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

Question

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

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

score 1 · Answer 1 · 2016-06-05T21:27:18.357

1

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.

edited Jun 05 '16 at 21:27

answered Jun 05 '16 at 21:13

so that means there's no difference in the proof ? – user1161 Jun 05 '16 at 21:19
@Jack You meant $\int_0^1 f^2$, right? – Clement C. Jun 05 '16 at 21:21
@ClementC.: Ah, yes. Edited it. Thanks! – Jun 05 '16 at 21:27

1 Answers1