这是一份Sydney悉尼大学MATH2023/MATH2923的成功案例

We have
$$
\Phi_{p}(f)=|\Omega|^{-\frac{1}{p}}|f|_{L^{p}(\Omega)} .
$$
By corollary $\Phi_{p}(f)$ viewed as a function of $p$ is increasing with
$$
\Phi_{p}(f) \leq|f|_{L^{\infty}(\Omega)}
$$ that
$$
|f|_{L^{\infty}(\Omega)} \leq \lim {p \rightarrow \infty} \Phi{p}(f)
$$
For $K \in \mathbf{R}$ let
$$
A_{K}:={x \in \Omega|| f(x) \mid \geq K} .
$$
The set $A_{k}$ is measureable since $f$ is and $\left|A_{K}\right|>0$ if $K<|f|_{L^{\infty}(\Omega)}$. Moreover,
$$
\Phi_{p}(f) \geq|\Omega|^{-\frac{1}{p}}\left(\int_{A_{K}}|f(x)|^{p} d x\right)^{1 / p} \geq|\Omega|^{-\frac{1}{p}}\left|A_{K}\right|^{\frac{1}{p}} K .
$$
Passing to the limit $p \rightarrow \infty$ we obtain
$$
\lim {p \rightarrow \infty} \Phi{p}(f) \geq K
$$
Because this holds for all $K<|f|_{L^{\infty}(\Omega)}$ we conclude
$$
\lim {p \rightarrow \infty} \Phi{p}(f) \geq|f|_{L^{\infty}(\Omega)}
$$

MATH2023/MATH2923 COURSE NOTES ：

Let $\left(f_{k}\right){k \in \mathrm{N}} \subset L^{P}(\Omega)$ be a Cauchy sequence. It suffices to show that $\left(f{k}\right)$ has a convergent subsequence. For every $i \in \mathrm{N}$ there is an integer $N_{i}$ so that
$$
\left|f_{n}-f_{m}\right|_{L P(\Omega)} \leq 2^{-i} \text { whenever } n, m \geq N_{i} .
$$
We construct a subsequence $\left(f_{k_{i}}\right) \subset\left(f_{k}\right)$ so that
$$
\left|f_{k_{i+1}}-f_{k_{i}}\right|_{L P(\Omega)} \leq 2^{-i}
$$
by setting $k_{i}:=\max \left{i, N_{i}\right}$. In order to simplify notation we will from now on assume that
$$
\left|f_{k+1}-f_{k}\right|_{L F(\Omega)} \leq 2^{-k}
$$
so that
$$
M:=\sum_{k \in \mathbf{N}}\left|f_{k+1}-f_{k}\right|_{L^{p}(\Omega)}<\infty
$$