# Maxima And Integration Proof

1. Feb 9, 2009

### JG89

1. The problem statement, all variables and given/known data

Let f be continuous and positive on [a,b] and let M denote its maximum value. Prove:

M = $$\lim_{n \rightarrow \infty} \sqrt[n]{\int_a^b (f(x))^n dx}$$

2. Relevant equations

3. The attempt at a solution

Writing out the integral as a limit of Riemann sums, we have $$\lim_{n \rightarrow \infty} \sqrt[n]{\int_a^b (f(x))^n dx} = \lim_{n \rightarrow \infty} \sqrt[n]{\frac{b-a}{n}[f^n(x_1) + f^n(x_2) + ... + f^n(\alpha) + ... + f^n(n)]}$$ where $$f(\alpha) = M$$.

This expression can be rewritten as $$\lim_{n \rightarrow \infty} \sqrt[n]{\frac{b-a}{n}} \sqrt[n]{f^n(x_1) + f^n(x_2) + ... + f^n(\alpha) + ... + f^n(n)}$$. Now, letting n tend to infinity, the nth root of (b-a)/n converges to 1 and so we need only evaluate the limit of the infinite series. Let all f values raised to the power of n, except for f(alpha) = M, be equal to $$\xi^n$$. Then we have $$\lim_{n \rightarrow \infty} \sqrt[n]{\int_a^b (f(x))^n dx} =\lim_{n \rightarrow \infty} \sqrt[n]{M^n + \xi^n}$$.

Now, since f is positive everywhere in the interval and remembering there are n terms under the nth root sign, then $$\sqrt[n]{M^n} = M < \sqrt[n]{M^n + \xi^n} \le \sqrt[n]{nM^n} = M\sqrt[n]{n}$$. Since the right hand side of the inequality converges to M (since the nth root of n converges so 1 as n goes to infinity), then the limit of the Riemann sums is also M and so we have $$M = \lim_{n \rightarrow \infty} \sqrt[n]{\int_a^b (f(x))^n dx}$$.

How does this proof look?