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Let $p,q$ reals such that $\frac{1}{p}+\frac{1}{q}=1$ and $1\leq p <\infty$. If $g\in L^q(X)$, we define $$ \mathcal{L}_g:L^p(X)\to \mathbb{R}, \quad \mathcal{L}_g(f)=\int_Xfgd\mu $$

I want to demonstrate that $\mathcal{L}_g$ is bounded by studying the cases $q=1\; (p=\infty)$, $p>1$ and $p=1$, but I'm having trouble with it.

For $q=1$ we can show that $||\mathcal{L}_g||\leq||g||_1$ and $||\mathcal{L}_g||\geq||g||_1$. For $p>1$ we have to use the Hölder inequality.

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  • $\begingroup$ Do you know what Hölder's inequality says Do you know the definition of "bounded" here? $\endgroup$ – David C. Ullrich Oct 21 '17 at 15:47
  • $\begingroup$ I know what Hölder inequality says. I'm not sure about the second question $\endgroup$ – Carlos Oct 21 '17 at 16:57
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    $\begingroup$ Well that's the problem then. HINT: You need to know the definitions! Trying to prove something about bounded operators without knowing what a bounded operator is is just stupid. $\endgroup$ – David C. Ullrich Oct 21 '17 at 18:19
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Note that we must have $p,q \in \langle 1, +\infty \rangle$ since the conjugate exponent of $1$ is $+\infty$.

For $f \in L^p(X)$ we have:

$$|\mathcal{L}_g(f)| = \left|\int_X fg\right| \le \int_X |f||g| \stackrel{\text{Hölder}}{\le} \sqrt[p]{\int_X |f|^p} \sqrt[q]{\int_X |g|^q} = \|f\|_p\|g\|_q$$

So $\mathcal{L}_g$ is bounded and $\|\mathcal{L}_g\| \le \|g\|_q$.

To obtain the reverse inequality consider $f = g^{\frac{q}p}$.

$g^{\frac{q}p}$ is in $L^p(X)$ since:

$$\left\|g^{\frac{q}p}\right\|_p = \sqrt[p]{\int_X |g|^q} = \|g\|^\frac{q}{p}_q < +\infty$$

We have $1 + \frac{p}q = q$ so:

$$\mathcal{L}_g\left(g^{\frac{q}p}\right) = \int_X |g|^{1+\frac{q}p} = \int_X |g|^q = \|g\|^q_q$$

Finally

$$\|\mathcal{L}_q\| \ge \frac{\left|\mathcal{L}_g\left(g^{\frac{q}p}\right)\right|}{\left\|g^{\frac{q}p}\right\|_p} = \frac{\|g\|^q_q}{\|g\|^\frac{q}p_q} = \|g\|_q$$

Thus, $\|\mathcal{L}_g\| = \|g\|_q$.

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