Does ∃ a continuous function $f: [0,1] \to [0, \infty)$ where $\int_0^1 f(x) \; dx = 1$ and $\lim_{n→∞}\int_0^1 f(x)^n \; dx = 0$? (Tifr gs 2022)

Does there exist a continuous function $f: [0,1] \to [0, \infty)$ where $\int_0^1 f(x) \; dx = 1$ and $\lim_{n→∞} \int_0^1 f(x)^n \; dx = 0$?

I have made an attempt and hope it will be an answer.
Using generalization of Steffensen inequality we can answer this question with the help of this inequality here it is ɡiven that $f$ is non zero and now consider another function $g(x)=1$, a constant map from the interval $[0,1]$. As $$\left(\int_0^1(f(x)g(x))dx\right)^p \leq\int_0^a(f(x))^p dx$$ holds for all $p\geq 1$, where $a=(\int_0^1g(x) dx)^p$. Now it is easy to check that $a=1$. Hence as here it is given that left side of the inequality is $1$ but right side is $0$, so no such function can exist.

Solution 1:

Suppose on the contrary that such a function $f$ exists.

Note that since $f$ is continuous, so is $f^n$. Now for any $t\in [0,1]$,if $f(t)\gt 0$, then there exists an interval $[p,q]\subset [0,1],p<q$ around $t$ such that $y\in [p,q]\implies f(y)^n\gt \frac 12 f(t)^n$.

$\int_0^1f(x)^n\,dx=\int_0^pf^n+\int_p^qf^n+\int_q^1f^n\ge \int_p^qf(x)^n\,dx\ge\frac 12f(t)^n (q-p)\ge0$, whence by squeeze theorem $f(t)^n\to 0$. Note that this is possible if and only if $f(t)\in (0,1)$.

It follows that for every $x\in [0,1], f(x)<1$.

$\implies \int_0^1f(z)\,dz\lt1 \tag 1$

But $(1)$ contradicts the given hypothesis and hence such a function can't exist.

Solution 2:

This this explanation of my comment.

Look the Holder's inequality. $\int\limits_0^1 |fg|\le\left[\int\limits_0^1 |f|^2\right]^{1/2}\left[\int\limits_0^1 |g|^2\right]^{1/2}$

Now take $g=1$ and $f$ as given. Then L.H.S become $1$, and R.H.S. become just $\left[\int\limits_0^1 |f|^2\right]^{1/2}$. So by squareing we get, $1\le \int\limits_0^1 |f|^2$.

Again, applying the holder's to obtain $\int\limits_0^1 |f|^2\le \left[\int\limits_0^1 |f|^4\right]^{1/2}$. So, $1\le \int\limits_0^1 |f|^4$.

Apply induction to obtain, $1\le\int\limits_0^1 |f|^{2n}$. Now, can $\int\limits_0^1 |f|^n\to0$?

Solution 3:

Four answers. "Always room for one more!"

This answer (comment, riff, elaboration) considers some generalizations and slightly different methods, all rudimentary. The answer for Lebesgue integrable functions is also "elementary" except for the fact that analysis students don't learn this stuff until late in their studies.

Problem I. Suppose that $f:[0,1[\to\mathbb R$ is nonnegative and Lebesgue integrable and that $\limsup_{n\to\infty} \int_0^1 [f(x)]^n\,dx < \infty$. Show that $$ \int_0^1 f(x)\,dx \leq 1 . \tag{1}$$

Problem II. Suppose that $f:[0,1[\to\mathbb R$ is nonnegative and Lebesgue integrable and that $$0=\liminf_{n\to\infty} \int_0^1 [f(x)]^n\,dx \leq \limsup_{n\to\infty} \int_0^1 [f(x)]^n\,dx < \infty.$$ Show that $$ \int_0^1 f(x)\,dx < 1 .\tag{2}$$

Proof for Problem I: Consider the set $E=\{x\in [0,1]: f(x)>1 \}$. This is a measurable set and is the union of the sequence $E_m=\{x\in [0,1]: f(x)> \frac{m+1}{m} \}$ for $m=1,2,3, \dots$. But each $E_m$ has measure zero.

Why? If $m(E_m)>0$ then $\int_0^1 [f(x)]^n\,dx \geq [\frac{m+1}{m}]^nm(E_m) \to \infty $ as $n\to \infty$ which contradicts the assumption that $\limsup_{n\to\infty} \int_0^1 [f(x)]^n\,dx < \infty$.

Consequently $m(E)=0$, $f(x) \leq 1$ almost everywhere in $[0,1]$ and (1) follows.

Proof for Problem I assuming $f$ is continuous: Consider the set $E=\{x\in [0,1]: f(x)>1 \}$. This is an open set and is the union of the sequence of open sets $E_m=\{x\in [0,1]: f(x)> \frac{m+1}{m} \}$ for $m=1,2,3, \dots$. But each $E_m$ contains no interval $(a,b)$. [An open set that contains no interval is empty.]

Why? If $(a,b) \subset (E_m)$ then $\int_0^1 [f(x)]^n\,dx \geq [\frac{m+1}{m}]^n (b-a) \to \infty $ as $n\to \infty$ which contradicts the assumption that $\limsup_{n\to\infty} \int_0^1 [f(x)]^n\,dx < \infty$.

Consequently each open set $E_m$ is empty implying that $E$ is empty. Hence $f(x) \leq 1$ everywhere in $[0,1]$ and (1) follows.

Proof for Problem II: We already know from problem I that $f(x) \leq 1$ almost everywhere. Consider the set $A=\{x\in [0,1]: f(x)=1 \}$. This is a measurable set and must have measure zero.

Why? If $m(A)>0$ then $\int_0^1 [f(x)]^n\,dx \geq m(A)>0$ which contradicts the assumption that $\liminf_{n\to\infty} \int_0^1 [f(x)]^n\,dx =0$.

Consequently $m(A)=0$, $f(x)<1$ almost everywhere in $[0,1]$ and (2) follows.

Proof for Problem II assuming that $f$ is continuous: We already know from problem I that $f(x) \leq 1$ everywhere. Consider the set $A=\{x\in [0,1]: f(x)=1 \}$. This is a closed set and must be nowhere dense, i.e., it contains no interval.

Why? If $A\supset (a,b)$ then $\int_0^1 [f(x)]^n\,dx \geq b-a>0$ which contradicts the assumption that $\liminf_{n\to\infty} \int_0^1 [f(x)]^n\,dx =0$.

Consequently there is at least one subinterval of $[0,1]$ on which $f$ is strictly less than $1$ and so (2) follows.