# If $f \colon [-1,1] \to \mathbb R$ satisfies $\vert f'' \vert \le k$ then $\vert f \vert \le \frac{k}{2}$

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 $f(-1)=f(1)=0$.

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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

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