Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
If your function is only continuous, how do you know for sure that it has a second derivative? –  Mercy Mar 30 at 1:01
    
Mistake in the post, sorry! Just corrected… –  Brontolo Mar 30 at 1:05

3 Answers 3

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

share|improve this answer
    
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 at 2:00
1  
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 at 2:03
    
@robjohn fixed typo and added a line .. thanks for pointing it out :) –  r9m Mar 30 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 at 10:17

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

share|improve this answer
    
Can you explain why $g(0)=0, g′(0)≤0$, and $g″(x)≤0$; therefore, $g(x)≤0$? –  Brontolo Mar 30 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 at 12:14
    
Perfect, clear thanks! –  Brontolo Mar 30 at 12:28

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.

share|improve this answer
    
clever idea (+1) –  robjohn Mar 30 at 8:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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