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I came across a question about how to prove that we can extend bounded linear operators (on normed spaces) of finite rank here: Extending a bounded linear operator of finite rank.

For clarity, I will copy/paste it below:

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 $$|\Lambda_j w|\le c||Tw||.$$

The accepted solution seems to be relying on the claim that $\Lambda_j$ is a linear functional on a finite dimensional space. But the question does not assume that $W$ is finite dimensional, only that $T(W)$ is. Am I missing something or is this "proof" not correct?

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You are making a fair point, but there is no harm done: the proof is correct, but (in my humble opinion) the person that posted this has made a small mistake (a typo, actually): it should have been $$a_j=\Lambda_j(Tw),\;\;\;w\in W$$ followed by

$|\Lambda_j(Tw)|\leq c\|Tw\|\leq c\|T\|\|w\|$, so that the functionals $\Lambda_j$ are defined on $T(W)$ which is finite dimensional (thus they are immediately continuous). The key is then to consider the functionals $\Lambda_j\circ T:W\to\mathbb{F}$ which are now immediately continuous as compositions of continuous maps and extend those on the entire domain.

I will add this as a comment to the answer in the original post too.

Edit: In the comments, OP asks what can we say about $\|T\|$ and $\|T'\|$ when $T$ is onto, where $T'$ denotes the extension. If $T$ is onto then we can assume without loss of generality that $Y=\mathbb{C}^n$ for some $n\geq1$, since $T$ is of finite rank. We equip $Y$ with the maximum norm, since all norms are equivalent. Let's look at the construction:

we begin with $T:W\to\mathbb{C}^n$ and we compose with the functional $\varepsilon_i:\mathbb{C}^n\to\mathbb{C}$ acting as $(z_1,\dots,z_n)\mapsto z_i$, so $\epsilon_i\circ T:W\to\mathbb{C}$ are bounded functionals. By Hahn-Banach we extend those to functionals $\varphi_i:X\to\mathbb{C}$ in a norm-preserving fashion, i.e. $\|\varphi_i\|=\|\varepsilon_i\circ T\|$. We now define $T':X\to\mathbb{C}^n$ as $T'(x)=(\varphi_1(x),\dots,\varphi_n(x))$. This is bounded, extends $T$ and observe that $$\|T'\|=\sup_{x\neq0}\frac{\|T'(x)\|}{\|x\|}=\sup_{x\neq0}\sup_{1\leq j\leq n}\frac{|\varphi_j(x)|}{\|x\|}=\sup_j\sup_x\frac{|\varphi_j(x)|}{\|x\|}=\sup_{1\leq j\leq n}\|\varphi_j\|=\sup_{1\leq j\leq n}\|\varepsilon_j\circ T\|$$ But $\|\varepsilon_j\circ T\|\leq\|T\|$, so $\|T'\|\leq \|T\|$. On the other hand, $T'$ extends $T$, so obviously $\|T\|\leq\|T'\|$.

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    $\begingroup$ It is also possible that $\Lambda_j : W \to \Bbb{F}$ are defined using $$Tw = \sum_{i=1}^n \Lambda_i(w)b_i, \quad w \in W$$ since the coefficients are unique and depend linearly on $w$. $\endgroup$ – mechanodroid Feb 4 at 22:13
  • $\begingroup$ @mechanodroid Since we have the expression $$Tw=\sum_{i=1}^na_ib_i $$ I think it makes more sense to define $\Lambda_j:TW\to\mathbb{C}$ as $\Lambda_j(Tw)=a_j$. Of course this is well-defined since $b_i$ are linearly independent. If we choose this route, then everything goes through smoothly. Instead OP (of the answer) defines the functionals $\Lambda_j$ on $W$, something that raises questions, at least. I thus believe that OP simply got confused and forgot the $T$ before the $W$. $\endgroup$ – JustDroppedIn Feb 4 at 22:19
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    $\begingroup$ I disagree, if we assume that the post makes sense as written (which knowing David C. Ullrich is a valid assumption), then the interpretation $\Lambda_i : W \to \Bbb{C}$ would be correct. This is precisely why he wrote $|\Lambda_j(w)| \le c\|Tw\|$: $c$ is not simply $\|\Lambda_j\|$ but the norm $\|b_j^*\|$ from my answer. Furthermore, say you are right, how do you finish the proof using your $\Lambda_j$? We can extend them to $Y$ and then what? $\endgroup$ – mechanodroid Feb 4 at 22:37
  • $\begingroup$ He basically confirmed the interpretation $\Lambda_i : W \to \Bbb{C}$ in a reply to ronalddb89. On the other hand, I agree that he got confused in a comment further below saying "@Anupam Any linear functoinal on a finte-dimensional space is bounded. 𝑇(𝑊) has finite dimension by hypothesis.", but notice that it was written almost three years later! So we can forgive him for that. Okay, my forensic analysis is over. :) $\endgroup$ – mechanodroid Feb 4 at 22:39
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    $\begingroup$ @mechanodroid No problem, of course I understand your approach as well (I was reading through it at this time) and it is a didactic one too, the dual basis is a smart tool to use, it is a +1 from me. $\endgroup$ – JustDroppedIn Feb 4 at 22:51
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"every linear functional on a finite-dimensional space is bounded" does not refer to the functionals $\Lambda_i$ which are indeed defined on $W$.

Rather, let $\{b_1^*, \ldots, b_n^*\}$ be the dual basis to the basis $\{b_1 ,\ldots, b_n\}$ of $T(W)$, i.e. those are functionals on $T(W)$ such that $b_j^*(b_i) = b_{ij}$.

Now $\Lambda_i$ are defined with $$Tw = \sum_{i=1}^n \Lambda_i(w)b_i, \quad w \in W$$ and they are evidently well-defined and linear. On the other hand, using the dual basis gives $$Tw = \sum_{i=1}^n b_i^*(Tw)b_i, \quad w \in W$$ so by comparison $\Lambda_i = b^*_i \circ T$. Now $b^*_i$ are bounded since $T(W)$ is finite-dimensional so $$|\Lambda_i(w)| = |b_i^*(Tw)| \le \|b^*_i\|\|Tw\| \le \|b^*_i\|\|T\|\|w\|.$$

Now extend the bounded functionals $\Lambda_i$ to bounded functionals on $X$...

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