Can This Limit Prove the Maximum Value of a Function on an Interval?

[JG89]

Homework Statement

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

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

Homework Equations

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?

 

[mighty2000]

Your proof looks good overall, but there are a few small things that could be clarified or improved upon.

First, it would be helpful to specify that the Riemann sums are being evaluated at the partition points x_1, x_2, ..., x_n, and that \alpha is a point in the interval [a,b] such that f(\alpha) = M. This will make it clear that you are using the fact that M is the maximum value of f on [a,b].

Second, when you rewrite the limit 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)}, it would be helpful to explain why this is allowed. This step relies on the fact that the Riemann sums converge to the integral, but it would be good to explicitly mention this.

Third, when you let all f values raised to the power of n, except for f(\alpha) = M, be equal to \xi^n, it would be helpful to explain why this is allowed. This step relies on the fact that \xi^n is a smaller term compared to M^n, and so it will not affect the limit.

Finally, it would be good to mention that the inequality \sqrt[n]{M^n} = M < \sqrt[n]{M^n + \xi^n} \le \sqrt[n]{nM^n} = M\sqrt[n]{n} follows from the fact that \xi^n \le M^n for all n, and the fact that the nth root function is increasing. This will make your argument clearer.

Overall, your proof is well-structured and clear. Just make sure to explain all of your steps and assumptions in more detail. Great job!