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We have well known inequality for convolution: $\|g\ast f\|_{X} \leq \|g\|_{Y} \|f\|_{Z}$ where $Y, X, Z$ are suitable Lebesgue spaces, e.g., see Young's inequality.

Let $f:\mathbb R^{2}\to \mathbb C, K:\mathbb R \to \mathbb C$ ( assume that $f\in \mathcal{S}(\mathbb R^2)$ (Schwartz space), $K\in \mathcal{S}(\mathbb R)$), We define convolution in the following sense: $$F(x,y) =\int_{\mathbb R} [K(x-z)-K(y-z)] f(z,z) dz$$

Can we expect to find norm linear spaces $(X, \|\cdot\|_{X}), (Y, \|\cdot\|_{Y}$ consists of functions on $\mathbb R^2$ and norm space $(Z, \|\cdot\|)$ consists of functions on $\mathbb R$ so that $$\|F\|_{X} \leq C \|K\|_{Z} \|f\|_{Y}$$?

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Your question is a little vague, but only the values of $f$ on the diagonal is used. Its more natural therefore to use $g(z) = f(z,z)$ and look for results for the one variable function $g$. Then your $F(x,y)$ is just

$$ F(x,y) = K*g(x) - K*g(y)$$

and for example, we have the crude estimate $$ |K*g(x) - K*g(y)| = \left| \int_y^x K'*g \right| \le \int_{\mathbb R }|K'*g|$$

This implies $$ \|F\|_{L^\infty(\mathbb R^2)} \le \|K'*g\|_{L^1} \le \|K\|_{W^{1,p}} \|g\|_{L^q}$$ for suitable $p,q$ via Young's inequality.

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  • $\begingroup$ Thanks. I'm trying to understand your second step: that is, ..=$\int_y^x K'\ast g$? ($K'$ is the derivative of $K$?) Please can you explain a bit? Do I need to use mean value theorem or so? Thanks. $\endgroup$ – Math Learner Nov 10 '18 at 0:15
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    $\begingroup$ @MathLearner yes, using $K*g'=K'*g = (K*g)'$ and integral mean value theorem (that is, fundamental theorem of calculus) $\endgroup$ – Calvin Khor Nov 10 '18 at 8:24
  • $\begingroup$ Thanks. I got it. Is there any way to get the estimate for $\|\hat{F}\|_{L^1(\mathbb R^2)}$ ($\hat{F}$ is the Fourier transform of $F$)? I'm curious to know this. Thanks. $\endgroup$ – Math Learner Nov 12 '18 at 15:54
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    $\begingroup$ @MathLearner the two variable fourier transform is not locally integrable, as a distribution if $(\xi,\eta)$ are the frequency variables it is $$\hat F (\xi,\eta) = \hat K(\xi) \hat g(\xi) \delta_0(\eta) - \hat K(\eta)\hat g(\eta) \delta_0(\xi)$$ $\endgroup$ – Calvin Khor Nov 12 '18 at 17:24
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    $\begingroup$ @MathLearner no, $\delta_0$ is not a function, it doesn't have an $L^1$ norm. Its the linear functional in $\mathcal S'$ defined by $$\mathcal S \ni \phi \mapsto \delta_0(\phi) := \langle \delta_0,\phi\rangle := \phi(0) \in \mathbb C $$ admittedly I used the sloppy physicist notation above, but the quantity $\delta_0 (\eta)$ cannot be interpreted in the normal way, it is defined only when it appears in an integral against a function in $\mathcal S$ $$ \int \delta_0(\eta) \phi(\eta) d\eta := \phi(0)$$ $\endgroup$ – Calvin Khor Nov 12 '18 at 18:46

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