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Let $X,Y$ normed $\mathbb{R}$-vector spaces. If $X$ is a dense subset of $Y$ and there is a continuous embedding $X \hookrightarrow Y$ then there holds: $Y'=\mathcal{L}(Y,\mathbb{R}) \hookrightarrow \mathcal{L}(X,\mathbb{R}) = X'$. I have just proven this claim.

Now, is there a simple example for $Y' \not\hookrightarrow X'$ if $X$ is not dense in $Y$?

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2 Answers 2

A clue can be a taken by inspecting where exactly in your proof you use the density.

A simple example would be to take $X$ finite dimensional and $Y$ having more dimension than $X$.

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In fact, every case where $X$ is not dense in $Y$ is a counterexample; the converse of your statement is also true. It's a consequence of the Hahn-Banach theorem that if $X$ is not dense in $Y$ then there exists a nonzero continuous linear functional $f \in Y'$ with $f(X) = 0$, and so the natural map $Y' \to X'$ is not injective.

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