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Let $f:[0,1]\rightarrow\mathbb{R}$ be a function whose second derivative $f''(x)$ is continuous on $[0,1]$. Suppose that f(0)=f(1)=0 and that $|f''(x)|<1$ for any $x\in [0,1]$. Then $$|f'(\frac{1}{2})|\leq\frac{1}{4}.$$

I tried to use mean value theorem to prove it, but I found no place to use condition of second derivative and I cannot prove this.

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  • $\begingroup$ Hint: there is a point $y_0$ such that $f^\prime(y_0)=0$ $\endgroup$ – Thomas Apr 28 '16 at 17:35
  • $\begingroup$ @Thomas Then if we apply mean value theorem, I can only get 1/2. Where should I use the continuity of second derivative? $\endgroup$ – Peter Liu Apr 28 '16 at 17:51
  • $\begingroup$ Sorry for the misleading comment ;-). +1 for the fun I had figuring this out at last. Would you mind telling me from where the problem orginates? (The continuity of the second derivative is not needed as far as I can tell). $\endgroup$ – Thomas Apr 30 '16 at 18:54
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A bit more tricky than I thought at first. The idea is easy, the calculation may look complicated. The idea is to find a second order polynomial with second derivate $=1$ which has the same values as $f$ for $x=0 $ and $x= \frac{1}{2}$, and then to show that this function $-f$ is convex, which allows to get an estimate for the derivative at the boundary.

First note that the assumptions and the claim are invariant with respect to multiplication by $-1$.

Now, since we may look at $-f$ instead of $f$, we may assume $f(\frac{1}{2})\le 0$

We will bound $f^\prime(\frac{1}{2})$ by looking at the comparison function

$$ h(x) = \frac{1}{2}x^2 + \left(2f(\frac{1}{2}) -\frac{1}{4}\right)x $$

You'll easily verify $h(0) = 0, \, h(\frac{1}{2}) =f(\frac{1}{2})$ and $h^{\prime \prime}(x) = 1$. So for $$\psi(x) = h(x)-f(x)$$ we have $\psi(0)= 0 = \psi(\frac{1}{2})$ and $ \psi^{\prime \prime}(x)=1- f^{\prime \prime}(x)>0$. This means $\psi$ is strictly convex with zero boundary values at $x= \frac{1}{2}$, which implies $\psi^\prime(\frac{1}{2})>0$

Since $h^\prime(x) = x+2f(\frac{1}{2})-\frac{1}{4}$ we have $h^\prime(\frac{1}{2}) = 2f(\frac{1}{2})+\frac{1}{4}$ and so

$$ \begin{eqnarray} 0< \psi^\prime(\frac{1}{2}) = 2f(\frac{1}{2})+\frac{1}{4} -f^\prime (\frac{1}{2}) \end{eqnarray} $$ which implies (using the assumption on the sign of $f$) $$ f^\prime (\frac{1}{2})< 2f(\frac{1}{2})+\frac{1}{4} \le \frac{1}{4}$$

To get a bound $f^\prime (\frac{1}{2})\ge- \frac{1}{4}$ with the same assumption $f(\frac{1}{2})\le 0$ the same trick is applied on the interval $[\frac{1}{2},1]$ using the comparison function $$h(x)=\frac{1}{2}(x-1)^2 - \left( 2f(\frac{1}{2})-\frac{1}{4}\right)(x-1) $$ If you now look at $\phi=h-f$ this is again convex with zero boundary values for $x= \frac{1}{2}$ and $x=1$, so $\phi^\prime(\frac{1}{2})< 0$ this time. The calculations are the same.

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  • $\begingroup$ A real good and detailed solution. $\endgroup$ – Peter Liu May 1 '16 at 16:55

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