**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 $f = e \in E$.