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I came across this Putname problem (2007, B2) the other day:

Suppose that $f: [0,1] \to \mathbb{R}$ has a continuous derivative and that $\int_0^1 f(x)\,dx = 0$. Prove that for every $\alpha \in (0,1)$, $$ \left| \int_0^\alpha f(x)\,dx \right| \leq \frac{1}{8} \max_{0 \leq x \leq 1} |f'(x)|. $$

I became interested in seeing if there was a solution using Fourier series. Here is what I tried:

For $x \in [0,1]$, let $F(x) = \int_0^x f(t) \, dt$. Extend $F$ to an odd function on $[-1,1]$, then extend $F$ to a periodic function on $\mathbb R$ with period $2$.

It follows that $F''$ is continuous, except possibly at $\mathbb Z$. We can represent $F''$ by its Fourier series $$ F''(x) \sim \sum_{n=1}^\infty a_n \sin \pi n x, \qquad \text{where } a_n = \int_{-1}^1 F''(x) \sin \pi n x \, dx $$

It follows that $$ F(x) = -\frac{1}{\pi^2} \sum_{n=1}^\infty \frac{a_n}{n^2} \sin \pi n x $$

Let $M = \max |F''(x)|$. My goal is to show that for all $x \in \mathbb R$, we have $|F(x)| \leq \frac{1}{8} M$, but I have been unable to do this. I've tried two different approaches, both of which gave me $|F(x)| < cM$ for some constant $c > \frac{1}{8}$.

1st approach. For each $n$, we have: $$ |a_n| = \left|\int_{-1}^1 F''(x) \sin \pi n x \, dx\right| \leq M \int_{-1}^1 |\sin \pi n x| \, dx = \frac{4}{\pi}M $$

So $$ |F(x)| = \left|\frac{1}{\pi^2} \sum_{n=1}^\infty \frac{a_n}{n^2} \sin \pi n x \right| \leq \frac{1}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2} \cdot \frac{4}{\pi}M = \frac{2}{3\pi}M $$

$\blacksquare$

I see that part of the problem with this approach is that the bound on the $a_n$ is too weak. We know by the Riemann-Lebesgue lemma that $\lim_{n \to \infty} a_n = 0$, so we should be able to get better bounds.

2nd approach. Let $u_n = \frac{1}{n^2}$ and $v_n = a_n \sin \pi n x$. By the Cauchy-Schwarz inequality, we have $$ \left| \sum_{n=1}^\infty u_n v_n \right|^2 \leq \left( \sum_{n=1}^\infty u_n^2 \right) \left( \sum_{n=1}^\infty v_n^2 \right) $$

We have $\sum u_n v_n = -\pi^2 F(x)$ and $\sum u_n^2 = \sum \frac{1}{n^4} = \frac{\pi^4}{90}$, and using Parseval's identity, we also have $$ \sum_{n=1}^\infty v_n^2 = \sum_{n=1}^\infty a_n^2 \sin^2\pi n x \leq \sum_{n=1}^\infty a_n^2 = \int_{-1}^1 F''(t)^2 \, dt \leq 2 M^2 $$

It follows that $$ |F(x)| \leq \sqrt{\frac{1}{90} \cdot 2M^2} = \frac{1}{3\sqrt{5}} M $$

$\blacksquare$

Is there any way for me to improve the constant factor?

share|improve this question
    
To use Fourier series, the function must have $\int_0^1 |f(x)|^2 \, dx < \infty$. We have a continuous derivative function, it is integrable and therefore square-integrable. –  john mangual Aug 30 '12 at 2:18
    
I am sorry, but I am confused. I am asking if there is a way to bound a function by its second derivative with Fourier series, not when a function has a Fourier series. –  Alan C Sep 1 '12 at 3:56

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