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.

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    $\begingroup$ 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
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    $\begingroup$ 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


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 $\sum_{k=1}^n |\xi_k(x)|\|e_k\|_1/n^2 \le c\|x\|_{M(1)}/n$.

  • $\begingroup$ 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

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.


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 $H_0$?

  • $\begingroup$ 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

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