Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $ f $ be a continuous, real-valued function on $[0, 1] $.

Show that $$\int_0^1 \int_0^1 |f (x)+f (y)| dx dy \ge \int_0^1 |f (x)| dx $$

I tried to dissect the square in triangles and use some symmetry but it didn't work.

Can you help?

share|improve this question
add comment

4 Answers

up vote 4 down vote accepted

Let us decompose $[0,1]^2$ as suggested by Martín-Blas by writing $P=\{x:f(x)>0\}$, $N=\{x:f(x)\le 0\}$ and write $g(x,y)=|f(x)+f(y)|$. Denote the Lebesgue measure on $[0,1]$ by $\lambda$. Then

$$\begin{align}\int_0^1 \int_0^1 |f(x)+f(y)| dx dy &= \int_{P\times P} g + \int_{P\times N} g + \overbrace{\int_{N\times P} g}^{=\int_{P\times N} g} + \int_{N\times N} g\\ &= 2\lambda(P) \int_{P} |f|+2\int_{P\times N} g + 2\lambda(N)\int_N |f|\tag{1}\\ \end{align}$$

Using the triangle inequality for integrals we get

$$\int_{P\times N} g\ge \left|\int_{P\times N} f(x)+f(y) d(x,y)\right|\tag{2} =\left| \lambda(N) \int_P |f| - \lambda(P) \int_N |f|\right| $$ Combining $(1)$ and $(2)$ we obtain

$$\begin{align} \int_0^1 \int_0^1 |f(x)+g(x)| dx dy &\ge 2\lambda(P)\int_P |f| + 2\left| \lambda(N) \int_P |f| - \lambda(P) \int_N |f|\right|+2\lambda(N)\int_N |f|\tag{3} \end{align} $$

We may assume without loss of generality that $\lambda(P)\ge \lambda(N)$, because if the inequality that we want to prove holds for some $f$, then it also holds for $-f$.

Now there are two cases.

Case 1: $\int_P |f|\ge \int_N |f|$. Use $(3)$, drop the absolute values around the middle term and remember that $\lambda(P)+\lambda(N)=1$ to conclude that $(3)$ is greater-equal

$$\int_P |f|+\int_N |f|=\int_0^1 f(x) dx$$

Case 2: $\int_N |f|>\int_P |f|$. Then we also have $\lambda(P)\int_N |f|\ge \lambda(N)\int_P |f|$. That allows us to evaluate the absolute value in the middle term of $(3)$ and again conclude as in Case 1.

share|improve this answer
    
I think this is a good method since it do not require a high regularity of $|f(x)|$. But can you show me how you deal with the absolute value of middle term in Case 1? What is more, in Case 2, the absolute value can be dropped by adding a negative mark of middle term. And how you get the conclusion you show in Case 1? Regards. –  Lion Feb 20 at 0:45
add comment

If $f(x)=0$ in $[0,1]$, the inequality is trivial to prove.

Now let $f(x)$ is not constant be zero in $[0,1]$. Consider the auxiliary function \begin{equation} F(\alpha)=\int_0^1\int_0^1\sqrt{f(x)^2+2\alpha f(x)f(y)+f(y)^2}dxdy \end{equation} Note that $F(1)=\int_0^1\int_0^1|f(x)+f(y)|dxdy$ and $F(0)\geq\int_0^1|f(x)|dx$. So what we going to prove is: \begin{equation} F(1)\geq F(0) \end{equation} Since $F(\alpha)$ is continuous about $\alpha$, from simple compute, we have: \begin{equation} F'(\alpha)=\int_0^1\int_0^1\frac{f(x)f(y)}{\sqrt{f(x)^2+2\alpha f(x)f(y)+f(y)^2}}dxdy \end{equation} Since $f(x)$ is not constant be zero in $[0,1]$, there must be $\inf\{f(x)^2+2\alpha f(x)f(y)+f(y)^2\}>0$ for $\alpha\in(0,1)$. What is more, if we assume $f(x)$ is bounded in $[0,1]$ (which is reasonable), then the $C=\sup\{f(x)^2+2\alpha f(x)f(y)+f(y)^2\}>0$ is existence. Thus, we have: \begin{equation} F'(\alpha)\geq C\int_0^1\int_0^1f(x)f(y)dxdy=C\int_0^1f(x)dx\int_0^1f(y)dy=C\left(\int_0^1f(x)dx\right)^2>0 \end{equation}

Thus, we can claim $F(0)\leq F(1)$ which is going to prove.

share|improve this answer
    
(+1) But I think you mean $C=\inf\{\cdots\}$. Also you should maybe say a word or two why you can interchange integral and derivative (e.g. Lebesgue's theorem for parameter integrals). –  Your Ad Here Feb 19 at 13:22
    
@TooOldForMath Yes, that is my mistake. I have edited my answer. –  Lion Feb 19 at 13:53
    
I don't understand why is reasonable to assume $f(x)$ is bounded . Can you enlighten me please ? –  user119228 Feb 19 at 22:46
    
@Julien More precisely, we can assume $f(x)$ absolutely integrable in $[0,1]$. Alternatively, you can use integral mean value theorem to get eliminate the denominator of the integrand. –  Lion Feb 20 at 0:23
    
@Lion Ah! Thank you for answering my question and it's a nice approach. (+1) –  user119228 Feb 20 at 1:01
add comment

Try this: $P=\{x\in[0,1]\,\vert\,f(x)\ge 0\}$, $N=\{x\in[0,1]\,\vert\,f(x)<0\}$ and decompose the domain: $$[0,1]\times[0,1]=(P\times P)\cup(P\times N)\cup(N\times P)\cup(N\times N).$$

share|improve this answer
    
Did that. Didn't help. –  user129798 Feb 19 at 10:57
add comment

lemma: let $x_{i}\in R$,then we have $$\sum_{1\le i,j\le n}|x_{i}+x_{j}|\ge n\sum_{k=1}^{n}|x_{k}|$$

This post have solution:How prove this inequality: $\sum_{i,j=1}^{n}|x_{i}+x_{j}|\ge n\sum_{i=1}^{n}|x_{i}|$?

then let $x_{i}\to f(x_{i}),x_{j}\to f(x_{j})$ $$\Longrightarrow \dfrac{1}{n^2}\sum_{i,j=1}^{n}|f(x_{i})+f(x_{j})|\ge\dfrac{1}{n}\sum_{i=1}^{n}|f(x_{i})|$$ $$\Longrightarrow \dfrac{1}{n^2}(n^2-n)\int_{0}^{1}\int_{0}^{1}|f(x)+f(y)|dxdy+\dfrac{2n}{n^2}\int_{0}^{1}|f(x)|dx\ge\dfrac{n}{n}\int_{0}^{1}|f(x)|dx$$ then we have $$\int_{0}^{1}\int_{0}^{1}|f(x)+f(y)|dxdy\ge\dfrac{n-2}{n-1}\int_{0}^{1}|f(x)|dx$$ let $n\to \infty$,then $$\int_{0}^{1}\int_{0}^{1}|f(x)+f(y)|dxdy\ge\int_{0}^{1}|f(x)|dx$$

share|improve this answer
    
why download?this solution is not true?I think this methods is nice ! +1 –  user94270 Feb 19 at 13:27
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.