Is a bounded integrable function square integrable?

Oct 2009
42
4
London
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:
Oct 2009
42
4
London
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?