Let $f:\mathbb{R}\to\mathbb{R}$ be continuous on [0, 1]; show that:
$$ \lim_{p \to +\infty} \left( \int_{0}^{1} |f|^{p} \right)^{\frac{1}{p}}=max|f| $$
My work:
I was able to show that:
$$ \left( \int_{0}^{1} |f|^{p} \right)^{\frac{1}{p}}\le \left( \int_{0}^{1} max(|f|)^{p} \right)^{\frac{1}{p}}=max|f| $$
with $ max|f|\in \mathbb{R} $ due to continuity and Weierstrass.
We can also see that, considering, for every x in [0, 1], $(f(x), 0)\in \mathbb{R}^{2}$:
$$ |f(x)|^{p}=\left( \left\| (f(x), 0) \right\|_{p} \right)^{p} $$
And there exists positive constants c, C such that:
$$ c \cdot max(f, 0)^{p}\le \left( \left\| (f(x), 0) \right\|_{p} \right)^{p}\le C\cdot max(f, 0)^{p} $$
I was thinking of applying Holder/Minkowski to bound the limit from below but I don't know how to continue.
