Let $f\colon[0,1]\to\mathbb R$ be continuous. Let $c_p=\left(\int_0^1|f(x)|^pdx\right)^{\frac{1}{p}}$. Then the limit $\lim_{p\to \infty}c_p$ is?
I know $\inf f(x) \leq c_p\leq \sup f(x)$ for $x\in [0,1]$. But no idea whether the sequence is increasing or decreasing and how to proceed to find the limit.
If $f$ is identically $0$, it's clear. Otherwise, consider $g$ defined by $g= f/\lVert f\rVert_\infty$ (as the supremum of $f$ on $[0,1]$, $\lVert f\rVert_\infty$ exists, is positive, and attained at some $x^\ast \in[0,1]$). We have that $g$ is continuous, and $|g| \leq 1$.
Then $$ \frac{c_p}{\lVert f\rVert_\infty} = \left(\int_{[0,1]} |g(x)|^p dx \right)^{1/p} \leq \left(\int_{[0,1]} 1 dx \right)^{1/p} = 1 \tag{1} $$ Further, for every $\varepsilon>0$, there exists some neighborhood $V_\varepsilon$ of $x^\ast$ of some size $\delta>0$ such that $g(x^\ast)\geq 1-\varepsilon$. Therefore, for all $p$, $$ \frac{c_p}{\lVert f\rVert_\infty} \geq \left(\int_{V_\varepsilon} |g(x)|^p dx \right)^{1/p} \geq (1-\varepsilon)|V_\varepsilon|^{1/p}= (1-\varepsilon)\delta^{1/p} \xrightarrow[p\to\infty]{} 1-\varepsilon \tag{2} $$ Since this holds for all $\varepsilon>0$, we get $$ \lim_{p\to\infty}\frac{c_p}{\lVert f\rVert_\infty} = 1 \tag{3} $$