# The Maximum of a Nonnegative Function as an Integral

Homework Statement
Let f: [a,b] -> R be continuous such that f(x) ≥ 0 for all x in [a,b]. Show that

$$\lim_{n \to \infty} \left( \int_a^b (f(x))^n \, dx \right)^{1/n} = \max\{f(x) : x \in [a,b]\}.$$

Relevant equations
The fundamental theorem of calculus and its corollaries.

The attempt at a solution
It is easy to show that the limit is less than or equal to the max of f. This doesn't rely on using the fact that f ≥ 0, so somehow this extra piece of info. turns 'less than or equal' into 'equal', but I have failed to determine why. Any tips?

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Dick
Homework Helper
Suppose M is the max. Pick any m such that 0<m<M. Define g_m(x) to be 0 if f(x)<m and m if f(x)>=m. The length of the set where g_m(x)=m is greater than zero, right? Call it d_m. g_m(x)<=f(x). Can you show if you put g_m(x) into your limit, you get m?

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Heuristically, you might think of the integral as a Riemann sum, that is

$$\int_a^b{dxf(x)^n}\approx\sum_{j=1}^N{\max{\{f(x):x\in[x_{j-1},x_{j}\}}^n(x_{j}-x_{j-1})}$$
where $\{x_j\}_{j=0,\dots,N}$ is a partition of [a,b] with $x_0=1,x_N=b$. Assume f assume its maximal value m in the interval $[x_{\tilde j-1},x_{\tilde j })$. Then you can factor out this m to obtain
$$m^n\left[x_{\tilde j}-x_{\tilde j -1}+\sum_{j=1,j\neq\tilde j}^N{\frac{\max{\{f(x):x\in[x_{j-1},x_{j}\}}^n}{m}(x_{j}-x_{j-1})}\right]$$
if you take take the nth root of this expression you get
$$m\left[x_{\tilde j}-x_{\tilde j -1}+\sum_{j=1,j\neq\tilde j}^N{\frac{\max{\{f(x):x\in[x_{j-1},x_{j}\}}^n}{m}(x_{j}-x_{j-1})}\right]^{1/n}$$
Can you show that the second factor (for fixed N) converges to 1?
Can you use this to prove the original question? (You will have to be careful with interchanging the two limit processes involved.)

Edit: Ok. Dick's method is a lot easier Suppose M is the max. Pick any m such that 0<m<M.
We have to assume that M > 0 for that to work. We can prove the case M = 0 easily.

Define g_m(x) to be 0 if f(x)<m and m if f(x)>=m. The length of the set where g_m(x)=m is greater than zero, right? Call it d_m. g_m(x)<=f(x). Can you show if you put g_m(x) into your limit, you get m?
What do you mean by "length of the set"? Do you mean the diameter of {x in [a,b] : g_m(x) = m}? Also if I put g_m(x) into my limit and get m, what then? Are you somehow going to convert m into M?

From
$$\left[\int_a^b{dx g_m(x)^n}\right]^{1/n}=m$$
and
$$f>g_m$$
for all 0<m<M it follows that
$$\left[\int_a^b{dx f(x)^n}\right]^{1/n}\geq m$$
Letting $m\to M$...

Letting $m\to M$...
I was just thinking that. Makes sense. Thanks.

I'm trying to prove that g_m(x) is integrable. I wanted to do this by showing that g has finitely many points of discontinuity but I'm not sure that is this is the case. Is there an easier way of proving that g_m(x) is integrable?

Hmm...I think the best way to show that g_m(x) is integrable is by using the definition together with the fact that f is integrable.

Are you talking about Riemann or Lebesgue integrability? If the latter is the case, it's enough to observe that
$$g_m(x)=m\chi_{\{f\geq m\}}$$
is essentially the indicator function of the set $d_m=\{x\in[a,b]:f(x)\geq m\}$. It is therefore integrable if $d_m$ is a Borel set, which is obvious from the continuity f. If you mean Riemann integrable the reasoning is essentially the same. Show that $d_m$ is a union of intervals, and so again $g_m$ is a simple step function which you hopefully know is integrable.

Dick