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On $C([0,1])$ define $||f||_p=(\int_0^1|f(t)|^pdt)^\frac{1}{p}$ for $1\leq p<\infty$ and $||f||_\infty=\sup_{t\in[0,1]}|f(t)|$. Show that $||f||_p\leq ||f||_\infty$

My approach is the following

\begin{align*} ||f||_p^p&=\int_0^1|f(t)|^pdt\\ &\leq|1-0|\cdot\sup_{t\in[0,1]}|f(t)|^p\\ &=(\sup_{t\in[0,1]}|f(t)])^p \\ &=||f||_\infty^p \end{align*}

However I was told, this is not good enough - that I need to consider some "special case", but I don't see what I need?

KJA
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    This looks OK. Your best option is to ask the person who made this commentary in the first place. The best thing would have been to ask immediately; don't be afraid, if you don't understand something, don't pretend you understood, ask. (I also pretend that I understood sometimes. It's never good) – Giuseppe Negro Dec 11 '18 at 10:56

1 Answers1

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It is almost correct (except that $\int_0^q$ should be $\int_0^1$). The only problem is that you are acting as if $\lVert f\rVert_\infty=\sup\lvert f\rvert$. Instead, $\lVert f\rVert_\infty$ is the essential supremum of $\lvert f\rvert$. Take that into account.

José Carlos Santos
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