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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.

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    $\begingroup$ @GPerez No, both absolute values are complex. $\endgroup$
    – Thomas Andrews
    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 $(\int C )^2 = (b-a)^2 C^2$. $\endgroup$
    – Eric Naslund
    Apr 13, 2015 at 18:50
  • $\begingroup$ Thanks for all of your comments. I corrected the question. $\endgroup$
    – SomeStrangeUser
    Apr 13, 2015 at 18:53
  • $\begingroup$ This can also be proven using Parsevals theorem. $\endgroup$
    – Eric Naslund
    Apr 13, 2015 at 18:59
  • $\begingroup$ Also look for "Jensen's Inequality". $\endgroup$
    – GEdgar
    Apr 13, 2015 at 20:21

1 Answer 1

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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 $f$

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