Limit of $L^p$ norm

Question

Could someone help me prove that given a finite measure space $(X, \mathcal{M}, \sigma)$ and a measurable function $f:X\to\mathbb{R}$ in $L^\infty$ and some $L^q$, $\displaystyle\lim_{p\to\infty}\|f\|_p=\|f\|_\infty$?

I don't know where to start.

Answer 1

Fix $\delta>0$ and let $S_\delta:=\{x,|f(x)|\geqslant \lVert f\rVert_\infty-\delta\}$ for $\delta<\lVert f\rVert_\infty$. We have $$\lVert f\rVert_p\geqslant \left(\int_{S_\delta}(\lVert f\rVert_\infty-\delta)^pd\mu\right)^{1/p}=(\lVert f\rVert_\infty-\delta)\mu(S_\delta)^{1/p},$$ since $\mu(S_\delta)$ is finite and positive. This gives $$\liminf_{p\to +\infty}\lVert f\rVert_p\geqslant\lVert f\rVert_\infty.$$ As $|f(x)|\leqslant\lVert f\rVert_\infty$ for almost every $x$, we have for $p>q$, $$ \lVert f\rVert_p\leqslant\left(\int_X|f(x)|^{p-q}|f(x)|^qd\mu\right)^{1/p}\leqslant \lVert f\rVert_\infty^{\frac{p-q}p}\lVert f\rVert_q^{q/p},$$ giving the reverse inequality.

Answer 2

Possibly another solution. Please let me know if there is an error.

In the following, we assume that $f \in L^p \cap L^{\infty}$ and thus $f\in L^q$ for all $q \in [p,\infty]$. The assumption of finite measure is not used.

First we show the result for simple functions. Suppose $f$ is simple such that $f \in L^q$ for $q \in [p,\infty]$. Let the standard representation of $f$ be, \begin{align} f = \sum_{1}^{n}a_j\chi_{E_j}. \end{align} Since $f \in L^q$, $\mu(E_j) < \infty$ for all $j \in \{1,\ldots,n\}$. Without loss of generality, assume that $a_j \neq 0$ and $\mu(E_j)>0$ for all $j$. Also, assume that $|a_j| \leq |a_n|$ for all $j$ and denote by $\eta = \sum_{1}^{n}\mu(E_j)$ (note that $0 < \eta < \infty$). Then, \begin{align} \left\|f\right\|_q &= \left(\sum_{1}^{n}|a_j|^q\mu(E_j)\right)^{1/q}\\ &= |a_n|\eta^{1/q}\left(\frac{\mu(E_n)}{\eta} + \sum_{1}^{n-1}\frac{|a_j|^q\mu(E_j)}{|a_n|^q\eta}\right)^{1/q}\\ \lim\inf_{q\rightarrow \infty} \left\|f\right\|_q &\leq \lim\inf_{q\rightarrow \infty}|a_n|(n\eta)^{1/q} = |a_n| = \left\|f\right\|_{\infty} \end{align} Also, since $|a_j|^q\mu(E_j) \leq \left\|f\right\|^q_q \implies |a_j|\mu(E_j)^{1/q} \leq \left\|f\right\|^q$ for all $j$, therefore, $\left\|f\right\|_{\infty} = |a_n| = \lim\sup_{q\rightarrow \infty}|a_n|\mu(E_n)^{1/q} \leq \lim\sup_{q\rightarrow \infty}\left\|f\right\|_q$. This concludes the result for the simple functions.

Now, let $f \in L^p \cap L^{\infty}$ be an arbitrary measurable functions. Then there exists sequence of simple functions $\{\phi_k\}$ such that $|\phi_1\| \leq \ldots \leq |f|$, $\phi_i \rightarrow f$. Let $q > \max\{p, 1\}$. Given $\epsilon > 0$, there exist $M, N \in \mathbb{N}$ such that for all $m \geq M$ and for all $n \geq N$ we have, \begin{align} \left\|f-\phi_m\right\|_q \leq \epsilon/2 \text{ and } \left\|f-\phi_n\right\|_{\infty} \leq \epsilon/2. \end{align} Since $\left\|\cdot \right\|_q$ and $\left\|\cdot \right\|_{\infty}$ are norms, by triangle inequality, we also have, \begin{align} |\left\|f\right\|_q-\left\|\phi_m\right\|_q| \leq \left\|f-\phi_m\right\|_q \text{ and } |\left\|f\right\|_\infty-\left\|\phi_n\right\|_\infty| \leq \left\|f-\phi_n\right\|_\infty. \label{folland7_1} \end{align} Now, take $K = \max\{M,N\}$, then for all $k \geq K$, we have, \begin{align} \left\|\phi_k\right\|_{q} - \epsilon/2 \leq \left\|f\right\|_{q} \leq \left\|\phi_k\right\|_{q} + \epsilon/2. \end{align} Take limit $q \rightarrow \infty$ (and use the above result for simple functions) to obtain, \begin{align} \left\|\phi_k\right\|_{\infty} - \epsilon/2 \leq \lim_{q\rightarrow \infty}\left\|f\right\|_{q} \leq \left\|\phi_k\right\|_{\infty} + \epsilon/2 \end{align} and then, \begin{align} \left\|f\right\|_{\infty} - \epsilon \leq \lim_{q\rightarrow \infty}\left\|f\right\|_{q} \leq \left\|f\right\|_{\infty} + \epsilon. \end{align} Since $\epsilon > 0$ is arbitrary, we conclude the result.