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Let $X$ and $Y$ be normed spaces and let $W$ be a subspace of $X$. Assume that $T$ is a bounded linear operator from $W$ to $Y$, that is of finite rank. Show that $T$ can be extended to a bounded linear operator $T'$ from $X$ to $Y$ such that $T'(X) = T(W)$.

I think the Hahn-Banach extension theorem is needed somewhere, but since that theorem deals with extensions of functionals, i have no idea where to go..

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Hint: Say $b_1,\dots,b_n$ is a basis for the finite-dimensional space $T(W)$. Then for every $w\in W$ there is a unique expansion $$Tw=\sum_{j=1}^n a_jb_j.$$Say $$a_j=\Lambda_jw\quad(w\in W).$$Since every linear functional on a finite-dimensional space is bounded, there exist $c$ so that $$|\Lambda_j w|\le c||Tw||.$$

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  • $\begingroup$ Thanks for your quick answer. I think i'm in the right direction now, but i'm not really sure about it. $\endgroup$ – ronalddb89 May 27 '16 at 15:05
  • $\begingroup$ Well think about it some - what I wrote is really a big hint... $\endgroup$ – David C. Ullrich May 27 '16 at 15:07
  • $\begingroup$ So $a_j = \Lambda_jw$. This is a bounded linear function to $\mathbb{F}$, so there exists an extension $\Lambda'_j : X \to \mathbb{F}$ by Hahn-Banach for every $j$. Then define $$T' \ : \ X \to T(W) \ : \ x \mapsto \sum_{j=1}^n\Lambda'_j(x)v_j.$$ This is a bounded linear operator from $X$ to $Y$, with $T'(X) = T(W)$. $\endgroup$ – ronalddb89 May 27 '16 at 16:12
  • $\begingroup$ Right. Note that $\Lambda_j$ is bounded by one of the inequalities in my answer, plus the fact that $T$ is bounded... $\endgroup$ – David C. Ullrich May 27 '16 at 16:22
  • $\begingroup$ It works. Thanks! $\endgroup$ – ronalddb89 May 27 '16 at 16:34

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