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Let $E$ be a dense linear subspace of a normed vector space $X$, and let $Y$ be a Banach space. Suppose $T_{0}\in\mathcal{L}(E,Y)$ is a bounded linear operator from $E$ to $Y$. Show that $T_{0}$ can be extended to $T\in\mathcal{L}(X,Y)$ (by continuity) without increasing its norm.

I have a dumb question: Given the Hahn-Banach theorem, what's to prove here? It seems to be the immediate consequence of that theorem. If I am wrong, please show me how to prove this. Thank you!

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    Hahn-Banach has nothing to do with the problem at hand (and one only speaks of functionALs if $Y$ is the ground field). The key words here are uniform continuity and completeness of $Y$.2012-03-14
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    @t.b. Thanks. I still need to think about this. I agree that I can not apply that theorem directly.2012-03-14
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    @t.b. $T_0$ is continuous. I just knew that it is also uniformly continuous. So for bounded linear operator, uniform continuity and continuity are equivalent?2012-03-15
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    Yes, a bounded linear operator is Lipschitz continuous by definition: $\|Tx_1 - Tx_2\|_Y \leq \|T\|\,\|x_1 - x_2\|_X$. Lipschitz continuity implies uniform continuity. The reason I phrased it the way I did is that it is a general fact that if $f_0: D \to Y$ is *uniformly* continuous where $D \subset X$ is dense in a metric space $X$ and $Y$ is a complete metric space then $f_0$ admits a *unique* extension to a (uniformly) continuous $f: X \to Y$. Applying this in the present situation you get the extension $T$ from this general fact and linearity of $T$ follows from uniqueness of the extension2012-03-15
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    @t.b Thanks a lot. I guess that general fact is base on Tietze extension theorem, which I can not use without proof. So I will probably still try azarel's idea.2012-03-15
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    Oh, no, no Tietze at all. It's exactly the same argument as the one azarel outlines. You'll see that you won't use that $T_0$ is linear when you define $T$ (or $f$ as azarel write), you'll only need that when verifying that it *is* linear... (and since nobody gave you a vote so far, here we go :))2012-03-15
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    @t.b. I figured out everything except one question: I need to use $x_n\rightarrow x\Rightarrow \lim ||x_n||=||x||$. I guess it should be OK, but I feel I have not seen such a property before.2012-03-15
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    That's a consequence of the [reverse triangle inequality](http://en.wikipedia.org/wiki/Triangle_inequality#Reverse_triangle_inequality) and the definition of $x_n \to x$: $|\|x_n\| - \|x\|| \leq \|x_n - x\| \to 0$, so $\|x_n\| \to \|x\|$.2012-03-15
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    Finally, problem solved. Thank you so much!2012-03-15

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Hahn-Banach only apply if $Y=\mathbb R$. For this particular problem you want to show that if $(x_n)$ converges to $x$ then $T_0(x_n)$ is a Cauchy sequence and then define $f(x)$ as the limit of the sequence. Finally you need to show that the map is a well-defined bounded linear function.

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    Thanks. I will try to figure out the details.2012-03-14