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Let $T>0$ be fixed and let $f$ be a real valued function such that $$\|f\|_{\infty}\le T,$$ where $\|\cdot\|_{\infty}$ denotes the sup-norm.

My question is the following: if $$\int_0^T f(x) dx \le \int_0^T |f(x)|^2 dx,$$

is that true $$\int_0^T f(x) dx \le \int_0^T |f(x)|^2 dx\iff 1\le \int_0^T |f(x)| dx \, ?$$

On the spot I would say that it is not true. Could someone please help me to understand under which assumptions on $f$ and/or $T$ does the previous equivalence hold true?

Thank you in advance!

$\bf{EDIT:}$ Also partial answer will be accepted (in particular, I am mostly interested in the condition "$\implies$"). Thank you.

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  • $\begingroup$ If $f$ is negative then the assumption is always satisfied. Moreover the conclusion implies $T\ge 1.$ $\endgroup$
    – Ryszard Szwarc
    Commented Nov 20, 2022 at 12:20
  • $\begingroup$ @RyszardSzwarc, thank you for the comment but I’m not sure I got it right. Could you please explain it better? And what happens if $f$ is positive? $\endgroup$
    – user603537
    Commented Nov 20, 2022 at 12:24
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    $\begingroup$ For example if $f(x)\equiv -{1\over NT}$ the assumption is satisfied as the left hand side is negative. Moreover $\int\limits_0^T|f(x)|\,dx={1\over N}.$ In case $f\ge 0,$ the condition $\int\limits_0^T|f(x)|\,dx\ge 1$ implies $T^2\ge 1,$ Hence $T\ge 1.$ Therefore the assumption $T\ge 1$ is necessary for the conclusion to hold. $\endgroup$
    – Ryszard Szwarc
    Commented Nov 20, 2022 at 12:30
  • $\begingroup$ @RyszardSzwarc do you mean that the equivalence I wrote is always satisfied if $T\ge 1$, regardless of the sign of $f$? $\endgroup$
    – user603537
    Commented Nov 20, 2022 at 12:40
  • $\begingroup$ No, in my previous comment I mentioned that $T\ge 1$ is necessary. $\endgroup$
    – Ryszard Szwarc
    Commented Nov 20, 2022 at 13:11

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