Let $(X, \left \| \cdot \right \|_X )$, $(X, \left \| \cdot \right \|_Y)$ two normed vector spaces with $X \subset Y$, by definition we have $X \hookrightarrow Y$ if $\left \| x \right \|_Y \leq C \left \| x \right \|_X$ for some $C \geq 0$ (general definition of continuous inclusions in normed spaces)

My problem is to show that $\mathcal{D}_K(\Omega) \hookrightarrow \mathcal{E}(\Omega)$ where there are the following notations:

The space of test functions is $\mathcal{D}(\Omega):=\cup_{K \in \mathcal{K}(\Omega)} \mathcal{D}_K(\Omega)$ where $K \subset \Omega$ is an arbitrary compact and obviously $\Omega \subseteq \mathbb{R}^n$ is an open nonempty. Moreover $\mathcal{D}_K(\Omega):=\lbrace \varphi \in \mathcal{E}(\Omega):$ supp$(\varphi) \subset K\rbrace$ where $\mathcal{E}(\Omega)$ represent the locally convex space of all $C^\infty$ functions with the topology defined by increasing sequence of semi-norms

$\displaystyle q_{j}(f)=\sum_{|\alpha| \leq j} \left \| D^\alpha f \right \|_{K_{j}}= \sum_{|\alpha| \leq j} \sup_{x \in K_j} |D^\alpha f|$.

Now, $\mathcal{E}(\Omega)$ and $\mathcal{D}_K(\Omega)$ are Frechét spaces and in particular is defined the Fréchet norm associated to the sequence of semi-norms. For example, and in the general case, for sequence of semi-norms $\lbrace p_n \rbrace$ the Frechet norm is $\left \| x \right \|:=\sum_{n=1}^\infty 2^{-n} p_n(x)/(1+p_n(x))$. The topology in $\mathcal{D}_K(\Omega)$ is defined by increasing sequence of semi-norms

$\displaystyle q_N(f)=\sum_{|\alpha| \leq N} \left \| D^\alpha f \right \|_{K_{N}} = \sum_{|\alpha| \leq N} \sup_{x \in K} |D^\alpha f|$

Since $\Omega$ is the union of an increasing sequence of the compact sets $\lbrace K_j \rbrace_{j \in \mathbb{N}}$, for any compact set $K=K_N \subset \Omega$ which corresponds to a fixed $N \in \mathbb{N}$, we have that $q_j(f) \leq q_N(f)$ for all $j \leq N$, since $K_j \subset K$. Therefore, in terms of Fréchet norm, follows the thesis.

However, I'm thinking that is most appropriate to define the continuous inclusion in locally convex spaces. I remembered also another result that extends the continuity condition for linear operators between normed spaces in locally convex spaces:

  1. "Let $\mathcal{P}$ and $\mathcal{Q}$ are two sufficient families (sufficient means that $p(x)=0$ for every $p \in \mathcal{P}$ implies $x=0$) of semi-norms for the locally convex space $E$ and $F$. A linear application $T: E \rightarrow F$ is continuous if and only if for every $q \in \mathcal{Q}$ there exist $p_j \in \mathcal{P}$ with $j \in J$ finite and a constant $C \geq 0$ so that $q(Tx) \leq C \max_{j \in J} p_j(x)$."

In particular, if the locally convex spaces $E$, $F$ are normable, by definition we have $\mathcal{P}=\lbrace \left \| \cdot \right \|_E \rbrace$ and $\mathcal{Q}=\lbrace \left \| \cdot \right \|_F \rbrace$ and for the previous result we find the condition of continuity for linear operators in normed spaces.

Then for this result, we say that $E \hookrightarrow F$ if the inclusion $\iota : E \rightarrow F$ is continuous, i.e. for every $q \in \mathcal{Q}$ there exist $p_j \in \mathcal{P}$ so that $q(x) \leq C \max_{j \in J} p_j(x)$ for some $C \geq 0$.

Consequently, in the above procedure, it is unnecessary to consider the Fréchet norm but remains correct for (1). It seems to me better and more elegant.

it's correct ?

  • $\begingroup$ I edited the question, it should be clearer. In particular for X and Y, it is only a general definition. $\endgroup$ – user288972 Nov 25 '15 at 11:12
  • $\begingroup$ Now I understand. I think the procedure is correct. $\endgroup$ – Emanuele Paolini Nov 25 '15 at 15:09
  • $\begingroup$ I'm sorry, but I have added some comments and it seems to me that the second justification is better, what do you think? $\endgroup$ – user288972 Nov 25 '15 at 16:02

To close the question, I replied to myself.

This is the way that generalizes the notion of continuity of linear operators in locally convex spaces, and in particular we have the continuous inclusions between locally convex spaces.

Here is a sample, but it all depends on the theorem (1)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.