6
$\begingroup$

If $H$, $H_0$ are two separable Hilbert spaces and $H$ is continuously and densly embedded in $H_0$, it is possible to construct a sequence of linear operators $$ P_n : H_0 \to H $$ such that for all $x \in H_0$ one has convergence $P_n x \to x$ in the $H_0$-norm.

The motivation is to generalize the idea of smoothing operators. For example, $H^2(\mathbb R)$ is densely embedded in $L^2(\mathbb R)$ and it is well known that one can approximate each $L^2$-function by an $H^2$-function. In this case $P_n$ could be a convolution with a smooth kernel with width $1/n$.

Can this be generalized to Fréchet spaces or even beyond? If $H$ and $H_0$ are Fréchet spaces rather than Hilbert spaces, is it always possible to construct a sequence of bounded operators $P_n$ as above?

In the Hilbert case one way to construct $P_n$ is by considering an unbounded, self-adjoint operator $A : D(A) \to H_0$, with $D(A)=H$, that represents the inner product via $$ \langle v,w \rangle_H = \langle Av, Aw \rangle_{H_0}\,. $$ If $\{ P_\Omega\}$ is the spectral measure accociated to $A$, then we can use $P_{[-n,n]}$ as our sequence of operators. It is less clear to me, what to with Fréchet spaces.

$\endgroup$
4
  • 1
    $\begingroup$ Suppose the embedding of $H$ into $H_0$ is compact. Then you are asking for the identity to be a strong limit of compact operators. I believe this is called the "compact approximation property" and there are Banach spaces that don't have it. Hopefully an expert can fill in the details or give references. $\endgroup$ Mar 7, 2017 at 16:19
  • $\begingroup$ Yes, I think you are right. It also fits with the answer given below: if we assume that our space has a Schauder basis, then it has the approximation property. $\endgroup$ Mar 8, 2017 at 2:59
  • 1
    $\begingroup$ A Banach or Frechet space $H_0$ is separable and has the BOUNDED approximation property (BAP) iff there is a sequence $P_n$ of continuous finite rank operators on $H_0$ that converges strongly to the identity. A simple perturbation argument show that you can take these operators to range in any dense subspace you want. On the other hand, if $H_0$ fails the analogous compact bounded approximation property and the inclusion from $H$ into $H_0$ is compact, then, as Nate pointed out, you cannot get a sequence that has the property you want. What else are you looking for? $\endgroup$ Mar 8, 2017 at 20:56
  • $\begingroup$ Thank you, now I am quite happy. I realized how to remove the requirement that the space admits a continuous norm and posted a writeup below. $\endgroup$ Mar 8, 2017 at 22:33

3 Answers 3

6
$\begingroup$

Here is a very simple method for separable Hilbert spaces (which easily generalizes to Frechet spaces with Schauder bases): Take an orthonormal basis $(e_k)_k$ in $H_0$ and choose $f_{n,k}\in H_1$ such that $\|e_k-f_{n,k}\|_0 \le 1/(n^2+k^2)$. Then define $P_n(x)=\sum_{k=1}^n \langle e_k,x\rangle_0 f_{n,k}$. The difference $\|P_n(x)-x\|_0$ is estimated just using the triangle inequality (so that this construction isn't bound to Hilbert spaces).

EDIT. I add some details for the Frechet case showing (at least under a mild additional assumption) that one does not need an absolute basis (which would exclude many Banach spaces, I recall the definitions below). Let $(\|\cdot\|_N)_{N\in\mathbb N}$ be an increasing sequence of seminorms giving the topology of the Frechet space $H_0$. A Schauder basis $(e_k)_k$ is a sequence such that that every $x\in H_0$ has a unique representation $x=\sum\limits_{k=1}^\infty \xi_k(x)e_k$. (The basis is called absolute if, for each $N$, there are $M$ and $c>0$ such that $\sum\limits_{k=1}^\infty |\xi_k(x)|\|e_k\|_N \le c\|x\|_M$ for all $x\in H_0$ -- this implies that the spaces is a projective limit of weighted $\ell^1$ spaces and excludes Hilbert spaces). By corollary 28.11 in the book Introduction to Functional Analysis of Meise and Vogt one has a slightly weaker condition for every Schauder basis: For every $N$ there are $M=M(N)$ and $c>0$ such that $\sup\lbrace|\xi_k(x)|\|e_k\|_N:k\in\mathbb N\rbrace \le c\|x\|_M$. In particular, the coefficient functionals $\xi_k$ (which are linear by the uniqueness) are continuous.

