Prove that $c_{m} \in[a, b],$ for all $m \geq 1, \lim _{m \rightarrow \infty} c_{m}$ exists and find its value.

Question

Let $f:[0,1] \rightarrow[0, \infty)$ be a continuous function. Let $$ a = \inf_{0 \leq x \leq 1} f(x) ~\text{ and }~ b = \sup_{0 \leq x \leq 1} f(x) . $$

For every positive integer $m$, define $$ c_{m}=\left[\int_{0}^{1}(f(x))^{m} d x\right]^{1 / m} $$ Prove that $c_{m} \in[a, b],$ for all $m \geq 1, \lim\limits_{m \rightarrow \infty} c_{m}$ exists and find its value.

My work: To show limiting value is definite I have to show the value of integration is a finite one! For doing the I am thinking to apply Leibniz rule.But I am not sure how to go step by step.Any hint or suggestion would be highly appreciated!

Answer 1

The estimates $a \leq c_n \leq b$ follow from $$ a^m = \int_0^1 a^m\, dx \leq \int_0^1 f(x)^m\, dx \leq \int_0^1 b^m\, dx = b^m. $$ Let us prove that $(c_m)$ converges to $b$.

If $b=0$ then also $a=0$, so that the claim follows from the first part.

Assume that $b>0$, let $\varepsilon \in (0, b)$, and let $I\subset [0,1]$ be a set where $f \geq b-\varepsilon$ (since we are assuming $f$ continuous, we can take a suitably small interval containing the maximum point of $f$).

Denoting by $L>0$ the length of the interval $I$, we have that $$ \int_0^1 f(x)^m\, dx \geq \int_I (b-\varepsilon)^m\, dx = (b-\varepsilon)^m L, $$ so that $$ c_m \geq (b-\varepsilon) L^{1/m}. $$ Since $L^{1/m} \to 1$ as $m\to +\infty$, we have that $$ \liminf_m c_m \geq b-\varepsilon. $$ Since $c_m\leq b$ for every $m$, the claim follows by the arbitrariness of $\varepsilon$:

Answer 2

This is meant as complementary to Rigel's answer since I wanted to remove emphasis from the continuous aspect of $f$ and emphasise the nature of the "$p$ norm"

When we take the "$p$ norm" of a function, i.e.:

$$ ||f||_p = \Big(\int_0^1 |f(x)|^p dx\Big)^{\frac{1}{p}} $$

As $p$ gets larger and larger, the part of the integral that contributes most are values the function takes near its max, because higher powers overemphasise these parts. Does it matter at all how the function behaves anywhere else apart from the max?

Suppose we know there is some finite interval $I$ of length $L > 0$ for which the function is equal to the max $b$, the "$p$ norm" on that interval is:

$$ ||f||_{p,I} = (L b^p)^{\frac{1}{p}} \rightarrow b$$

Where $||f||_{p,I}$ is the "$p$ norm" with integral restricted to $I$. We know that

$$ ||f||_{p,I} \leq ||f||_p \leq b $$

And so we must have $||f||_p \rightarrow b$

Rigel's proof then uses continuity to say that essentially we can keep finding an interval where $f$ is arbitrarily close to $b$ rather than equal to $b$ to make this proof work.

Because of this property of "$p$ norm" as $p \rightarrow \infty$, the max/supremum norm can also be called the infinity norm: https://en.wikipedia.org/wiki/Uniform_norm