, and let $\delta_1>0$ such $$|f(x+h+v)-f(x+h)-f'(x+h)v|\leq\epsilon|v|$$ for $|v|\leq\delta_1$ (which is possible by $f$ being differentiable at $x+h$). Take a non-zero $v\in B(0,\delta_1)$ with $h+v\in B(0,\delta)$. Then $$|f(x+h+v)-f(x+h)-f'(x)v|\leq\epsilon|v|$$ by applying the inequality in (1) to $x$. Thus $|f'(x)v-f'(x+h)v|\leq 2\epsilon|v|$ by the triangle inequality. Dividing by $|v|$, we have $|f'(x)-f'(x+h)|\leq 2\epsilon$, so since this holds for any $h\in B(0,\delta)$, $f'$ is continuous at $x$.

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maybe this is of help:

For the '<-'-direction: Set $t = 0$, then for all $0 \neq h \in B(0, \gamma): |f(x+h) - f(x) - l*h| \leq \epsilon * |h|$ which is equivalent to $\frac{|f(x+h) - f(x) - l*h|}{|h|} \leq \epsilon$. Now get rid of the absolute value function to see that: $\frac{f(x+h) - f(x)}{h} - l \in (-\epsilon,\epsilon)$. Since this is true for all $\epsilon > 0$, we find $\frac{f(x+h) - f(x)}{h} \to_{h \to 0} l$.

This should also give you an intuition for the '->'-direction, $l$ should be the value of the derivative at x.

Greetings.

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  • $\begingroup$ This definitely helps, but I don't see how the derivative should be continuous which I also need to proof. The other direction I still don't see :( $\endgroup$ – DeanTheMachine Jun 28 '15 at 11:32
  • $\begingroup$ The other direction is very confusing, but what I'm trying now is to subtract $(f(x+h)-f(x))/h$ and $(f(x+t)-f(x))/t$ , but I don't feel like I'm allowed to do this $\endgroup$ – DeanTheMachine Jun 28 '15 at 11:39

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