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EDIT 2: I just posted my revised proof, where I used two Taylor expansions, and subtracting both equations to get something that's pretty close to what I want. What do you think? Please see below. Thanks,

EDIT: I think that, with the help of Joey Zou and Claudeh5's attempts at a solution (please see below), I am pretty close to an answer. At present there are some things of concern:

a) Joey Zou's more technical proof seems to rely on $f''$ being continuous or at least Riemann-integrable. Unfortunately, I don't think we can assume that.

b) Claudeh5's cute one-line proof almost does the job. It uses Taylor expansion, the Lagrange Remainder, and centering the expansion about $x=a$, and evaluating the series at $x=b$. However, the estimate is not quite good enough.

Any hints or comments are welcome. Thanks,

The problem statement is:

Assume that $f(x)$ has second derivatives on [a,b], and $f′(a)=f′(b)=0$.

Prove that there exists a point $c∈[a,b]$ such that $f′′(c)≥\frac{4}{(a−b)^2}|f(b)−f(a)|$.

Using Claudeh5's approach, here is my revised proof, which is perhaps closer to the desired upper bound:

By Taylor expansion, and the Lagrange remainder, we have that

$$f(b) = f(a) + \frac{f''(\psi_1)}{2!} (b-a)^2$$

Now again, but centering the expansion about $x=b$ gives us

$$f(a) = f(b) + \frac{f''(\psi_2)}{2!} (a-b)^2$$

Note that the first derivatives vanish at $x=a$ and $x=b$, by assumption.

Now subtracting the two Taylor series gives

$$2[f(b)-f(a)]= \frac {[f''(\psi_1) - f''(\psi_2)]}{2!}(a-b)^2$$

$$\implies \frac {4[f(b)-f(a)]}{(a-b)^2}= [f''(\psi_1) - f''(\psi_2)]$$

$$\implies \frac {4[f(b)-f(a)]}{(a-b)^2} \le max \{f''(\psi_1), f''(\psi_2)\}$$

$$=:f''(c)$$

I am not confident about the last two lines of my proof.

Am I on the right track? I feel a bit closer now to achieving the upper bound ...

Thanks,

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    $\begingroup$ Hint: $f(b)-f(a)=\int_a^b f'(u) du= \int_a^b \int _a^u f''(v) dv du$ $\endgroup$ – Claudeh5 Nov 25 '15 at 4:23
  • $\begingroup$ Hi @Claudeh5, I just posted my attempt, and noticed your comment. hmmm....why do you have double-integration? I will think about it now...thanks for your hint, $\endgroup$ – User001 Nov 25 '15 at 4:27
  • $\begingroup$ Hi @Claudeh5, I agree with your derivation, but how does it help? Thanks :-) $\endgroup$ – User001 Nov 25 '15 at 4:34
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    $\begingroup$ @LebronJames should the inequality be on $|f''(c)|$ and not $f''(c)$? It seems false as currently stated. $\endgroup$ – Joey Zou Nov 25 '15 at 4:52
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    $\begingroup$ "This seems to show that the first derivative is constant on the interior (a,b)" No !!!! (very big fault) and $\psi$ is depending of a and b. $\endgroup$ – Claudeh5 Nov 25 '15 at 4:55
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Let $M = \sup\limits_{x\in[a,b]}{|f''(x)|}$. Since $f'(a) = 0$, we have, for $a\le x\le b$, $$ |f'(x)| = \left|f'(a) + \int\limits_{a}^{x}{f''(x)\text{ d}x}\right|\le\int\limits_{a}^{x}{|f''(x)|\text{ d}x}\le M(x-a). $$ Since $f'(b) = 0$, we similarly have $$ |f'(x)| = \left|f'(b) + \int\limits_{x}^{b}{f''(x)\text{ d}x}\right|\le\int\limits_{x}^{b}{|f''(x)|\text{ d}x}\le M(b-x). $$ It follows that \begin{align*} |f(b)-f(a)| &= \left|\int\limits_{a}^{b}{f'(x)\text{ d}x}\right|\\ &\le\int\limits_{a}^{\frac{a+b}{2}}{|f'(x)|\text{ d}x} + \int\limits_{\frac{a+b}{2}}^{b}{|f'(x)|\text{ d}x}\\ &\le\int\limits_{a}^{\frac{a+b}{2}}{M(x-a)\text{ d}x} + \int\limits_{\frac{a+b}{2}}^{b}{M(b-x)\text{ d}x} \\ &= M\frac{(b-a)^2}{4}. \end{align*} Furthermore, I claim that equality cannot be achieved. Note that $f'$ is continuous (as $f''$ exists), and since $|f'(x)|\le M(x-a)$, the only way for the equality $$\int\limits_{a}^{\frac{a+b}{2}}{|f'(x)|\text{ d}x} = \int\limits_{a}^{\frac{a+b}{2}}{M(x-a)\text{ d}x}$$ to hold is if $|f'(x)| = M(x-a)$ for all $x\in\left(a,\frac{a+b}{2}\right)$. Similarly, for the other equality to hold, we need $|f'(x)| = M(b-x)$ for all $x\in\left(\frac{a+b}{2},b\right)$. But it is impossible for $f'$ to be differentiable at $\frac{a+b}{2}$ while satisfying these two equalities. Hence, the inequality is strict, so $$|f(b)-f(a)| < M\frac{(b-a)^2}{4}\implies M > \frac{4}{(b-a)^2}|f(b)-f(a)|. $$ Since $M$ is the supremum over all values of $|f''(x)|$ on $[a,b]$, it follows that there exists $c\in[a,b]$ such that $|f''(c)|\ge\frac{4}{(b-a)^2}|f(b)-f(a)|$.


