Show that $2 \int f^2 \leq \int |f'| \cdot \int |f|$ Let $f(x)$ be a continuously differentiable function defined on closed interval $[0, 1]$ for which$$\int_0^1 f(x)\,dx = 0.$$How do I show that$$2 \int_0^1 f(x)^2\,dx \le \int_0^1 |f'(x)|\,dx \cdot \int_0^1 |f(x)|\,dx?$$
 A: I am assuming these are real-valued functions. Define$$g(x) = \int_0^x f(t)\,dt,$$and note that$$g(0) = g(1) = 0.$$Integrating by parts,$$\int_0^1 f(x)^2\,dx = g(x)f(x)\big|_0^1 - \int_0^1 g(x)f'(x)\,dx = -\int_0^1 g(x)f'(x)\,dx.$$A straightforward estimate then shows that$$\int_0^1 f(x)^2\,dx \le \sup_{0 \le x \le 1} \left|g(x)\right| \int_0^1 \left|f'(x)\right|\,dx.$$Now, what we have to do is to show that$$\sup_{0 \le x \le 1} \left| g(x)\right| \le {1\over2} \int_0^1 \left|f(x)\right|\,dx.$$Since$$g(x) = \int_0^x f(t)\,dt = -\int_x^1 f(t)\,dt,$$we can say that for all $x \in [0, 1]$, we have$$\left|g(x)\right| \le \min\left(\left|\int_0^x f(t)\,dt\right|,\,\left| \int_x^1 f(t)\,dt\right|\right) \le \min\left( \int_0^x \left|f(t)\right|\,dt,\, \int_x^1 \left|f(t)\right|\,dt\right).$$But the two integrals we are taking the minimum add up to$$\int_0^1 \left|f(t)\right|\,dt.$$If we have $a$, $b > 0$ and $a + b = c$, then $\min(a, b) \le c/2$. We conclude that$$\left|g(x)\right| \le {1\over2} \int_0^1 \left|f(t)\right|\,dt,$$and we are done.
A: Here is a slightly different point of view:
First, we note that $f(x)=f(0)+\int_0^x f'(t)\,dt$ and hence (here we also use that $\int_0^1f(x)\,dx=0$)
$$
\int_0^1 (f(x))^2\,dx=\int_0^1 f(x)\biggl[f(0)+\int_0^x f'(t)\,dt\biggr]\,dx=\int_0^1\int_0^x f(x)f'(t)\,dt\,dx.\tag{1}
$$
Changing order of integration in the right-hand side of $(1)$, using the fact that $\int_0^1 f(x)\,dx=0$, and then changing order of integration again, we find that
$$
\begin{aligned}
\int_0^1 (f(x))^2\,dx&=\int_0^1f'(t)\int_t^1 f(x)\,dx\,dt\\
&=-\int_0^1f'(t)\int_0^tf(x)\,dx\,dt\\
&=-\int_0^1\int_x^1f(x)f'(t)\,dt\,dx
\end{aligned}
$$
Thus, by adding these two formulas, and using the triangle inequality,
$$
\begin{aligned}
2\int_0^1(f(x))^2\,d&=\biggl|\int_0^1\int_0^x f(x)f'(t)\,dt\,dx-\int_0^1\int_x^1f(x)f'(t)\,dt\,dx\biggl|\\
&\leq \int_0^1\int_0^x |f(x)|\,|f'(t)|\,dt\,dx+\int_0^1\int_x^1|f(x)|\,|f'(t)|\,dt\,dx\\
&=\int_0^1|f(x)|\,dx\cdot\int_0^1|f'(t)|\,dt,
\end{aligned}
$$
and we are done.
