If f is bounded ($\vert f(x)\lvert \leq M$), and $f\in L^1(R)$, then

Question

I’m doing this problem:

If f is bounded ($\vert f(x)\lvert \leq M$), and $f\in L^1(R)$, then: $f\in L^2(R)$ with $\lvert\lvert f\lvert\lvert_{L^2(R)}\leq M^{1/2}\lvert\lvert f\lvert\lvert_{L^1(R)}^{1/2}$.

I’ve tried to prove the equivalent: $\int \lvert f\lvert^2\leq M\int\lvert f\lvert$,

And if we let $\lvert g \lvert =\lvert f \lvert^2\in L^1$, We have $\lvert \lvert g \lvert \lvert_1\leq M^{1/2}[\int\lvert g\lvert^2]^{1/2}$.

And I don’t know how to continue. Any ideas?

Thank you!

Answer 1

I think you started correctly and overcomplicated the situation. Since

$$f(x)^2 \le M |f(x)|$$

you get

$$\int f(x)^2 \le \int M|f(x)| = M \int |f(x)|.$$

In norm notation,

$$\|f\|_2^2 \le M \|f\|_1.$$