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It is known that $H^{s,p}(\mathbb{R}^n)\subset \mathcal{C}(\mathbb{R}^n)\cap L^{\infty}(\mathbb{R}^n)$ when $s>n/p$, with $H^{s,p}$ Sobolev space. What is the norm that is considered in space $\mathcal{C}(\mathbb{R}^n)$ for a function to be continuous? sup-norm?

eraldcoil
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    There is no natural norm on this space. You use the sup norm on the balls of radius $1,2,3...$ to generate a locally convex topology. – Kavi Rama Murthy Jun 21 '20 at 12:44
  • Note that a topological vector space (over the reals or the complex numbers) is always path connected (via paths $t\mapsto tv$). It follows that any metric generating the topology, and in particular any norm, must not attain $\infty$, which the $\sup$ norm does in this space. The supremum norm does define a nice topological group structure (it is translation-invariant, and even commutes with scalar multiplication), but it does not make scalar multiplication continuous. – tomasz Jun 21 '20 at 15:50

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The space $\mathcal{C}(\mathbb{R}^n)$ is not normable. It does admit a complete translation invariant metric (in fact many, all equivalent) for which the sum and scalar multiplication are continuous. The topology it generates is that of convergence in compact sets (as Prof. Kavi Rama Murthy mentioned)

Here is one such metric:

$$\rho(f,g)=\sum^\infty_{n=1}\frac{1}{2^n} (\lVert f-g\rVert_n\wedge1)$$

where $\|f-g\|_n=\sup_{x\in B(0;n)}|f(x)-g(x)|$, $B(0;r)$ being the ball in $\mathbb{R}^n$ centered at $0$ and of radius $r>0$.

Under this metric, $\mathcal{C}(\mathbb{R}^n)$ is a Frechet space. A good place to read about this and many other important examples (the space of distributions, the Schwatz space, among others) is Chapter 1 of Rudin's functional analysis.

Mittens
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  • I see. But, if $u\in H^{s,p}$ with $s>n/p$. In what sense the function $u$ is continuous? that is, with what norm? – eraldcoil Jun 21 '20 at 19:38
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    This means that $u\in H^{s,p}$ has a version $u^*$ that is bounded and continuous in $\mathbb{R}^n$. Just like the $f\in L_1$ and $\hat{f}\in L_1$ implies that $f$ has a version that is continuous. – Mittens Jun 21 '20 at 19:51