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

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

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

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


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 ...


  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ "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

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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @LebronJames you bring up good points. I have edited my answer accordingly. $\endgroup$ – Joey Zou Nov 26 '15 at 1:28

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...

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.