# Unit ball of Banach space remain closed in embedding

Let $Y\subset X$ and $X$ and $Y$ be Banach spaces with

$$\|y\|_X \leq C \|y\|_Y$$

for all $y\in Y$. Show that the unit ball of $Y$ is closed in $X$. In other words, I want to show that $Y$ semi-embeds into $X$.

I am sure this is elementary but I am quite stuck.

If $Y =X$ one can use the open mapping theorem to show that the embedding has a continuous inverse to conclude the result. If $Y$ is a proper subset I am not sure how to prove this. I would like to invert the embedding operator on the unit ball but it seems that the image of the embedding is not a Banach space so I cannot naively apply the bounded inverse theorem.

I can consider a sequence $x_n\to x\in X$ where $x_n \in T(B_1)$ and $T\colon Y\to X$ is the inclusion operator. From the assumptions it follows that $T$ is bounded and $x_n \in B_{C}$ (ball of size $C$) in $X$. Therefore, $x\in B_{C}$, but it is not obvious to me that $x$ is in the image of $T$ and that $T^{-1}x \in B_1$.

Thanks in advance for any insights!

This is not true. Try $Y = C([0,1])$ and $X = L^p(0,1)$ for any value $p \in [1,\infty)$.
If you add the assumption that $Y$ is reflexive, your statement becomes true: Let $\{y_n\}_{n \in \mathbb N} \subset Y$ be a sequence with $\|y_n\|_Y \le 1$ and $y_n \to x$ in $X$. Since $Y$ is reflexive, there is a subsequence (without relabeling) and $y \in Y$ with $y_n \rightharpoonup y$ in $Y$. Since the embedding is linear and continuous, it is also weakly continuous. Hence, $y_n \rightharpoonup y$ in $X$. Together with the strong convergence, this implies $y = x$. Hence, $x$ belongs to the (embedded) unit ball of $Y$.
You can also argue similarly in other, specific situations. For example, the statement is also true in $Y = L^\infty(0,1)$ and $X = L^p(0,1)$ for any $p \in [1,\infty)$: The convergence $y_n \to x$ in $L^p(0,1)$ implies pointwise a.e. convergence (along a subsequence). Hence, $x$ is bounded pointwise a.e. by $1$.