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Let's consider $H^\infty(\mathbb{R})$ to be the intersection of all Sobolev spaces $H^s$ for $s\geq0$, that is, $$ H^\infty(\mathbb{R}):=\bigcap_{s\geq 0}H^s(\mathbb{R}). $$ I am wondering some trivial questions about this space, like for example, is this space different from the space of Schwartz functions $\mathcal{S}$? Or maybe do we have an inclusion like $$ H^\infty\subset\mathcal{S} \quad \hbox{or} \quad \mathcal{S}\subset H^\infty? $$ If not, I was wondering if even possible to prove that any function $f\in H^\infty$ belongs to $f\in L^1$. This last question arises to me because I know that by Sobolev's embedding we have that $f$ belongs to any $L^p$ space for $p\geq 2$, but what about $p<2$? Since we have a "super" regularity, I guess this doesn't sound crazy right? Finally, does $f\in H^\infty$ implies (for example) exponential decay?

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First, since the spaces are nested, $$ H^\infty(\mathbb R) = \bigcap_{k=0}^\infty H^k(\mathbb R).$$ Secondly, we always have $\mathcal S \subset H^k$ for any $k\ge0$, and therefore $\mathcal S \subset H^\infty$. The reverse inclusion is not true: one counterexample is $$ f(x) := \frac1{\sqrt{1+x^2}}\in H^\infty\setminus \mathcal S.$$ This is easy to see because it's clearly a smooth function in $L^2$, and the derivatives $f^{(n)}$ decay faster than $f$ itself, so $f$ belongs to all $H^k$ spaces. It's also true that $f\notin L^1$, so this proves $H^\infty\not\subset L^1$.

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  • $\begingroup$ Sorry, maybe one silly question. I kept thinkig about $f\in H^\infty$ with $f\notin L^1$. I understand your counter example, but I was wondering why the following argument cannot hold. Consider the $L^1$ norm of $f$ and then, by Holder: $$\int \vert f\vert\leq \Vert (1+x^2)^{-1}\Vert_{L^2}\left(\int (1+x^2)f^2\right)^{1/2}.$$ Then, I believe that, by Plancharel's Theorem I should be able to write the last integral in terms of its fourier transform, and the Fourier transform of a polynomial writes as derivatives. Since $f\in H^\infty$, all derivatives of $f$ are in $L^2$. Why is this wrong? $\endgroup$
    – W2S
    Aug 12, 2020 at 18:21
  • $\begingroup$ I mean, I see taking $f$ equals to the function in your counter example we would get $+\infty$ in the right-hand side. My question is why this Plancharels argument fails :o $\endgroup$
    – W2S
    Aug 12, 2020 at 18:27
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    $\begingroup$ @W2S the first term in the RHS should be L1 or L2 of the square root but not important. The issue is that we have all derivatives of $f$, not of $\hat f$. $\endgroup$ Aug 12, 2020 at 20:39

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