# Tagged Questions

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### $L^{p}(\mathbb R)\subset L^{1}(\mathbb R) \ast L^{p}(\mathbb R), (1< p< \infty)$?

Let $\mathbb T$ be a circle group. In 1939, Salem, has shown that, every member of $L^{1}(\mathbb T)$ can written as a product(convolution) some other two members of $L^{1}(\mathbb T),$ that is, ...
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By Schwartz-inequality and Riesz–Fischer theorem, one can deduced that, ($\ast$ stands for a usual convolution ) $$L^{2}(\mathbb T) \ast L^{2}(\mathbb T) = A(\mathbb T)(:= \{f\in L^{1}(\mathbb T): ... 1answer 91 views ### Let f \in M^{1,1} (\mathbb R) (Feichtinger's algebra /Modulation Space). Can we say Fof\in M^{1,1}(\mathbb R); F is an entire function? The Modulation space ( Feichtinger's algebra),$$S_{0} (\mathbb R) = M^{1, 1}(\mathbb R): = \{ f\in L^{2}(\mathbb R) : V_{g}(f) \in L^{1}(\mathbb R^{2}) \}; where $V_{g}f (x, w)$ is the short- ...
Let $L$ be the Banach algebra of $L^1$-functions from $\mathbb{R}$ to $\mathbb{C}$ with $L^1$-norm and convolution as algebra multiplication. Assume that we knew that the homomorphisms from $L$ to ...