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I have proved the following theorem in an earlier part of the question:

Let $p,q \geq 1$ be such that $\frac{1}{p} + \frac{1}{q} = 1$. Show that:

$$\|fg\|_1 \leq \|f\|_p \|g\|_q$$. I proved this statement using Young's Inequality. I am now asked to show the following:

If instead, $p,q,s \geq 1$ be such that $\frac{1}{p} + \frac{1}{q} = \frac{1}{s}$, show that if $f \in L^p(\mathbb{R}), g \in L^q(\mathbb{R})$ then $fg \in L^s(\mathbb{R})$ and that $$\|fg\|_s \leq \|f\|_p \|g\|_q$$. I feel as though I should prove the inequality first;

I attempted the prove it from scratch, following similar methods for the first inequality I stated, however I fell quickly into trouble with the $1/s$ term, and I'm really not sure how to go about it. I feel as though it should be doable directly using the first inequality, but again I am not sure how to prove it.

Any help would be appreciated, thank you.

edit: using $\frac{1}{p/s} + \frac{1}{q/s} = 1$ we have from the original holders inequality:

$\|fg\|_1 \leq \|f\|_{p/s} \|g\|_{q/s}$, but $$\|f\|_{p/s} = \left( \int_\mathbb{R} |f|^{p/s} \right)^{s/p}$$, similary for $g$. I was having torubles manipulating these to get the required inequality.

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  • $\begingroup$ What can you say about $1/(p/s)+1/(q/s)$ $\endgroup$
    – sinbadh
    Feb 3, 2016 at 22:50
  • $\begingroup$ It is equal to 1, this was my first attempt. I will edit my original post. $\endgroup$
    – math
    Feb 3, 2016 at 22:52
  • $\begingroup$ Look at $\int \lvert f\rvert^s \lvert g\rvert^s$. $\endgroup$ Feb 3, 2016 at 22:59

2 Answers 2

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If $f \in L^p$, then $|f|^s \in L^{p/s}$. Using $\frac{1}{p/s} + \frac{1}{q/s} = 1$ and Hölder's inequality yields the result:

$$\|fg\|_s = \||f|^s|g|^s\|_1^{1/s} \le \||f|^s\|_{p/s}^{1/s} \cdot \||g|^s\|_{q/s}^{1/s} = \|f\|_p \|g\|_q$$

You can easily verify the two equalities I've used by plugging in the definition of the occuring $L^r$-norms.

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  • $\begingroup$ Hey, thank you for the answer - I just am stuck on the last part of the quesiton, i.e. showing that $fg \in L^s(\mathbb{R})$. Seeing as we have the inequality just proven, and that $\|f\|_p, \|g\|_q < \infty $ is it justified to say that $\|fg\|)s < \infty$? $\endgroup$
    – math
    Feb 3, 2016 at 23:13
  • $\begingroup$ This just follows from the standard Hölder inequality, as $|fg|^s$ is in $L^1$. $\endgroup$
    – Dominik
    Feb 3, 2016 at 23:20
  • $\begingroup$ Could elaborate, if $|fg|^s$ is in $L^1$ then why does that mean $fg \in L^s$? $\endgroup$
    – math
    Feb 3, 2016 at 23:24
  • $\begingroup$ Just write down what both definitions mean, then you should see it. $\endgroup$
    – Dominik
    Feb 4, 2016 at 6:32
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almost there. Let $F = f^s$, $G = g^s$, apply your result so $||FG||_{1} <= ||F||_{p/s}||G||_{q/s}$ from here, you should get your result by subsituting F and G with $f^s$ and $g^s$

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  • $\begingroup$ Hey, thank you for the answer - I just am stuck on the last part of the quesiton, i.e. showing that $fg \in L^s(\mathbb{R})$. Seeing as we have the inequality just proven, and that $\|f\|_p, \|g\|_q < \infty $ is it justified to say that $\|fg\|)s < \infty$? $\endgroup$
    – math
    Feb 3, 2016 at 23:13
  • $\begingroup$ @math The inequality proven works for measurable functions. $\endgroup$
    – Eric Thoma
    Feb 3, 2016 at 23:35

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