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

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

Does it follow that $F = E$?

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

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