Verification of alternative proof of $\lim_{p\to \infty}\|u\|_p=\|u\|_\infty$

Question

I have to show that $$\lim_{p\to \infty}\|u\|_p=\|u\|_\infty$$ Suppose $u\in L^\infty (E) $ for measurable $E \subset \mathbb{R}^d$ having finite measure.

I come up with this proof: $$\Big| \|u\|_p - \|u\|_\infty \Big| \leq \Big| \|u\|_\infty \ m(E)^{1/p} - \|u\|_\infty \Big| = \|u\|_\infty\ \Big| m(E)^{1/p} - 1 \Big| \to 0 $$ as $p \to \infty.$

Is this correct?

Answer 1

It is true that $\| u \|_p \leq \| u \|_\infty m(E)^{1/p}$ but that fact can't just propagate outside the absolute value. (Note that $1 \leq 2$ but $|1-3| \geq |2-3|$.)

To actually do this, start by proving it for simple functions. In this case the proof is quite straightforward, and I will omit it.

To finish the problem, it suffices to show that for $\varepsilon>0$ you can find a simple function $s_\varepsilon$ and $P \geq 1$ such that for all $p \geq P$:

1. $|\| u \|_p - \| s_\varepsilon \|_p|<\varepsilon/2$
2. $| \| s_\varepsilon \|_p - \| s_\varepsilon \|_\infty|<\varepsilon/2$.