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Let $f_1, f_2, f_3\in C_0^\infty(\mathbb{R}^2)$, $\psi\in C_0^\infty(\mathbb{R})$, and $\psi_t(x)=t\psi(tx)$ for all $t>0$. Show that if $\tfrac{1}{p_1}+\tfrac{1}{p_2}+\tfrac{1}{p_3}=1$ for some $1\le p_1,p_2,p_3\le\infty$, then $$\left|\int_{\mathbb{R}^3}f_1(x,y)f_2(y,z)f_3(z,x)\psi_t(x+y+z)dxdydz\right|\le\lVert f_1\rVert_{L^{p_1}(\mathbb{R}^2)}\lVert f_2\rVert_{L^{p_2}(\mathbb{R}^2)}\lVert f_3\rVert_{L^{p_3}(\mathbb{R}^2)}\lVert\psi\rVert_{L^1(\mathbb{R})}$$ for all $t>0$.

See https://en.wikipedia.org/wiki/Bump_function and http://mathworld.wolfram.com/Lp-Space.html for notation explanation.

My first instinct is to use Holder's Inequality, but the pairings of the variables on the LHS and the additional $\psi$ terms make it difficult to apply. I don't have any other ideas.

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  • $\begingroup$ @MostafaAyaz I don't think that this is really a duplicate. Read the last paragraph of the question carefully. $\endgroup$ – Arnaud D. Feb 19 '18 at 13:05
  • $\begingroup$ Sorry....Thanks for the feedback... $\endgroup$ – Mostafa Ayaz Feb 19 '18 at 15:47
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For brevity, I will only consider the case of two variables below, and write $p,q$ instead of $p_1,p_2$. The three variable case is left as an exercise to the reader.

We have \begin{align*} &\left| \int f(x) g(y) \psi_t (x+y) d(x,y)\right| \\ &\leq \int |f(x)| |\psi_t(x+y)|^{1/p} |g(y)| |\psi_t (x+y)|^{1/q} d(x,y)\\ & \leq \left(\int |f(x)|^p |\psi_t (x+y)| d(x,y)\right)^{1/p} \left(\int |g(y)|^q |\psi_t (x+y)| d(x,y)\right)^{1/q}. \end{align*} Now, we use Fubini's theorem and the substitution $z=x+y$ to estimate the first integral. The second one is analogous. We have \begin{align*} &\int |f(x)|^p |\psi_t (x+y)| d(x,y)\\ &=\int |f(x)|^p \int |\psi_t (x+y)| dy dx\\ &= \int |f(x)|^p \int |\psi_t (z)| dz dx\\ &= \|f\|_{L^p}^p \|\psi_t\|_{L^1}. \end{align*} Finally note that $\|\psi_t\|_{L^1}=\|\psi\|_{L^1}$.

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Thanks PhoemueX. Since I'm TeXing this anyways, here's the solution for three variables.

By Holder's Inequality,

$$\left|\int_{\mathbb{R}^3}f_1(x,y)f_2(y,z)f_3(z,x)\psi_t(x+y+z)d(x, y, z)\right|\le\int_{\mathbb{R}^3}|f_1(x,y)\psi_t(x+y+z)^{1/p_1}||f_2(y,z)\psi_t(x+y+z)^{1/p_2}||f_3(z, x)\psi_t(x+y+z)^{1/p_3}|d(x, y, z)\le\left(\int_{\mathbb{R}^3}|f_1(x,y)^{p_1}\psi_t(x+y+z)|d(x, y, z)\right)^{1/p_1}\left(\int_{\mathbb{R}^3}|f_2(y,z)^{p_2}\psi_t(x+y+z)|d(x, y, z)\right)^{1/p_2}$$ $$\left(\int_{\mathbb{R}^3}|f_3(z,x)^{p_3}\psi_t(x+y+z)|d(x, y, z)\right)^{1/p_3}$$ Let $x, y\in\mathbb{R}$ and $u=x+y+z$ for all $z\in\mathbb{R}$. As $z$ ranges through $\mathbb{R}$, $u$ ranges through $\mathbb{R}$. $du=dz$. $|f_1(x, y)^{p_1}\psi_t(x+y+z)|$ is nonnegative and measurable on $\mathbb{R}^3$. By Tonelli's Theorem, $$\int_{\mathbb{R}^3}|f_1(x,y)^{p_1}\psi_t(x+y+z)|d(x, y, z)=\int_{\mathbb{R}^2}\left(|f_1(x,y)|^{p_1}\int_{\mathbb{R}}|\psi_t(x+y+z)|dz\right)d(x,y)=\int_{\mathbb{R}^2}\left(|f_1(x,y)|^{p_1}\int_{\mathbb{R}}|\psi_t(u)|du\right)d(x,y)=\int_{\mathbb{R}^2}|f_1|(x,y)|^{p_1}\lVert\psi_t\rVert_{L^1(\mathbb{R})}=\lVert f_1\rVert_{L^{p_1}(\mathbb{R}^2)}^{p_1}\lVert\psi_t\rVert_{L^1(\mathbb{R})}$$ Similarly, $$\int_{\mathbb{R}^3}|f_2(x,y)^{p_2}\psi_t(x+y+z)|d(x, y, z)=\lVert f_2\rVert_{L^{p_2}(\mathbb{R}^2)}^{p_2}\lVert\psi_t\rVert_{L^1(\mathbb{R})}$$ and $$\int_{\mathbb{R}^3}|f_3(x,y)^{p_3}\psi_t(x+y+z)|d(x, y, z)=\lVert f_3\rVert_{L^{p_3}(\mathbb{R}^2)}^{p_3}\lVert\psi_t\rVert_{L^1(\mathbb{R})}$$ so $$\left|\int_{\mathbb{R}^3}f_1(x,y)f_2(y,z)f_3(z,x)\psi_t(x+y+z)d(x, y, z)\right|\le(\lVert f_1\rVert_{L^{p_1}(\mathbb{R}^2)}^{p_1}\lVert\psi_t\rVert_{L^1(\mathbb{R})})^{1/p_1}(\lVert f_2\rVert_{L^{p_2}(\mathbb{R}^2)}^{p_2}\lVert\psi_t\rVert_{L^1(\mathbb{R})})^{1/p_2}(\lVert f_3\rVert_{L^{p_3}(\mathbb{R}^2)}^{p_3}\lVert\psi_t\rVert_{L^1(\mathbb{R})})^{1/p_3}=\lVert f_1\rVert_{L^{p_1}(\mathbb{R}^2)}\lVert f_2\rVert_{L^{p_2}(\mathbb{R}^2)}\lVert f_3\rVert_{L^{p_3}(\mathbb{R}^2)}\lVert\psi\rVert_{L^1(\mathbb{R})}$$

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