Extensions of bounded linear operators of finite rank (understanding a proof)

Question

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?

Answer 1

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 $\|T\|\leq\|T'\|$.

Answer 2

"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 $X$...