Problem: Given a continuous function $f(x)\geq 0$ defined on $[a,b]$, let $$ M_n = \left[\int_a^b (f(x))^ndx\right]^{\frac{1}{n}}. $$ Show that $M_n \rightarrow \sup_{x \in [a,b]} f(x)$ as $n \rightarrow \infty$.
My Attempt: I had a hard time trying to find a way to show the equality so I just tried to use squeeze theorem, let me know if this approach works/ if there's a much better way to approach the problem. Thanks in advance!
Because $f(x)$ is continuous on a compact set it is bounded by the min-max theorem; and in particular there are values on the interval for which it attains it's supremum and infimum. So if we let $m = \inf_{x \in [a,b]} f(x)$ and $M = \sup_{x \in [a,b]} f(x)$ then $m \leq f(x) \leq M$. It follows that we can bound $M_n$ as $$ M_n \leq \left[\int_a^bM^ndx\right]^{\frac{1}{n}}. = \left[M^n \int_a^b dx \right]^{\frac{1}{n}} = \left[M^n(b-a)\right]^{\frac{1}{n}} = M(b-a)^\frac{1}{n}. $$ Note also that because $f(x) \geq 0$ we can equivalently bound $f$ as $-M \leq f \leq M$ and moreover $-M^n \leq f^n \leq M^n$. Note that we know that the product of Riemann integrable functions is also riemann integrable; so $f^n \in \mathscr{R}$, and so we can conclude that for any partition $P$ of $[a,b]$ that $$ -M^n(b-a) \leq L(P,f^n) \leq \overline{\int_{a}^{b}} f^ndx = \int_a^b f^ndx. $$ (Here the overlined integral denotes the upper Riemann integral) Note above the inequalities are because $f^n$ is bounded, the rectangle is less than any lower sum which is itself bounded by an upper Riemann integral, and because $f^n \in \mathscr{R}$ the upper integral is the integral. From this it follows that \begin{align*} (M^n(a-b))^\frac{1}{n} &\leq \left[\int_a^b f^n dx\right]^\frac{1}{n} \\ M (a-b)^\frac{1}{n} &\leq M_n. \end{align*} Now note that if we combine the two inequalities we get that $$ M (a-b)^\frac{1}{n} \leq M_n \leq M(b-a)^\frac{1}{n}. $$ and taking $n\rightarrow \infty$ across the inequalities gives that $$ M\leq M_n \leq M, $$ and so we conclude that $\lim_{n \rightarrow \infty} M_n = M = \sup_{x \in [a,b]} f(x)$.