Supremum of a function as the limit of an integral

Question

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.

Any help would be helpful

If you know some measure theory take a look at this https://math.stackexchange.com/questions/242779/limit-of-lp-norm .You do not need continuity here, but there might be a simpler proof if you impose it. — Dayton, Aug 23 '19 at 21:55

Feng · Accepted Answer · 2019-08-23T22:51:08.163

4

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 $0<\epsilon<M$ 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 $$\lim_{n\rightarrow\infty} \left(\int_a^b f(x)^n dx\right)^{1/n} = M=\sup \{ f(x) :a \leq x \leq b \}.$$

edited Aug 23 '19 at 22:51

answered Aug 23 '19 at 22:37

Feng

13,920

It should be $M(b-a)^{1/n}$ in your first inequality. The proof however, still works. – uniquesolution Aug 23 '19 at 22:41
@uniquesolution Thanks! Fixed now. – Feng Aug 23 '19 at 22:51

Supremum of a function as the limit of an integral

1 Answers1