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In this post it is said that if $f : \mathbb R \to \mathbb R$ is differentiable at $a$ then there exists a continuous function $\varphi$ defined on an interval $[-\epsilon, \epsilon]$ such that $\varphi(0) = 0$ and $$ f(a + h) = f(a) + f'(a)h + \varphi(h)h $$ for all $h \in (-\epsilon, \epsilon)$.

Now my question: Is see that $\varphi$ in some sense resembles the error term, and that as $$ \varphi(h) = \frac{f(a + h) - f(a)}{h} - f'(a) $$ we have $\varphi(h) \to 0$ as $h \to 0$, hence it is continuous at $0$ (which implies that $f$ is continuous at $a$). But is $\varphi$ necessarily continuous on the entire interval $[-\epsilon, \epsilon]$? I cannot prove that...

EDIT: Found this post, which is somehow related.

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StefanH
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  • Yes because f and the identity map are continuous, and quotient of two continuous functions is continuous. – Singh Mar 17 '16 at 18:25
  • @Singh: We can just suppose that $f$ is continuous at the point where it is differentiable, not on the whole of $[-\epsilon,\epsilon]$. – StefanH Mar 17 '16 at 18:29
  • Your equality implies $f$ is continuous on $(a-\epsilon,a+\epsilon)$, but differentiability at $a$ does not imply this. So the answer is no under these assumptions. Almost any way you can think of to strengthen the assumptions changes the answer to yes. – Ian Mar 17 '16 at 18:29
  • @Ian Yes, thank you that would certainly not hold. – StefanH Mar 17 '16 at 18:30
  • Thnx Stefan I misunderstood. – Singh Mar 17 '16 at 18:30
  • @Ian Almost anyway? Even supposing continuity of $f$ on its entire domain would not be enough, as this does not imply continuity of $\varphi$ on the endpoints $-\epsilon$ and $\epsilon$... (equality is just asserted in the interior of the interval) – StefanH Mar 17 '16 at 18:33
  • @Stefan In the linked post there is no actual need to include the endpoints of the same interval, because the result is local, so you can just replace $\epsilon$ by $\epsilon/2$. The problem really occurs when $f$ is not continuous in any neighborhood of $a$. – Ian Mar 17 '16 at 18:34
  • @Ian You are right, I just thought about fixed $\epsilon$'s, but the statement just supposes for some $\epsilon$. – StefanH Mar 17 '16 at 18:36

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Check the function given by $f(x) = x^2$ if $x \in \mathbb{Q}$ and $f(x) = x^3$ if $x\not\in \mathbb{Q}$ at $a=0$.

Catalin Zara
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  • Nice example... Catalin – Singh Mar 17 '16 at 18:43
  • I see, as $h^3 / h \le f(h)/h \le h^2 / h$ for all $-1 < h < 1, h \ne 0$ this limit (i.e. the derivative at $0$) goes to zero, hence it is differentiable at $0$, but $f$ is not continous at any point $\ne 0$ as it always does jumps of size $x^3 - x^2$ for some $x \ne 0$ in each interval. – StefanH Mar 17 '16 at 18:49