What can be said about a complex valued, continuous function $f$, defined on $[0,1]$, such that: $$ \int_0^1{|f|^2}=\left|\int_0^1{f}\right|^2 $$ I encountered this form as part of an exercise. Obviously, the above holds for any constant $f$, and it seems intuitive that the converse also holds (i.e. that if the above equality is true, then $f$ is constant), but I could not prove it.

Any help will be appreciated.

  • 1
    $\begingroup$ @GPerez No, both absolute values are complex. $\endgroup$ Apr 13, 2015 at 18:13
  • $\begingroup$ What you are trying to prove is false. The $L^2$ norm does not equal the $L^1$ norm in general for constant functions. Notice that $\int C^2 =(b-a)C^2$ whereas $(\int C )^2 = (b-a)^2 C^2$. $\endgroup$ Apr 13, 2015 at 18:50
  • $\begingroup$ Thanks for all of your comments. I corrected the question. $\endgroup$ Apr 13, 2015 at 18:53
  • $\begingroup$ This can also be proven using Parsevals theorem. $\endgroup$ Apr 13, 2015 at 18:59
  • $\begingroup$ Also look for "Jensen's Inequality". $\endgroup$
    – GEdgar
    Apr 13, 2015 at 20:21

1 Answer 1


This is a case of equality in the Cauchy-Schwarz Inequality $$\left|\int_{0}^1 f\overline g\right|^2\leq \int_0^1 |f|^2\int _0^1|g|^2 $$ where $g=1$ is a constant function. And the equality holds if and only if $f$ and $g$ are dependent,i,e $f$ is a scalar multiple of $g$.

Note that when we change the bounds to $a,b$, the equality in question is not true for constants functions $f$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.