Let $X$, $Y$ be two Banach spaces, and $X \hookrightarrow Y$ a continuous embedding. Let $A \subset X$ be a closed subspace. Let $\overline{A}$ be the closure of $A$ in $Y$ under the norm of $Y$. Then is it the case that $\overline{A} \cap X = A$?
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No. For example, let $X = L^\infty([0,1])$, $Y = L^1([0,1])$, with the natural inclusion map. Take $A = C([0,1])$ which is closed in $X$. But $A$ is dense in $Y$, so $\overline{A} = L^1([0,1])$, and $\overline{A} \cap X = X$.