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For an arbitrary vector space $V$ over $\mathbb{F}$, consider continuous linear maps $f: V \to \mathbb{F}$ where continuity is defined as sequential continuity, i.e. if $\phi_j \to \phi$ in $V$ then $f(\phi_j) \to f(\phi)$ in $\mathbb{F}$.

A distribution $u \in \mathcal{D}'(\mathbb{R}^n)$ is a continuous linear functional on $C_C^\infty(\mathbb{R}^n)$ with the above definition of continuity.

In the space $C_C^\infty(\mathbb{R}^n)$, we say that a sequence $\lim_{j \to \infty} \phi_j = \phi$ if there is a compact set $K$ such that $\phi_j \equiv 0$ outside $K$ for all $j$ and all derivatives of $\phi_j$ converge uniformly to the corresponding derivatives of $\phi$.

Now, in the text I'm reading it claims that, with the above definitions, the continuity of $u$ is equivalent to: for all $K$ compact there exists $m$ and $C > 0$ such that for all $\phi \in C_C^\infty(\mathbb{R}^\infty)$ with $\textrm{supp}\, \phi \subset K$, $$ |u(\phi)| \leq C\sum_{|\alpha| \leq m} \sup_{\mathbb{R}^n}|\partial^\alpha \phi|. $$

I'm having some trouble figuring out how these definitions are equivalent. In particular, the definitions of continuity are given in terms of a sequence $\phi_j$ and limiting value $\phi$, but in this equivalent formulation we don't have any sequences. Are they folded into the constant $C$?

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For a linear functional $u$ on normed spaces, continuity is just $||u(x)||\le C||x||$. Sequences don't have to enter into this particular formulation, but continuity defined in terms of sequences can be shown to be equivalent to the condition just stated (e.g., $||u(x_n)-u(x)||=||u(x_n-x)||\le C||x_n-x||$ so $u(x_n)\to u(x)$ if $x_n\to x$ in the norm topology).

Now, $C_c^\infty(\mathbb{R}^n)$ is not normable with the usual topology used in distribution theory. Nonetheless, it's not surprising that you can find inequalities that don't involve sequences but are equivalent to continuity for linear functionals on $C_c^\infty(\mathbb{R}^n)$.

One construction of the topology on $C_c^\infty(\mathbb{R}^n)$ (found e.g. in Rudin Functional Analysis 6.2) is based on sets of norms on every $C_K^\infty(\mathbb{R}^n)$ for $K$ compact. These norms are like the RHS of your displayed inequality. That inequality, with Rudin's definitions, is proved in 6.8(b) which in turn relies on 6.6. I can't really reproduce here the details of the proof since it's too long. More generally, however, in functional analysis, the trick of constructing a topology with a set of norms (or seminorms) is common, and the spaces so constructed behave a lot like normed spaces.

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  • $\begingroup$ Thank you for the reference. $\endgroup$
    – user83387
    Jan 20 '16 at 6:01
  • $\begingroup$ For an additional reference, see Chapter 5 of Reed & Simon's Methods of Mathematical Physics Vol. I. There you will see that $C_c^\infty(\mathbb{R}^n)$ does have an inductive limit topology, which one gets by taking the "limit" of the locally convex topologies on $C^\infty(K)$ as you let the compact sets $K$ grow to $\mathbb{R}^n$. The space of distributions is then identified as the topological dual of this TVS. $\endgroup$ Jan 20 '16 at 8:10

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