Easy exercise (hint) Real Analysis

Question

I've been stuck for a while with this problem. I suppose is something very easy, but I cannot figure out yet the correct approach. I'd really appreciated not a complete solution just some hints (integrals can't be use yet).

If $f$ is twice differentiable function with $f(0)=0$ and $f(1)=1$, and $f'(0)=f'(1)=0$, then $|f''(x)|\ge 4$ for some $x\in[0,1]$.

Sketch-proof: We shall show that there is some $x\in[0,1/2]$ such that $f''(x)\ge 4$ or else $f''(x)\le-4$ for some $x\in [1/2,1]$. Suppose to the contrary that for all $x\in[0,1/2]$ we have $f''(x)< 4$ and for all $x\in [1/2,1]$ we have $f''(x)>-4$.

1) Let $y\in (0,1/2]$ then by the MVT we have

\begin{align}f'(y)=f'(y)-f'(0)=f''(\xi)(y-0)<4y\\ f'(y)<4y \end{align}

Since $\xi\in (0,y)\subset [0,1/2]$. Hence for $y\in [0,1/2]$ we thus have $f'(y)\le4y$. Now we apply the MVT again. Let $x\in (0,1/2]$ then we have:

\begin{align}f(x)-f(0)=f'(\xi)(x-0)<4\xi x<4x^2\\ f(x)<4x^2\end{align} Since $\xi\in (0,x)$. Thus $f(1/2)<1$

2) Now for $y\in [1/2,1]$, using the MVT we get

\begin{align} f'(1)-f'(y)=f''(\xi)(1-y)\ge -4(1-y)\\ f'(y)\le 4(1-y)\end{align}

So for $x\in [1/2,1]$

\begin{align} f(1)-f(x)=f'(\xi)(1-x)\le 4(1-\xi)(1-x)\\ 1-4(1-\xi)(1-x)\le f(x)\\ 1+2(\xi-1)\le f(1/2) \end{align}

Since $\xi \in [1/2,1]$

And in this point is where I'm stuck. I think that maybe the exercise should be say $|f''(x)|\ge 2$. In that way at least we get a very beautiful and symmetric way to do it. Anyway thanks for any help :)

Answer 1

Here's an idea that tries to skirt using integrals. Maybe it will be useful. Define $g$ on $[0,1/2]$ by $g(x)=2x^2$. Observe that $g(0)=g'(0)=0$, $g''(x)=4$ for all $x$, and $g(1/2)=1/2$. Look at the function $g-f$ and its derivatives, and after you've made a conclusion about $f$ on $[0,1/2]$ repeat the same type of thing on $[1/2,1]$. Let me know if you have any questions.

If you want to keep trying to find a proof using MVT (I was unsuccessful there), one thing you could try is to note that $f$, $f'$, and $f''$ are all continuous on $[0,1]$, a closed and bounded interval, so they attain maximum and minimum values. I think this is most useful for $f'$, since by comparing the maximum value of $f'$ and the values $f'$ takes on the endpoints you can make a statement about $f''$.

Answer 2

Since f''(x) is continuous on a closed interval, we will assume by way of arguing a contradiction that | f''(x) | <= 4 - e for some positive e.

Now apply the integral form of Taylor's theorem at a=0 to obtain:

f(x) = f(0) + f'(0) x + integral from 0 to x of t f''(t) dt = 0 + 0 + integral from 0 to x of t f''(t) dt <= (4-e) * integral from 0 to x of t dt = (4-e) x^2 / 2

In particular we get that f(1/2) <= (4-e)/8 < 1/2.

Now apply Taylor's theorem once again at a=1 to get f(x) = f(1) + f'(1) (x-1) + integral from 1 to x of (t-1) f''(t) dt. = 1 + 0 + integral from x to 1 of (1-t) f''(t) dt >= 1 - (4-e) (1-x)^2 /2.

In particular f(1/2) >= 1 - (4-e)/8 > 1/2.

This contradiction proves the theorem.