# Is a bounded integrable function square integrable?

#### aukie

If $$\displaystyle f \in L^1(E)$$ is bounded, and E is of finite measure, i.e $$\displaystyle m(E)<\infty$$, (m here is the lebesgue measure but it can be arbitrary).

Is $$\displaystyle f \in L^2(E)$$?

Is it enough to say:
since $$\displaystyle |f|<M$$ for some real M.
$$\displaystyle \int _E \left|f\right|^2dm<\int _EM^2dm=M^2\int _Edm=M^2m(E)<\infty$$

But what I dont get is that this argument only uses the fact f is bounded, not the fact that f is integrable.

I know that $$\displaystyle L^2(E) \subseteq L^1(E)$$ since E is of finite measure. I wonder if this has something to do with the answer?

Last edited:

#### aukie

Perhaps if i write $$\displaystyle |f|^2=|f||f|<M|f|$$ then

$$\displaystyle \int _E \left|f\right|^2dm<\int _EM|f|dm=M\int _E|f|dm<\infty$$

then it would be true that if f is integrable then it is also square integrable... does this seem plausible?