Edit: as pointed out by Lebron James, it is not necessarily true that $f''$ is Riemann integrable, or even bounded. Now

  1. If $f''$ is not bounded, then we automatically get our result.

  2. Otherwise, $M=\sup\limits_{x\in[a,b]}{|f''(x)|}$ exists and is finite. We can still show that $|f'(x)|\le M(x-a)$ and $|f'(x)|\le M(b-x)$ using the mean value theorem. The rest of thep proof should still work.

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    $\begingroup$ I should point out that the whole shtick with proving that the inequality is strict and whatnot can be avoided if we assume that $f''$ is continuous, since then the supremum is just a maximum. $\endgroup$ – Joey Zou Nov 25 '15 at 5:38
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    $\begingroup$ a very nice answer! $\endgroup$ – David Holden Nov 25 '15 at 5:42
  • $\begingroup$ Hi @JoeyZou, what's the motivation for choosing (a+b)/2 as your upper limit in one integral and the lower limit in the second integral? Thanks, $\endgroup$ – User001 Nov 25 '15 at 5:54
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    $\begingroup$ @LebronJames good question! The idea is that I got two upper bounds for $|f'(x)|$ (namely $M(x-a)$ and $M(b-x)$), and the midpoint $(a+b)/2$ is where these upper bounds are equal. Essentially, I wanted to bound by the smaller of the two upper bounds, and $M(x-a)$ is smaller for $x<(a+b)/2$, while $M(b-x)$ is smaller for $x>(a+b)/2$. $\endgroup$ – Joey Zou Nov 25 '15 at 6:00
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    $\begingroup$ @LebronJames you bring up good points. I have edited my answer accordingly. $\endgroup$ – Joey Zou Nov 26 '15 at 1:28
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hint2 : with Taylor-Lagrange formula : $f(b)=f(a) +(b-a) f'(a)+ \frac{(b-a)^2}2 f''(d)$ ($d$ \in ]a,b[$)

so here, $f(b) - f(a) = \frac{(b-a)^2}2 f''(d)$ and if $M=\max_{d \in[a,b]} |f''(d)|$, $\exists c$ with $|f''(c)| = M$.

so, $|f(b)-f(a)| = \frac{(b-a)^2}2 |f''(d)| \le \frac{(b-a)^2}2 |f''(c)|$

but we have only 2, not 4...

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  • $\begingroup$ Oo, I took a peak at your first line, mentioning Taylor / Lagrange remainder. I will proceed from there - very cool hint! BTW, what do those brackets mean? Do you mean $d \in [a,b]$, as in, $d$ in the closed interval? Thanks @claudeh5 $\endgroup$ – User001 Nov 25 '15 at 5:25
  • $\begingroup$ Hi @claudeh5, very nice work. btw, the second derivative isn't assumed to be continuous, so there may not be a max attained by $f''$ on the interval ... $\endgroup$ – User001 Nov 25 '15 at 5:39
  • $\begingroup$ hmm...how to get the 4 ... so, so close ... $\endgroup$ – User001 Nov 25 '15 at 5:40
  • $\begingroup$ Hi @Claude5, we are so close.... :-) I am revisiting your Taylor-Lagrange remainder strategy now, coupled with Joey Zou's write-up (see above) to see if we can put together something that works :-) $\endgroup$ – User001 Nov 26 '15 at 0:49
  • $\begingroup$ Hi @claudeh5, please see up my updated question, where I used Taylor expansion, like you did, but twice -- and subtracting the equations. Now I have the coefficient "4" appear! :-) Not quite done yet, though....almost there.... $\endgroup$ – User001 Nov 26 '15 at 1:27

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