# Parseval identity $\int_0^1|f(x)|^2dx=\sum\limits_{n\in\mathbf Z}|\hat f_n|^2$ weaker condition

Parseval identity $\int_0^1|f(x)|^2dx=\sum\limits_{n\in\mathbf Z}|\hat f_n|^2$ holds for square integrable $f$, what if the condition is dropped ?

I have two questions, in both of which I have to prove the Parseval's equality but

• for the first $f$ is supposed to be $C^1$-regular and periodic
• for the other in $L^2$ also periodic.

I don't know what the difference is. Is the first condition weaker and imply the other?

What if the condition is dropped?

If you drop the condition $f\in L^2((0,1))$ completely, then the integral $\int_0^1f(x)^2\ dx$ would not even make sense.

Is the first condition weaker than the other one?

No. It is stronger. Note that continuous functions on a closed interval are always square integrable.

• so that means there's no difference in the proof ? Jun 5, 2016 at 21:19
• @Jack You meant $\int_0^1 f^2$, right? Jun 5, 2016 at 21:21
• @ClementC.: Ah, yes. Edited it. Thanks!
– user9464
Jun 5, 2016 at 21:27