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 $$\|F\|_{X} \leq C \|K\|_{Z} \|f\|_{Y}$$?


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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