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.

Let $f \colon [-1,1] \to \mathbb R$ be a twice differentiable function s.t. $f(-1)=f(1)=0$ and there exists $k>0$ s.t. $\vert f''(x) \vert \le k$ for every $x \in [-1,1]$. Show that $$ \max_{[-1,1]}\vert f \vert \le \frac{k}{2}. $$

The book suggests: take $x_0\in (-1,1)$ such that $\vert f(x_0)\vert$ is maximum and use Taylor.

Following this hint, I write: $$ f(x_0+h)=f(x_0) + \frac{f''(\xi)}{2}h^2 $$ The linear term, $f'(x_0)=0$ (since the point is in the interior of $[-1,1]$ and is a maximum or a minimum for $f$).

Now how can I conclude? I can't see how to use the hypotesis $f(-1)=f(1)=0$.

share|improve this question
    
You are missing an integral term in your Taylor expansion. –  vanna Sep 14 '12 at 17:07
    
@vanna Am I? I used Taylor expansion with Lagrange remainder. Did I have to add an integral remainder? My book doesn't mention it... –  Romeo Sep 14 '12 at 17:09
    
Sorry about that I took $\xi$ for $x_0$, your expression is correct. –  vanna Sep 14 '12 at 17:13
add comment

2 Answers 2

up vote 4 down vote accepted

Take $x_0$ such that $|f(x_0)$ is maximized, and note that $f'(x_0)=0$. Using Taylor, we have $$f(x_0+h)=f(x_0)+f'(x_0)h+R(h)\text{ where } |R(h)|\leq \left|\frac{\max_{x\in [-1,1]}|f''(x)|}{2}h^2\right|$$ and since $f'(x_0)=0$ this simplifies to $f(x_0+h)=f(x_0)+R(h)$. If we take $h$ such that $|h|\leq 1$ and $x_0+h=\pm 1$ (which is always possible), we get $0=f(x_0)+R(h)$ so $$|f(x_0)|=|R(h)|\leq\left|\frac{\max_{x\in [-1,1]}|f''(x)|}{2}h^2\right|\leq \frac{k}{2}|h|^2\leq \frac{k}{2}$$ which completes the proof.

share|improve this answer
    
Feel so stupid... it was so easy! Thank you very much. –  Romeo Sep 14 '12 at 17:22
1  
No problem. +1 to your question for doing almost all the work yourself, BTW. –  Alex Becker Sep 14 '12 at 17:24
add comment

Try choosing $h$ so that $x_0+h$ is either $-1$ or $1$ (whichever results in the smaller $h$).

share|improve this answer
add comment

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.