# 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.

• We are not here to do your homework question. Show your attempt, then we will discuss about your problem. Don't mind Sir, put your effort. Thanks. Dec 14, 2021 at 13:06
• This is really just Holder's inequality that you've used. Are you just asking if this is a valid argument? Since $f$ is continuous and nonnegative I think you can make a more elementary argument. Dec 14, 2021 at 14:59
• Use holder's inequality and induction on $n$, to prove that $\int\limits_0^1 f^{2n}\ge 1$ Dec 14, 2021 at 15:14
• @B.S.Thomson sir I cannot understand what you said can you explain in detail Dec 14, 2021 at 18:03
• Oh ok got the point Dec 14, 2021 at 18:41

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 $$\int\limits_0^1 |f|^n\to0$$?

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 $$[p,q]\subset [0,1],p around $$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.

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.

This is an elementary proof.

Assume that such function exists and let $$f(x)$$ be such a function.

It's clear that f(x) can't be Always equal to 1 because in that cause you would have :

$$\lim_{n\to+\infty}{\int_0^1{{f(x)}^n dx}}=1$$

You can therefore prove that there exist $$x_0 \in [0,1] : f(x_0) > 1$$

To do it assume that there isn't such $$x_0$$, then $$f(x) \leq 1 \, \forall x \in [0,1]$$ but because of what I said before you can't have $$f(x)$$ always equal to 1 so there exist $$x_1 \in [0,1] : f(x_1) < 1$$

(To make things simpler I assume that $$x_1 \in (0,1)$$ )

Now because of continuity, chosing any $$0 < \epsilon < 1 - f(x_1)$$ there exists $$\delta > 0$$ such that $$\forall x \in (x_1 - \delta,x_1 + \delta) \, f(x) < f(x_1) + \epsilon$$

So because $$f(x) \leq 1$$ you have

$$\int_0^1{f(x)dx} = \int_0^{x_1-\delta}{f(x)dx} + \int_{x_1-\delta}^{x_1+\delta}{f(x)dx} + \int_{x_1 + \delta}^{1}{f(x)dx} \leq (x_1 - \delta) + (f(x_1) + \epsilon)2\delta + (1-x_1-\delta) = 1 - 2(1 - f(x_1) - \epsilon)\delta < 1$$

wich contradict the hypothesis, therefore $$x_0$$ have to exist

If such $$x_0$$ exists then (assuming that \$x_0 \in (0,1) for making things easier) by continuity you have that

$$\exists \delta > 0 : \forall x \in (x_0-\delta,x_0+\delta) \, f(x) > 1$$

(You can let $$\epsilon = \frac{f(x_0)-1}{2})$$ and the use the definiton of continuity to prove It)

Because $$f(x) \geq 0$$ you have that $$\forall n \in \mathbb{N}$$ $${f(x)}^n \geq 0$$

$$\forall x \in (x_0 - \delta, x_0 + \delta) \, {f(x)}^n > 1$$

There you have that

$$\int_0^1{{f(x)}^n dx} = \int_0^{x_0 - \delta}{{f(x)}^n dx} + \int_{x_0 - \delta}^{x_0 +\delta}{{f(x)}^n dx} + \int_{x_0 +\delta}^1{{f(x)}^n dx} \geq 2\delta > 0$$

Therefore because $$\delta$$ is constant, in particolar It doesn't depend on n, you have that

$$\lim_{n\to+\infty}{\int_0^1{{f(x)}^n dx}} \geq 2\delta > 0$$

(The limit exists because I assumed that It was equal to 0 so I also assumed that It exists)

But clearly I've got a contraddiction because I assumed that the limit was equal to 0 and I proved that it's greater than 0. Therefore my assumption was wrong and therefore such function can't exists.

(For both $$i=0$$ and $$i=1$$ I assumed that $$0 \leq x_i - \delta < x_i + \delta \leq 1$$ I can do that assumption because If $$x_i \in (0,1)$$ then the inequality Is true by chosing $$\delta$$ enough small, If $$x_i = 0,1$$ then the proof if pretty similar.)

If $$f(x_0)> 1$$ then there exists an interval $$L$$ of length $$|L|>0$$ such that $$x_0\in L\subset [0,1]$$ and $$\forall x\in L\,(f(x)\ge (1+f(x_0)/2).$$ Then $$\int_0^1f(x)^ndx\ge \int_Lf(x)^ndx\ge \int_L((1+f(x_0)/2)^ndx =|L|((1+f(x_0)/2)^n$$ and the last term above $$\to\infty$$ as $$n\to\infty$$ because $$(1+f(x_0)/2>1$$ and $$|L|>0.$$

If $$f(x_1)<1$$ then there exists an interval $$M$$ of length $$|M|>0$$ such that $$x_1\in M\subset [0,1]$$ and $$\forall x\in M\,(f(x)\le (1+f(x_1)/2).$$ Then $$1=\int_Mf(x)dx+\int_{[0,1]\setminus M}f(x)dx\le$$ $$\le |M|(1+f(x_1)/2+(1-|M|)\max \{f(x):x\in [0,1]\}$$ which implies $$\max \{f(x):x\in [0,1]\}>1$$ because $$(1+f(x_1)/2<1$$ and $$|M|\le 1,$$ implying there exists $$x_0$$ with $$f(x_0)>1$$.

And the case $$\forall x (f(x)=1)$$ is of course trivial.

However I do like your use of Holder's Inequality for a quick one-line proof, which applies to any essentially-bounded non-negative measurable $$f$$.