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Prove that for all $f\in C^2([0,1])$ with $f(0)=f(1)=0$ and $|f''(x)| \le 1$ $$|f(x)| \le \frac{1}{2}x(1-x)$$ $\forall x \in [0,1]$.

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If your function is only continuous, how do you know for sure that it has a second derivative? – Mercy King Mar 30 '14 at 1:01
Mistake in the post, sorry! Just corrected… – Brontolo Mar 30 '14 at 1:05
You have only accepted an answer in 2/20 questions asked. You should remember to accept the answer you though was most useful to you. – Winther Jul 14 '15 at 23:59
up vote 3 down vote accepted

For some $\xi_k\in[0,1]$, $$ f(1)=f(0)+f'(0)+\frac12f''(\xi_1) \quad\text{and}\quad f(0)=f(1)-f'(1)+\frac12f''(\xi_2) $$ implies $$ |f'(0)|\le\frac12 \quad\text{and}\quad |f'(1)|\le\frac12 $$ Let $g(x)=f(x)-\frac12x(1-x)$ and $h(x)=f(x)+\frac12x(1-x)$.

$g(0)=0$, $g'(0)\le0$, and $g''(x)\le0$; therefore, $g(x)\le0$.

$h(0)=0$, $h'(0)\ge0$, and $h''(x)\ge0$; therefore, $h(x)\ge0$.

Thus, $$ \overbrace{-\frac12x(1-x)\le}^{h(x)\ge0}f(x)\overbrace{\le\frac12x(1-x)}^{g(x)\le0} $$

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Can you explain why $g(0)=0, g′(0)≤0$, and $g″(x)≤0$; therefore, $g(x)≤0$? – Brontolo Mar 30 '14 at 10:09
@TheMaker94: Mean Value Theorem on $[0,1]$. Since $g'(0)\le0$ and $g''(x)\le0$, we have $g'(x)\le0$. Since $g(0)=0$ and $g'(x)\le0$, $g(x)\le0$. – robjohn Mar 30 '14 at 12:14
Perfect, clear thanks! – Brontolo Mar 30 '14 at 12:28

We have, $\displaystyle (1-x)\int_0^x tf''(t)\,dt -x \int_x^1 (t-1)f''(t)\,dt = -f(x)$, for $x \in [0,1]$ (just apply integration by parts on LHS to ease the simplification).

Since, $t\ge 0$ for $t\in [0,1]$ and $(t-1) \le 0$ for $t \in [0,1]$, taking modulus on both sides,

$|f(x)| = \displaystyle \left|(1-x)\int_0^x tf''(t)\,dt -x \int_x^1 (t-1)f''(t)\,dt\right|$

$\le \displaystyle \left|(1-x)\int_0^x tf''(t)\,dt\right| + \left|-x \int_x^1 (t-1)f''(t)\,dt\right|$

$\le \displaystyle \sup\limits_{t\in[0,1]}|f''(t)|. \bigg( (1-x)\int_0^x t\,dt -x \int_x^1 (t-1)\,dt \bigg)$

$= \dfrac{x(1-x)}{2}.\sup\limits_{t\in[0,1]}|f''(t)| \le \dfrac{x(1-x)}{2}$.

Aliter: Define $g(t)=f(t)-\dfrac{t(t-1)}{x(x-1)}f(x)$, on $[0,1]$.

Then, $g(0)=g(x)=g(1)=0$.

Applying Rolle's Theorem twice on $(0,1)$, $\exists \alpha \in (0,1)$ such that $g''(\alpha)=0$.

That is $g''(\alpha) = f''(\alpha) - \dfrac{2}{x(x-1)}f(x)=0$

or, $|f(x)|=\dfrac{x(1-x)}{2}|f''(\alpha)| \le \dfrac{x(1-x)}{2}$.

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How do you pull $\sup|f''(t)|$ out of the integrals while maintaining the subtraction between the integrals? Suppose $f''(t)=1$ on $[0,x]$ and $f''(t)=-1$ on $[x,1]$ (or a continuous function very close to that). – robjohn Mar 30 '14 at 2:00
I think if you add the absolute values of the integrals you get what you want. That is, $$\frac12x^2(1-x)+\frac12x(1-x)^2=\frac12x(1-x)$$ – robjohn Mar 30 '14 at 2:03
@robjohn fixed typo and added a line .. thanks for pointing it out :) – r9m Mar 30 '14 at 2:08
Nice way the second one! I thought that it would be good to do with Taylor Theorem (therefore the title) but I found no way. – Brontolo Mar 30 '14 at 10:17

We want to prove that for each $x$:$$ \exists c\ \ f(x) =\frac 12 f''(c) x(1-x) $$

We want to find such a $c$ via the Rolle theorem (or via the mean value theorem, but we can always go back to the Rolle version).

We can already apply the Rolle theorem, which gives an annulation for $f'$.

Let us modify $f$ to go to $0$ once more: assuming $0<x<1$,

$$ g(u) := f(u) + A_xu(1-u)\\ g(x) = 0\Leftarrow A_x = - \frac{f(x)}{x(1-x)} $$

Now $g(1)=g(x) = g(0)$ hence, applying several times the Rolle theorem:

$$ \exists c \ \ 0=g''(c) = f''(c) - \frac{f(x)}{x(1-x)}(-2)\\ f(x) = -\frac 12 x(1-x)f''(c) $$

NB: the error in the sign does not change the final inequality.

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clever idea (+1) – robjohn Mar 30 '14 at 8:09

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