# Lebesgue Integration problem

Can I ask a homework question here?

Assume that $f \in L^q(\mathbb R^d)$ for some $q < \infty$ . show that

$\mathrm{lim}_{p \to \infty}||f||_p = ||f||_{\infty}$

$p$ conjugate of $q$

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"Can I ask a homework question here?" - as long as you add the homework tag. Please do so. –  Ｊ. Ｍ. Aug 13 '12 at 12:07
Please, make the question clear. You wrote "$p$ is the conjugate of $p$" but this doesn't make sense. –  Romeo Aug 13 '12 at 12:18
Thank you for your comment –  Park Aug 13 '12 at 12:45
I think that you should remove "$p$ conjugate of $q$". This is senseless since $p$ is a dummy parameter in $\lim_{p\to \infty}\lVert f\rVert_p$ while $q$ is fixed. –  Giuseppe Negro Aug 13 '12 at 12:52

In order your question to be correct you need to require that for some $q$ we have $f\in L^p(\mathbb{R}^d)$ for all $p\geq q$. To solve this problem use the following approach

1) Take $\{p_n:n\in\mathbb{N}\}$ such that $\lim\limits_{n\to\infty} p_n=+\infty$ and $\lim\limits_{n\to\infty}\Vert f\Vert_{p_n}=\liminf\limits_{p\to\infty}\Vert f\Vert_p:=a$.

2) Using Chebyshev inequality prove that $\lim\limits_{n\to\infty}\mu(\{x\in\mathbb{R}^d:|f(x)|> a+\varepsilon\})=0$ for all $\varepsilon>0$.

3) Conclude that $\Vert f\Vert_\infty\leq a$.

4) Prove that $\Vert f\Vert_p\leq \Vert f\Vert_q^{q/p}\Vert f\Vert_\infty^{1-q/p}$ for all $p> q$ and then conclude $\limsup\limits_{p\to\infty}\Vert f\Vert_p\leq\Vert f\Vert_\infty$.

5) The rest is clear.

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+1 for 5) the rest is clear. :) –  t.b. Aug 13 '12 at 13:11
@t.b. Thanks :) –  Norbert Aug 13 '12 at 13:12
Could you explain the first sentence of your answer? The question seems correct as soon as $f\in L^q$ for one value of $q$. –  Did Aug 13 '12 at 13:23
@did As far as I know $L^p(\mathbb{R}^d)$ is not contained in $L^q(\mathbb{R}^d)$. Even more there are examples when function belongs to $L^p$ for only one $p$. Hence the requirement –  Norbert Aug 13 '12 at 13:29
The usual interpretation that $\|f\|_p=+\infty$ when $f$ is not in $L^p$, with the usual topology on $[0,+\infty]$, saves the day. –  Did Aug 13 '12 at 13:38

Let $\mu:=|f|^qd\lambda$.

• Check that $f\in L^{\infty}(\lambda)$ if and only if $f\in L^{\infty}(\mu)$.
• Show that $\lVert f\rVert_{L^p(\lambda)}=\lVert f\rVert_{L^{p-q}(\mu)}^{\frac{p-q}q}$.
• Use this to show the result when $f\in L^{\infty}$.
• When $f\notin L^{\infty}$, use the sets $B_n:=\{x, |f(x)|\geq n\}$.
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This solves the question under the hypothesis that $\|f\|_q$ is finite for some fixed positive finite $q$.

Assume without loss of generality that $\|f\|_\infty=1$. Then:

• $\|f\|_p^p=\int |f|^p\leqslant\int |f|^q=\|f\|_q^q$, hence $\|f\|_p\leqslant\|f\|_q^{q/p}$, where $\|f\|_q^{q/p}\to1$ when $p\to\infty$.
• In the other direction, by definition of $\|f\|_\infty=1$, for every $u\lt1$ there exists a measurable set $A$ of positive finite Lebesgue measure $m$ such that $|f|\geqslant u$ uniformly on $A$. Thus, $\|f\|_p^p\geqslant\int_A u^p=mu^p$ and $\|f\|_p\geqslant m^{1/p}u$, where $m^{1/p}u\to u$ when $p\to\infty$.

Thus, for every $u\lt1$, $u\leqslant\liminf\limits_{p\to\infty}\|f\|_p\leqslant\limsup\limits_{p\to\infty}\|f\|_p\leqslant1$, hence $\lim\limits_{p\to\infty}\|f\|_p=1$.

Now, show that the without loss of generality step at the beginning of the post does not make all the rest a proof by intimidation.

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