Show that the seminorm $|\cdot|_p$ satisfies the second norm axiom, a question about $(\cdot)^p$ and $(\cdot)^{1/p}$, when $p = \infty$

Question

The problem:

I have been asked to show that the seminorm $|f|_p$ of a measurable function $f : E\to [-\infty,\infty]$, \begin{equation} |f|_p = \left(\int_E |f|^p \rm d\mu\right)^{1/p} \in[0,\infty] \end{equation} satisfies the second axiom in the definition of the norm: \begin{equation} \lVert cx\rVert = \lvert c \rvert \lVert x\rVert, \end{equation} for all $c\in\mathbb R$ and $x \in X$, where $X$ is a vector space.

Now, for the proof to be complete, I need to consider both cases, $1 \leq p < \infty$ and $p = \infty$. The first case is not really and issue, since for all $p\in [1,\infty)$, \begin{align} |cf|_p &= \left(\int_E |cf|^p \rm d\mu\right)^{1/p}\\ &= \left(\int_E |c|^p |f|^p \rm d\mu\right)^{1/p}\\ &= \left(|c|^p\int_E |f|^p \rm d\mu\right)^{1/p} && (\text{Homogeneity of integration})\\ &= |c|^{p\frac 1 p}\left(\int_E |f|^p \rm d\mu\right)^{1/p}\\ &= |c|\left(\int_E |f|^p \rm d\mu\right)^{1/p}\\ &= |c| |f|_p \end{align}

What I'm struggling with is the idea of an infinite root, $(\cdot)^{1/p}$, especially since the target of $f$ is $[-\infty,\infty]$. I could end up with things like $\infty^{1/\infty}$ or $(-\infty)^{1/\infty}$, which I don't know how to deal with.

How should I approach this issue?

Answer 1

The $L^{\infty}$ norm is actually defined as $$ \lVert f \rVert_{\infty} = \inf \{ \lambda \geq 0 : \lvert f(x) \rvert \leq \lambda\ \text{almost everywhere} \}$$ so you don't need to try and make sense of what an $\infty$ root is. Because it's not of the same form, the $p = \infty$ case is usually treated separately in proof that the $p < \infty$ case.

Answer 2

The seminorm $|\cdot|_p$ as $p\to\infty$ is the supremum norm $|\cdot|_\infty$ defined by,

$$|f|_\infty=\sup\{|f(x)|\colon x\in E\}$$

When $f$ is continuous, we have,

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

To see this, note that,

$$|f|_p:= \left(\int_E |f|^p \rm d\mu\right)^{1/p}=(|f(c)|^p\mu(E))^{1/p}=|f(c)|\left(\mu(E)\right)^{1/p}\to |f(c)|\leq\sup_{c\in E}|f(x)|=:|f|_\infty$$

as $p\to\infty$ for some $c\in E$ by the MVT for integrals.