Proof that the "maximum norm" on the vector space of continuous functions is the limit of the $L^p$ norm Let $C([0,1])$ denote the real vector space of the continuous functions from $[0,1]$ to $\mathbb R$. On this vector space, it is common to define the $L^p$ norm for $p\ge1$ by
$$
\lvert\lvert f\rvert\rvert_p =\left(\int_0^1 \lvert f\rvert ^p\right)^{1/p} \, ,
$$
and define the $L^\infty$ norm by
$$
\lvert \lvert f\rvert \rvert_{\infty}=\max_{t\in[0,1]}\lvert f(t)\rvert \, .
$$
Presumably, the latter norm deserves its name because $\lvert\lvert f\rvert\rvert_\infty=\lim_{p\to\infty}\lvert\lvert f\rvert\rvert_p$ for every $f\in C([0,1])$, but I am unable to find a proof of this fact. How does one going about proving this?
 A: $$ 0 \leq m:=\min_{t\in[0,1]}\lvert f(t)\rvert \leq M:=\max_{t\in[0,1]}\lvert f(t)\rvert. $$
Let $\ a\in [0,1]\ $ be a point such that $\ \vert f(a)\vert = M.$
Let $\ 0< \varepsilon < 1.\ $ By the definition of $\ m\ $ and the fact that $\ f\ $ is continuous, there exists $\ c \in [0,1],\ c\neq a,\ $ such that either $\ f: [c,a]\to [M(1-\varepsilon) + m\varepsilon, M]\ $ is increasing, or $\ f: [a,c]\to [ M, M(1-\varepsilon) + m\varepsilon ]\ $ is decreasing. Suppose the former, WLOG.
Then,
$$ (a-c) \left( M(1-\varepsilon) + m\varepsilon \right) ^p \leq \int_c^a \lvert f(x)\rvert ^p dx \leq \int_0^1 \lvert f(x)\rvert ^p dx \leq ((1-0)\cdot M)^p  \quad \forall\ p\in\mathbb{N}, $$
and so:
$$ (a-c)^{1/p} ( M(1-\varepsilon) + m\varepsilon ) \leq \left( \int_0^1 \lvert f(x)\rvert ^p dx \right)^{1/p} \leq M  \quad \forall\ p\in\mathbb{N}, $$
which, as $\ p \to\infty,\ $ gives:
$$ M(1-\varepsilon) + m\varepsilon \leq \lvert \lvert f\rvert \rvert_{\infty} \leq M. $$
Since $\ \varepsilon\ $ was arbitrary, letting $\ \varepsilon \to 0^+\ $ gives the result as a sandwich.
