Extending a bounded linear operator of finite rank

Question

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..

David C. Ullrich · Accepted Answer · 2021-02-05T13:51:27.523

3

Hint: Say $b_1,\dots,b_n$ is a basis for the finite-dimensional space $T(W)$. Let $a_1,\dots,a_n:T(W)\to\Bbb F$ be the coordinate functionals: $$v=\sum_1^na_j(v)b_j\quad(v\in T(W)).$$Note that $a_j$ is bounded, since $T(W)$ has finite dimension.

So $a_j\circ T:W\to\Bbb F$ is bounded; let $\alpha_j:X\to\Bbb F$ be a bounded linear extension of $a_j\circ T$. Define $S:X\to Y$ by $$Sx=\sum_1^n \alpha_j(x)b_j.$$Then $S$ is linear and bounded, and if $w\in W$ then $$Tw=\sum a_j(Tw)b_j=\sum\alpha_j(w)b_j=Sw.$$

edited Feb 05 '21 at 13:51

answered May 27 '16 at 14:43

David C. Ullrich

89,985

Thanks for your quick answer. I think i'm in the right direction now, but i'm not really sure about it. – ronalddb89 May 27 '16 at 15:05
Well think about it some - what I wrote is really a big hint... – David C. Ullrich May 27 '16 at 15:07
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 $T'(X) = T(W)$. – ronalddb89 May 27 '16 at 16:12
Right. Note that $\Lambda_j$ is bounded by one of the inequalities in my answer, plus the fact that $T$ is bounded... – David C. Ullrich May 27 '16 at 16:22
It works. Thanks! – ronalddb89 May 27 '16 at 16:34
How $|\Lambda_j(w)|\leq c|T(w)|$? – Anupam Feb 01 '19 at 07:22
@Anupam Any linear functoinal on a finte-dimensional space is bounded. $T(W)$ has finite dimension by hypothesis. – David C. Ullrich Feb 01 '19 at 12:19
@DavidC.Ullrich $\Lambda_j$ is defined on $W$ and not on $T(W)$ ! – S. Maths May 27 '19 at 18:53
@DavidC.Ullrich I think you should edit this and say that the functionals $\Lambda_j$ are defined on $T(W)$ and $a_j=\Lambda_j(Tw)$. Everything will work fine after that and we still get $|a_j|\leq c|Tw|$. – Just dropped in Feb 04 '21 at 22:08

Extending a bounded linear operator of finite rank

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