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Let $f\in C^1([a,b])$ have 2nd-order derivative in $(a,b)$. Prove that there exists $c\in (a,b)$ such that $$f(b)-2 f\left(\frac{a+b}{2}\right)+f(a)=\frac{1}{4} (b-a)^2 f''(c)$$

Jonathan Y.
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user103354
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  • Thanks for your zeal! I am a college student and this is a question that I come across in a reference book. But it didn't provide the proof. The main point that sucks is the existence of (a+b). – user103354 Nov 07 '13 at 11:19
  • See my answer to a related question. I'm sure I have seen several similar questions like this before.. – achille hui Nov 07 '13 at 11:54

2 Answers2

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Hint: As you say, generally $a+b\not\in[a,b]$, but $\frac{a+b}{2}$ divides $[a,b]$ to two. Consider, then, the function $g\in C^1([a,\frac{a+b}{2}])$ defined $g(x) = f(x+\frac{b-a}{2})-f(x)$.

Added (some more explicit remarks, please consider before reading everything)

  1. One notes that $g$ indeed is (twice-)differentiable on $(a,\frac{a+b}{2})$ and $$g^\prime(x) = f^\prime(x+\frac{b-a}{2})-f^\prime(x).$$

  2. One also notes $g(\frac{a+b}{2}) - g(a) = f(b)-2f(\frac{a+b}{2})+f(a)$.

  3. The mean value theorem implies that there exists some $c\in(a,\frac{a+b}{2})$ such that $$g(\frac{a+b}{2}) - g(a) = \frac{b-a}{2}g^\prime(c) = \frac{b-a}{2}\left(f^\prime(c+\frac{b-a}{2})-f^\prime(c)\right).$$

  4. This last part completes the problem:

    Another application of the mean value theorem for $f^\prime$ over the interval $[c,c+\frac{b-a}{2}]\subset[a,b]$, yields $\tilde{c}\in(c,c+\frac{b-a}{2})$ such that $$f^\prime(c+\frac{b-a}{2})-f^\prime(c) = \frac{b-a}{2}f^{\prime\prime}(\tilde{c}).$$

Jonathan Y.
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Let $p = (a + b)/2$ and $h = (b - a)/2$ so that $a = p - h, b = p + h$ then we have $$f(b) - 2f\left(\frac{a + b}{2}\right) + f(a) = f(p + h) - 2f(p) + f(p - h)$$ Let $g(x) = f(p + x) - 2f(p) + f(p - x)$ then $g(0) = 0$ and $g'(x) = f'(p + x) - f'(p - x)$ so that $g'(0) = 0$. It follows from Taylor's theorem that $$g(h) = g(0) + hg'(0) + \frac{h^{2}}{2}g''(c')$$ where $c' \in (0, h)$. Now $g''(x) = f''(p + x) + f''(p - x)$ and hence $g''(c') = f''(p + c') + f''(p - c')$ and since derivatives follow intermediate value theorem, there is a value $c \in [p - c', p + c']\subset (a, b)$ such that $f''(c) = \{f''(p + c') + f''(p - c')\}/2 = g''(c')/2$.

It follows that $$g(h) = h^{2}f''(c)$$ and putting value of $h = (b - a)/2$ we are done.

Paramanand Singh
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  • How does Taylor's theorem justify the first equality ($g(h)$)? – Jonathan Y. Nov 07 '13 at 13:49
  • Taylor's theorem says that $f(a + h) = f(a) + hf'(a) + \cdots + \dfrac{h^{n - 1}}{(n - 1)!}f^{(n - 1)}(a) + \dfrac{h^{n}}{n!}f^{(n)}(a + \theta h)$ where $\theta \in (0, 1)$. I have put $n = 2$ and $a = 0$. – Paramanand Singh Nov 07 '13 at 13:53
  • To Jonathan Y.: By the way I would also like to see your hint expanded into an answer, because I couldn't complete that hint into answer myself and had to think of a bit long answer. – Paramanand Singh Nov 07 '13 at 13:55
  • That's probably because I had a typo. $\frac{a+b}{2}$ should've been $\frac{b-a}{2}$ in the definition of $g$ (it isn't even well-defined the other way around). Either way, a full solution is now included. – Jonathan Y. Nov 07 '13 at 14:46
  • I should have noted this typo, but couldn't and was led astray. Thanks for a very good solution (+1). – Paramanand Singh Nov 07 '13 at 16:12
  • Much appreciated (as was your spotting an inconsistency). – Jonathan Y. Nov 07 '13 at 16:24