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Prove that, $\lim_{h\to 0}\frac{f(a+h)-f(a-h)}{2h}=f'(a)$ if $f'(x)$ is continuous at $a.$

This was how the question was presented in a regional book focusing upon Mean Value Theorems and no other details were given.

My solution is as follows:

We have, $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{h\to 0}\frac{f(a-h)-f(a)}{-h}=\lim_{h\to 0}\frac{f(a)-f(a-h)}{h}.$$

We note that, $$\lim_{h\to 0}\frac{f(a+h)-f(a)-f(a-h)+f(a)}{h}=\lim_{h\to 0}\frac{f(a+h)-f(a-h)}{h}=2f'(a)\implies \lim_{h\to 0}\frac{f(a+h)-f(a-h)}{2h}=f'(a),$$ as required.

However, the solution given in the book is, as follows:

From Lagrange's Mean Value Theorem, we have, $f(a+h)=f(a)+hf'(a+\theta h),\theta\in (0,1)$ and $f(a)=f(a-h)+hf'(a-\theta ' h),\theta '\in (0,1).$

Adding these relations and rearranging we have, $$f(a+h)-f(a-h)=h(f'(a+\theta h)+f'(a-\theta ' h))\implies \frac{f(a+h)-f(a-h)}{2h}=\frac 12[f'(a+\theta h)+f'(a-\theta ' h)].$$

Taking limit $h\to 0$ on both sides, we have, $$\lim_{h\to 0}\frac{f(a+h)-f(a-h)}{2h}=\frac 12\lim_{h\to 0}[f'(a+\theta h)+f'(a-\theta ' h)]=\frac 12\times 2f'(a)=f'(a).$$

However, I feel the approach given in the book has a major flaw. First of all, how did they apply Lagrange's Mean Value Theorem just like that without checking whether $f$ is continuous at $[a,a+h]$ and $[a-h,a].$ Also, in order to apply this theorem we need to have $f$ being differentiable in $(a,a+h)$ and $(a-h,a).$ They never guaranteed such claims holding true. I think, this makes the proof given in the book incorrect.

Also, the information that, $f'(x)$ is continuous at $a$ seems totally useless and unnecessary!

Lastly, is my solution at the beginning a valid one?

Matthew Towers
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Thomas Finley
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  • @MartinR I am sorry but I wanted to prove the thing that's required to prove mentioned in the beginning of the OP. The assumptions made on $f$ are nothing except the only one mentioned at the beginning of the OP. – Thomas Finley Sep 23 '23 at 14:38
  • @MartinR Based on the question as posed, is my solution valid? Also, doesn't the approach in the book much nonsensical? – Thomas Finley Sep 23 '23 at 14:39
  • If the statement is “If $f$ is diffentiable at $a$ then $\lim_{h\to 0}\frac{f(a+h)-f(a-h)}{2h}=f'(a)$” then your proof is correct (and similar to the proof in the thread that I linked to). – Martin R Sep 23 '23 at 14:50
  • That “problem from the book” does not even make sense to me. You could say “Prove ... for all functions $f$ such that $f'$ is continuous at $a$” which would be correct, but with unnecessary hypotheses. – Martin R Sep 23 '23 at 14:53
  • @MartinR Thanks a ton! I got a sigh of relief! But If $f$ was not differentiable at $a$ then $f'(a)$ as given in the question makes no sense and is absurd, isn't it? Also, when they say, "...such that $f'(x)$ is continuous at $a$" I think, this implies $f(x)$ is everywhere differentiable in the domain of $f$ as otherwise, the symbol $f'(x)$ in this line, wouldn't make any sense. – Thomas Finley Sep 23 '23 at 14:53
  • @MartinR I just rephrased the problem in OP. I think this might clear the confusion. – Thomas Finley Sep 23 '23 at 14:55
  • "Also, the information that, f′(x) is continuous at a seems totally useless and unnecessary!" It's necessary. You use this fact in the first line of your proof. – Alborz Sep 23 '23 at 15:11
  • @Alborz: I don't think so. $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ holds just by the definition of the derivative at the point $a$. It is not needed that $f'$ is continuous. – Martin R Sep 23 '23 at 15:19
  • @Alborz I agree with MartinR and I think it follows naturally from the definition of derivatives. – Thomas Finley Sep 23 '23 at 15:20
  • Related: Prove that the symmetric derivative of a function exists whenever the derivative exists AND Symmetric Derivative is it right? AND Show that differentiability at a point implies symmetric differentiability at a point AND (for mature readers) this MSE answer. – Dave L. Renfro Sep 23 '23 at 15:41
  • @DaveL.Renfro Sorry, but till now, I have never encountered the phrase "Symmetric Derivative". – Thomas Finley Sep 23 '23 at 15:42
  • I have never encountered the phrase "Symmetric Derivative" --- I wonder whether textbook authors, even before the internet, didn't mention "symmetric derivative" when giving this as an exercise (something I've noticed is often the case) in order to prevent students from simply looking up a proof in real analysis texts that included a proof (e.g. go to the library where real analysis texts are shelved and look for "symmetric derivative" in the index of various real analysis texts). – Dave L. Renfro Sep 23 '23 at 15:51

1 Answers1

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I'll assume that all this is about a function $f: I \to \Bbb R$ for some interval $I \subset \Bbb R$ and $a$ is an interior point of $I$.

$\lim_{h\to 0}\frac{f(a+h)-f(a-h)}{2h}=f'(a)$ holds whenever $f$ is differentiable at $a$. Only the definition of $f'(a)$ is needed, as your proof correctly demonstrates.

If (as in your book) it is said that $f'$ is continuous at $a$ then it is probably meant that $f'$ exists in a neighborhood of $a$ (and is continuous at $a$), which in turn implies that $f$ is continuous in a neighborhood of $a$. In that case, the mean-value theorem can be applied to $[a-h, a]$ and $[a, a+h]$ for sufficiently small $h$ and the “proof from the book” is valid as well.

But again, that condition ($f'$ is continuous at $a$) is not need for the conclusion.

Martin R
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  • Just one more question: How do you say, " $f'$ is continuous at $a$ then this implies that $f'$ is differentiable in a neighborhood of $a$ "? – Thomas Finley Sep 23 '23 at 15:56
  • @ThomasFinley: I was about to say that a function (here: $f'$) can only be continuous at a point if it is defined in a neighbourhood of that point. But actually that is not correct. I have adjusted the formulation in my answer. – Martin R Sep 23 '23 at 16:14
  • Thanks! It's perfect now. – Thomas Finley Sep 23 '23 at 16:18