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I'm practicing some old exams for my Functional Analysis test and i'm stuck on the following question:

Let $X$ be any Banach space with $Y \subset X$, $Y \neq X$ and $Y$ dense in $X$. Show that the identity operator on $Y$ cannot be extended to a continuous function from $X$ to $Y$.

A hint says that the Hahn-Banach extension theorem is an essential argument for the proof.

I came up with the following proof:

Suppose that the identity operator $I_Y$ on $Y$ can be to a continuous function $f$ from $X$ to $Y$. Let $x \in X-Y$ and $\{y_n\} \subset Y$ such that $y_n \rightarrow x$; this sequence exists since $Y$ is dense in $X$. Now because of the continuity of $f$ we have $$ Y \ni f(x) = f(\lim_{n\to\infty}y_n) = \lim_{n\to\infty} f(y_n) = \lim_{n\to\infty} I_Y(y_n) = x,$$ and hence $Y=X$.

I'm a bit confused since the proof seems right but it does not use the Hahn-Banach extension. Am i missing something here?

ronalddb89
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    Your proof seems fine to me. The question is strange. – user159517 May 30 '16 at 19:37
  • Looks good to me. – copper.hat May 30 '16 at 19:40
  • Alright, thanks! Thought so already, but the hint confuses me. – ronalddb89 May 30 '16 at 19:40
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    The last limit is in Y norm, white you only know that $y_n\to x$ in $X$. – Jorkug May 31 '16 at 17:17
  • @Jorkug, could you please elaborate? – ronalddb89 May 31 '16 at 19:16
  • $\lim_{n\to\infty} I_Y(y_n) = x$ does not have to hold, because this limit is meant in $Y$, while $y_n$ were chosen to converge to $x$ in the norm of $X$. Your argument just shows that $y_n$ has different limits in $X$ and in $Y$. – Jorkug Jun 01 '16 at 12:07
  • Allright, and how do we proceed? I have no idea how to solve this then. – ronalddb89 Jun 02 '16 at 09:59
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    Now that I reread the question, it never says that $Y$ has its own norm... So I got the question wrong and your initial proof seems right. Could also prove with Hahn-Banach: let $\phi$ be a functional that separates $x\in X\backslash Y$ from $f(x)$, then $\phi\circ f -\phi\equiv 0$ on a dense set and is continuous on $X$, but isn't zero. – Jorkug Jun 02 '16 at 18:14

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