We construct $P_n$ under the additional assumptions that $\|\cdot\|_1$ is a norm and not only a seminorm (I am quite optimistic that this can be removed). For $n,k \in\mathbb N$ choose $f_{n,k}\in H_1$ with $\|e_k-f_{n,k}\|_n\le \|e_k\|_1/n^2$ and set, as previously, $P_n(x)=\sum\limits_{k=1}^n\xi_k(x)f_{n,k}$. These are continuous linear operators $H_0\to H_1$, and for each $x\in H_0$, $N\in\mathbb N$, and $n\ge N$ we have $$ \|P_n(x)-x\|_N \le \sum_{k=1}^n |\xi_k(x)|\|f_{n,k}-e_k\|_N + \|\sum_{k=n+1}^\infty\xi_k(x)e_k\|_N. $$ The second term tends to $0$ and (since $n\ge N$) the first term can be estimated by $\sum_{k=1}^n |\xi_k(x)|\|e_k\|_1/n^2 \le c\|x\|_{M(1)}/n$.

$\endgroup$
4
  • $\begingroup$ Thank you, this is very helpful. When generalizing the construction to Fréchet spaces, I seem to require an absolute basis, in order to control the terms $\langle u_k, x \rangle$ in the sum; here $u_k \in H_0^\ast$ are the coordinate functionals of the basis $(e_k)_k$. $\endgroup$ Mar 8, 2017 at 2:56
  • $\begingroup$ Another remark. Uniqueness of the basis representation is not need and one can argue with atomic decompositions instead. The article arxiv.org/abs/1212.0969 of Bonet, Galbis, Fernandez, and Ribera contains some information about that. $\endgroup$ Mar 8, 2017 at 10:06
  • $\begingroup$ Corollary 1.5 of the mentioned article says that a Frechet space has an atomic decomposition if and only if it has the bounded approximation property. This fits very well with Nate's comment above. $\endgroup$ Mar 8, 2017 at 10:14
  • $\begingroup$ I tried to remove the requirement that $\|\cdot\|_1$ is a norm, both in your construction and in Matthew's construction below, but I was not able to do this. One has to estimate things with norms $\| \cdot\|_n$ with $n$ higher and higher and looses too much control in the process. In any case this has been very helpful! $\endgroup$ Mar 8, 2017 at 18:26
3
$\begingroup$

Thanks to Jochen, Matthew and Bill, this is a detailed proof for Fréchet spaces.

Proposition. Let $E$ be a separable Fréchet space with the bounded approximation property, $F$ a topological vector space, continuously and densely embedded in $E$. Then there exists a sequence of continuous linear maps $P_n : E \to F$, such that $$ \forall x \in E\,:\, P_n x \to x \text{ in }E\,. $$

Proof. Let $(x_n)_{n \in \mathbb N}$ be a countable, dense sequence in $E$ and $(\|\cdot\|_n)_{n \in \mathbb N}$ an increasing fundamental system of seminorms. We assumed that $E$ has the bounded approximation property, hence there exists an equicontinuous sequence of linear maps $T_n : E \to E$ with finite rank that converge to $\operatorname{Id}_E$, uniformly on compact sets. By passing to a subsequence we can assume that $$ \| T_n x_j - x_j \|_n \leq \frac 1n \text{ for }j \leq n\,. $$ Due to equicontinuity there exists for each $m$, an $N_m \in \mathbb N$ and $C_m>0$ such that $$ \forall n \in \mathbb N\,,\; \forall x \in E\,:\, \| T_n x \|_m \leq C_m \| x \|_{N_m}\,. $$

For each $n$, the space $T_n(E)$ is finite dimensional. Let $n'=n'(n)$ be such that $\|\cdot\|_{n'}$ is a norm on $T_n(E)$. We can construct a map $S_n : T_n(E) \to F$ with $$ \| S_ny - y \|_{n} \leq \frac 1n \| y \|_{n'}\,, $$ for all $y \in T_n(E)$. To see that this is possible choose a basis $y_1, \dots, y_m$ of $T_n(E)$ and note that it is sufficient to define $S_n(y_i) \in F$, such that $\| S_n(y_i) - y_i \|_{n}$ is small enough. This is possible, because $F$ is dense in $E$. Define $P_n = S_n T_n$.

We have to show convergence $P_n x \to x$. Fix $x \in E$ and a seminorm $\|\cdot\|_m$. For $n$ and $k$ satisfying $m ,\,N_m,\,N_{m'} \leq n$ and $k \leq n$ we have \begin{align*} \| P_n x &- x \|_m \leq \|S_n T_n(x-x_k) -T_n(x-x_k) \|_m + \| T_n(x-x_k) \|_m + \\ &\qquad\qquad\qquad + \|S_n T_n x_k - T_n x_k \|_m +\| T_n x_k - x_k\|_m + \| x_k - x\|_m \\ &\leq \frac 1n \| T_n(x-x_k) \|_{m'} + C_m \| x - x_k \|_{N_m} + \frac 1n \| T_n x_k \|_{m'} + \frac 1n + \|x_k - x \|_m \\ &\leq \frac {C_{m'}}{n} \| x-x_k \|_{N_{m'}} + C_m \| x - x_k \|_{N_m} + \frac {C_{m'}}n \left( \| x\|_{N_{m'}} + \| x-x_k \|_{N_{m'}} \right) + \\ &\qquad\qquad + \frac 1n + \|x_k - x \|_m \,. \end{align*} We see that by choosing $n$ large enough and $\|x - x_k\|_n$ small enough we can achieve convergence.

