How to prove that an $L^2$ function is also an $L^1$ function? I have a function $f(t)$ defined on $[-a,a]$ that belongs to $L^2$. How do I prove that $f$ also belongs to $L^1$?
In general this fact is not true.
Is the cauchy-schwarz inequality the only way?
If a it's equal to pi this fact will change something?

As a comment said, we use the Cauchy-Schwarz inequality:
$$\|f\|_{L^1([-a,a])}=\|f\cdot1\|_{L^1([-a,a])}\leq\|f\|_{L^2([-a,a])}\|1\|_{L^2([-a,a])}=\|f\|_{L^2([-a,a])}(2a)^{1/2}<\infty$$
by the given fact that $\|f\|_{L^2([-a,a])}<\infty$.
But perhaps a more fundamental way to prove it is Jensen's inequality with the function $\varphi(x)=x^2$.
