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Young's inequality tells us that: $L^{1}\ast L^{p} \subset L^{p}, (1\leq p \leq \infty)$

My Question: What are examples of functions $F:L^{1}(\mathbb R)\to L^{p}(\mathbb R)$ with the property $F(f\ast g)= f\ast F(g)$ for all $f,g \in L^{1}$? Can we characterize it?

Inquisitive
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    Every such operator commutes with translations: If we let $f_\varepsilon$ be an approximation of unity, we get $F(T_x g) \leftarrow F(f_\varepsilon \ast T_x g)=F(T_x f_\varepsilon \ast g) = (T_x f_\varepsilon) \ast F(g) = T_x (f_\varepsilon \ast F(g)) \to T_x F(g)$. This strongly indicates that $F$ is given by convolution with some distribution, see also http://math.stackexchange.com/questions/948649/what-can-we-say-about-operators-in-lp-that-commute-with-translation. EDIT: Here, I of course assumed that $F$ is linear. – PhoemueX Aug 25 '15 at 08:05
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    In the comment to the answer I linked above, @WillieWong links to http://link.springer.com/article/10.1007%2FBF02547187 . Theorem 1.2 in that paper shows that every bounded translation invariant operator $A : L^p \to L^q$ is given by convolution with a distribution $T \in \mathcal{S'}(\Bbb{R}^d)$, at least on the set of $C_c^\infty$ functions. So essentially you want to characterize all tempered distributions for which convolution with this distribution extends to a bounded linear operator $L^1 \to L^p$. Hoermander calls this space $L_1^p$. – PhoemueX Aug 25 '15 at 08:11
  • @PhoemueX; thanks a lot; If I do slight change: $F:L^{1}\to C_0$such that $F(f\ast g)= g\cdot F(f)$; then can we say some thing about $F$? – Inquisitive Aug 25 '15 at 08:46
  • Yes, in that case, the map $L^1 \to \Bbb{K}, f \mapsto F(f)(0)$ is a bounded linear functional on $L^1$. Hence, there is $g \in L^\infty$ with $F(f)(0) = \int f \cdot g , dy$ for all $f \in L^1$. Now, use translation invariance to obtain $F(f)(x) = F(T_{-x}f)(0) = \int T_{-x}f (y) \cdot g(y) , dy = \int f(x+y) \cdot g(y) , dy = \int f(z) \cdot g(z-x) , dz = (f \ast g^{\sim})(x)$ with $g^{\sim} (x) = g(-x)$. Hence, $F$ is given by convolution with an $L^\infty$ function. – PhoemueX Aug 25 '15 at 09:52
  • Related: http://math.stackexchange.com/questions/61509/bounded-linear-operators-that-commute-with-translation – Tomasz Kania Aug 28 '15 at 20:22

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