Basic questions about the sobolev space $H^\infty(\mathbb{R})$

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?

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 $$H^\infty\not\subset L^1$$.
• 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?
• 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
• @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 $\hat f$. Aug 12, 2020 at 20:39