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Does Hölder's Inequality implies that $\int_{-\infty}^\infty |f(t) \,g(t)|\,\mathrm dt \leq \int_{-\infty}^\infty |f(t)|\,\mathrm dt \sup_t |g(t)|$ ?

I am thinking in the case $p=1$ and $q=\infty$ so $1=\frac{1}{p}+\frac{1}{q}$ holds, but I am not sure if the norms $||\cdot||_\infty$ and $||\cdot||_1$ are well interpreted from what it is said on Wikipedia here and here, and also if the inequality is valid on unbounded domains as $(-\infty;\,\infty)$.

Also please this related question I had left here. Thanks you very much.

Eric Wofsey
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Joako
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1 Answers1

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Yes, it works ... and the proof is quite immediate: just notice that $|f(t)\,g(t)|≤ (\sup_s{|g(s)|})\ |f(t)|$, and now since $\sup_s{|g(s)|}$ is just a constant independent of $t$, you can put it out of the integral ...

LL 3.14
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  • Is not clear for me since they are under a integral and $g(t)$ could be of the form $e^{-iwt}$ so $|e^{-iwt}|=1$, and then a sum of complex modulated values will have magnitude always less than the sum of the unmodulated values magnitudes, ¿is that right? – Joako Oct 20 '21 at 01:24
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    I do not understand your problem but let me add details then ... $C = \sup |g|$ is real. So since $|f(t)g(t)|≤ |f(t)|,|g(t)| ≤ C, |f(t)|$, we deduce $$ ∫ |f(t) g(t)|,\mathrm d t \leq ∫ C |f(t)|,\mathrm d t = C ∫ |f(t)|,\mathrm d t $$ At which step do you have a difficulty here? – LL 3.14 Oct 20 '21 at 05:04