TITLE: Show $|f(1)|<\frac{\sqrt{5}}{2}\|f\|_{L^2([0,1])} $
QUESTION [6 upvotes]: Let $f(x)$ be a twice continuously differentiable real-valued function on the interval [0,1]. If
$$f''(x) +xf(x) = 0 \text{, } f'(0)=0 \text{, and } \int_0^1 f(x)dx = 0$$
show that
$$ |f(1)|\leq \frac{\sqrt{5}}{2}\|f\|_{L^2([0,1])}$$
So far I have the following:
Since $f''(x) = -xf(x) $ we have that
$$f'(x) = \int_0^x f''(t)dt +f'(0) = \int_0^x -tf(t)dt$$
I see that we will need to use Cauchy-Schwarz at sometime, but I haven't gotten the $\sqrt{5}/2$ out of it yet. Any hints would be appreciated!

REPLY [6 votes]: By integration by parts, we have
$$0 = \int_0^1 f(x)dx=1f(1)-0.f(0)-\int_0^1xf'(x)dx=f(1)+ \int_0^1x \left( \int_0^x tf(t)dt\right)dx$$
$$\implies
 f(1) = -\int_{
\left\{
\begin{align}
0 \le x \le 1 \\
0 \le t \le x \\
\end{align}
\right\}
}xtf(t)dtdx
=-\int_{\{0 \le t\le x\le1\}}xtf(t)dxdt
$$
Change the order of integration
$$\implies
 f(1) = -\int_0^1 \left(\int_t^1 xdx \right) tf(t) dt = -\frac{1}{2}\int_0^1 (1-t^2)tf(t) dt \tag{1}
$$
Applying the Cauchy-Schwarz inequality in $\mathcal{L}_2$  to $(1)$
$$|f(1)| \le \frac{1}{2}\|f\|_{L^2([0,1])} \sqrt{\int_0^1(1-t^2)^2t^2dt} = \frac{1}{2}\|f\|_{L^2([0,1])}  \sqrt{\frac{8}{105}}< \frac{\sqrt{5}}{2}\|f\|_{L^2([0,1])}$$
The equality doesn't occur.
Q.E.D