# Supremum of a function as the limit of an integral

I came across a problem and I really don't know where to start from. It states that:

$$\lim_{n\rightarrow\infty} \left(\int_a^b f(x)^n dx\right)^{1/n} = \sup \{ f(x) :a \leq x \leq b \}$$

with $$f : [a,b] \rightarrow \mathbb R$$ being a continuous , nonnegative function.

I tried using the median value theorem, saying that:

$$\int_a^b f(x)^n dx = f(\xi)^n(b-a)$$, for some $$\xi \in [a,b]$$, and then concluded that the limit was $$f(\xi)$$.

However, I couldn?t find any other relationship between it and the other values of $$f$$.

I also tried using the definition of the supremum of a set , but I can't even prove that it is an upper bound.

Let $$M=\sup \{ f(x) :a \leq x \leq b \}$$. Then clearly, $$\left(\int_a^b f(x)^n\,dx\right)^{1/n}\leq M(b-a)^{\frac1n}\to M.$$ Since $$f$$ is continuous, there exists $$x_0\in[a,b]$$ such that $$f(x_0)=M$$ and for any $$0<\epsilon there exists $$(c,d)\subset[a,b]$$ such that $$f(x)>M-\epsilon$$ for $$x\in(c,d)$$. Hence $$\left(\int_a^b f(x)^n\,dx\right)^{1/n}\geq \left(\int_c^d f(x)^n\,dx\right)^{1/n}\geq (M-\epsilon)(d-c)^{\frac1n}\to M-\epsilon.$$ Therefore $$\lim_{n\rightarrow\infty} \left(\int_a^b f(x)^n dx\right)^{1/n} = M=\sup \{ f(x) :a \leq x \leq b \}.$$
• It should be $M(b-a)^{1/n}$ in your first inequality. The proof however, still works. – uniquesolution Aug 23 '19 at 22:41