# On dense embedding of Banach spaces

Disclaimer: When I came up with this question yesterday, I suspected it to be trivial (trivially true or trivially false). Then it kept me awake several hours tonight... (I still hope, though, this is just due to my ignorance.)

Question. Let $$E,F$$ be Banach space and suppose that $$E$$ embeds densely and continuously into $$F$$ (so we consider $$E$$ as a subspace of $$F$$ from now on). Assume that there exists a constant $$M \in (0,\infty)$$ with the following property:

For each $$f \in F$$ we can find a sequence $$(e_n)$$ in $$E$$ that converges to $$f$$ with respect to $$\|\cdot\|_F$$ and that satisfies $$\|e_n\|_E \le M \|f\|_F$$.

Does it follow that $$E = F$$?

Remark. I first thought the answer should be yes due to some application of the open mapping theorem: clearly, it suffices to show that $$\|\cdot\|_E$$ and $$\|\cdot\|_F$$ are equivalent on $$E$$, and by the open mapping theorem this is true iff $$\|\cdot\|_F$$ is complete on $$E$$; but I wasn't able to prove that latter property.

Am I overlooking some simple argument, or a simple counterexample?

Edit. It is probably worthwhile to note the following fact:

As observed by Nate Eldredge in a (now deleted) comment, it is easy to see that the answer is "yes" if $$E$$ is reflexive: Given $$f \in F$$ and $$(e_n) \subseteq E$$ as above, we can choose a subsequence of $$(e_n)$$ that converges weakly (in $$E$$) to a vector $$e \in E$$. For each $$f' \in F'$$ this implies that $$\langle f', f\rangle = \langle f', e\rangle$$, so $$f = e \in E$$.

Great question! What you need is Sandy Grabiner's approximation lemma:

Lemma: Let $$E$$ and $$F$$ be Banach spaces, let $$T \in B(E,F)$$, and let $$M > 0$$ and $$0 < r < 1$$. Suppose that for each $$y \in [F]_1$$ there exists $$x_0 \in [E]_M$$ with $$\|y - Tx_0\| \leq r$$. Then for each $$y \in F$$ there exists $$x \in E$$ with $$\|x\| \leq \frac{M\|y\|}{1-r}$$ and $$Tx = y$$.

I think this immediately implies that $$F = E$$ in your question. You can probably prove the lemma yourself very easily (hint: geometric series), but for the sake of completeness a reference is: Theorem 3.35 of my book Measure Theory and Functional Analysis.

It's a great lemma and deserves to be better known. For instance, both the open mapping theorem and Tietze's extension theorem follow easily from it.

• Ha! Thanks a lot, you just saved my day! I thought I'd loose my mind... Apr 2, 2020 at 14:21
• @JochenGlueck: you are welcome! Apr 2, 2020 at 14:37
• Probably it was when I was a graduate student when I first saw what is called the "little open mapping theorem". It says that if $T$ is in $L(X,Y)$ and the closure of $T\mathring{B}_X(1)$ contains $\mathring{B}_Y(r)$, then $T\mathring{B}_X(1)$ contains $\mathring{B}_Y(r)$. The first step is of course the proof of what you call Sandy's approximation lemma. The open mapping theorem is an immediate consequence of the LOMT. A non linear version of this was useful when I was working on Lipschitz quotient mappings $20+$ years ago. After proving the LOMT in class, an easy HW problem is... Apr 2, 2020 at 15:35
• "Every separable Banach space is isometrically isomorphic to a quotient of $\ell_1$." Apr 2, 2020 at 15:41
• There is a version for complete metric groups of this lemma in Tougeron's book Ideaux de Fonctions Differentiables which I can't check now because of Corona. Apr 5, 2020 at 18:29

After Nik Weaver answered the question and Bill Johnson pointed out in a comment that what one needs is actually a part of the usual proof of the open mapping theorem, I thought about it once again, and I think it is worthwhile to explicitly remark that the following more general assertion is true:

Theorem. Let $$E$$, $$F$$ be Banach space and suppose that $$E$$ embeds densely and continuously into $$F$$ (so we may consider $$E$$ as a subspace of $$F$$). Suppose that for each $$f \in F$$ there exists a sequence $$(e_n)$$ in $$E$$ which converges to $$f$$ with respect to $$\|\cdot\|_F$$ and which is bounded with respect to $$\|\cdot\|_E$$. Then $$E = F$$.

Discussion. The point here is that, in constrast to the statement in the original question, we do not assume a priori that the bound $$\sup_n \|e_n\|_E$$ is uniform with respect to $$\|f\|_F$$.

Proof. Actually, this is almost exactly the proof of the open mapping theorem, but with a (very) small additional perturbation argument. The details are as follows:

Let $$j$$ denote the embedding of $$E$$ into $$F$$. Let $$B_E(e,r)$$ denote the open ball in $$E$$ with radius $$r$$ and center $$e$$, and likewise for $$F$$. It follows from the assumption that $$\bigcup_{n \in \mathbb{N}} \overline{ j B_E(0,n)}^F = F.$$ Due to Baire's theorem we can thus find an integer $$n \in \mathbb{N}$$ such that $$\overline{ j B_E(0,n)}^F$$ has non-empty interior in $$F$$, i.e. it contains a ball $$B_F(f,\varepsilon)$$.

Since $$j(E)$$ is dense in $$F$$, we can choose $$e \in E$$ such that $$j(e)$$ is closer than $$\varepsilon / 2$$ to $$f$$ (with respect to $$\|\cdot\|_F$$). (This is the "perturbation argument" mentioned above - well, "argument" is quite an exaggeration...) Hence, $$\overline{ j B_E(0,n)}^F \supseteq B_F(f,\varepsilon) \supseteq B_F(j(e), \varepsilon/2).$$ Now, we proceed again as in the proof of the open mapping theorem: We can write each $$x \in B_F(0, \varepsilon/2)$$ as $$x = (x+j(e)) - j(e)$$, and the vector in brackets is in $$B_F(j(e), \varepsilon/2)$$ and can thus be approximated (wrt $$\|\cdot\|_F$$) by a sequence $$(j(e_k))$$ with $$\|e_k\|_E < n$$. Since $$j(e_k - e)$$ approximates $$x$$ (wrt $$\|\cdot\|_F$$), we obtain $$x \in \overline{ j B_E(0,n + \|e\|)}^F$$.

We proved that the $$F$$-closure of $$j B_E(0,n + \|e\|)$$ contains $$B_F(0, \varepsilon/2)$$, so it follows from the small open mapping theorem (see Bill Johnson's comment to Nik Weaver's answer) that $$j B_E(0,n + \|e\|)$$ itself contains an open ball in $$F$$ centred at $$0$$. Hence, $$j(E) = F$$.