$\endgroup$
1
$\begingroup$

Edit: So I think my real mistake was in the claim that "if $H_0$ is separable then we can use a sequence". As Bill Johnson implictly points out, you can always find a net $P_\alpha:H_0\rightarrow H$.

Just to correct the argument (though Martins now gives it in more generality)... If $H_0$ has the bounded approximation property, then there is an absolute constant $\lambda>0$ so that for $x_1,\cdots,x_n\in H_0$ there is a finite-rank operator $T:H_0\rightarrow H_0$ with $\|T\|\leq \lambda$ and $\|T(x_i)-x_i\| \leq \epsilon$ for each $i$.

In our case, $\iota:H\rightarrow H_0$ is a continuous map with dense range. For $x_1,\cdots,x_n \in H_0$ and $\epsilon>0$ we find a finite-rank $T$ with $\|T\|\leq\lambda$ and $\|T(x_i)-x_i\| \leq \epsilon$ for each $i$.

As $T(H_0)$ is finite dimensional and $\iota$ has dense range, we can find a linear map $S: T(H_0) \rightarrow H$ so that that $\|\iota S(x) - x\| \leq \epsilon$ for all $x$ in the unit ball of $T(H_0)$. [Proof: If $M\subseteq H_0$ is finite-dimensional, with linear basis $m_1,\cdots,m_n$, then as all norms are equivalent on $M$, if we can ensure that $\|\iota S(m_i)-m_i\|$ is very small, then $\|\iota S(x)-x\|$ will be small uniformly on the unit ball of $M$. But this follows as $\iota$ has dense range and we can choose each $S(m_i)$ completely freely.]

Then $\| \iota ST(x_i) - x_i\| \leq \| \iota ST(x_i) - T(x_i) \| + \| T(x_i) - x_i \| \leq \epsilon \|T(x_i)\| + \epsilon$ $\leq \epsilon^2 + \epsilon\|x_i\| + \epsilon$. So $ST : H_0 \rightarrow H$ approximates the identity in the $H_0$ norm.

  • If we want a net, we just let $(x_i)_{i=1}^n$ run through all finite subsets of $H_0$, and observe that we didn't use the bound on $T$, so as Bill Johnson suggests, we could just take $T$ to be a projection onto the span of the $(x_i)$, no condition on $H_0$ needed.

  • If $H_0$ is separable, let $(x_k)$ be a countable dense subset, and let $S_nT_n$ be chosen as above for $(x_i)_{i=1}^n$ and $\epsilon=1/n$. If $x\in H_0$ with $\|x - x_i\| \leq \epsilon$ for some $i\leq n$, then \begin{align*} & \| \iota S_nT_n(x) - x \| \leq \| \iota S_nT_n(x) - x \| \\ & = \| \iota S_nT_n(x-x_i) - T_n(x-x_i) + \iota S_n T_n(x_i) - T_n(x_i) + T_n(x) - x \| \\ &\leq \epsilon\|T_n(x-x_i)\| + \epsilon\|T_n(x_i)\| + \|T_n(x-x_i) - (x-x_i) + T_n(x_i) - x_i \| \\ &\leq \epsilon^2\lambda + \epsilon(\epsilon+\|x_i\|) + \epsilon\lambda + \epsilon \\ &\leq \epsilon^2\lambda + \epsilon(2\epsilon+\|x\|) + \epsilon\lambda + \epsilon. \end{align*} Without the BAP you seemingly cannot control $\|T(x-x_i)\|$ for example.

Remark 1: Having the "compact approximation property" doesn't seem to help. By definition, this means we can only choose $T$ to be compact not finite-rank. Then the image of the unit ball of $H_0$ under $T$ is a compact set, but I don't know how to form the equivalent of $S$. That is, how do you (linearly) distort a compact set from $H$ into $H_0$?

$\endgroup$
2
  • $\begingroup$ Thank you for the answer. I checked the details for Fréchet spaces. I can generalize your argument to show that if $H_0$ has the bounded approximation property and admits a continuous norm, then one can construct the sequence of operators $P_n$. In that case the family $(P_n)$ is also equicontinuous. This fits nicely with Jochen's comment, that existence of atomic decompositions is equivalent to the bounded approximation property. $\endgroup$ Mar 8, 2017 at 18:20
  • $\begingroup$ @MatthewDaws: That is not the definition of the AP. Moreover, if $E$ is a finite dimensional subspace of a locally convex space $X$, then there is a continuous projection from $X$ onto $E$. $\endgroup$ Mar 8, 2017 at 20:45

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.