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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?

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Lion
    Feb 20 '14 at 0:45
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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.

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  • $\begingroup$ (+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). $\endgroup$
    – J.R.
    Feb 19 '14 at 13:22
  • $\begingroup$ @TooOldForMath Yes, that is my mistake. I have edited my answer. $\endgroup$
    – Lion
    Feb 19 '14 at 13:53
  • $\begingroup$ I don't understand why is reasonable to assume $f(x)$ is bounded . Can you enlighten me please ? $\endgroup$
    – user119228
    Feb 19 '14 at 22:46
  • $\begingroup$ @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. $\endgroup$
    – Lion
    Feb 20 '14 at 0:23
  • $\begingroup$ @Lion Ah! Thank you for answering my question and it's a nice approach. (+1) $\endgroup$
    – user119228
    Feb 20 '14 at 1:01
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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: $$[0,1]\times[0,1]=(P\times P)\cup(P\times N)\cup(N\times P)\cup(N\times N).$$

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  • $\begingroup$ Did that. Didn't help. $\endgroup$
    – user129798
    Feb 19 '14 at 10:57
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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 $$\int_{0}^{1}\int_{0}^{1}|f(x)+f(y)|dxdy\ge\int_{0}^{1}|f(x)|dx$$

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  • $\begingroup$ why download?this solution is not true?I think this methods is nice ! +1 $\endgroup$
    – user94270
    Feb 19 '14 at 13:27

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