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

Question

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?

Answer 1

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 $H^\infty\not\subset L^1$.