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Let X be a normed space over $\mathbb{C}$ and Y be a finite-dimensional space over $\mathbb{C}$ with supremum norm. If K is a subspace of X and $f:K\rightarrow Y$ is a bounded linear operator, prove that f can be extended to a bounded linear operator F from X into Y such that $||F|| =||f||$.

For each $x \in K$, $f(x) \in Y$. Let $\lbrace e_1,\cdots , e_n \rbrace$ be a basis for Y then $f(x)=\sum_{i=1}^n a_i e_i$, then $$||f||=\sup \lbrace |f(x)|\; ; \; ||x||_X=1\rbrace=sup|\sum_{i=1}^n a_i e_i|\leq \sum_{i=1}^n a_i \sup{e_i}$$

Then I put $F=f(x)$ when $x \in K$ and $F(x)=0$ otherwise.

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  • $\begingroup$ If you define it like that it will not be linear. Take $x \in K$ and $y \notin K$. Then $x+y \notin K$. But $0=F(x+y)\neq F(x)+F(y) =F(x)$. $\endgroup$ – Maik Pickl Jan 13 '17 at 13:26
  • $\begingroup$ You can assume that $Y=\mathbb{C}^n$, then you consider projections $\pi_j:\mathbb{C}^n\to\mathbb{C}$, then you apply Hahn-Banach for each composition $\pi_j\circ f$ and finally you just combine results with inclusions $i_j:\mathbb{C}\to\mathbb{C}^n$. $\endgroup$ – freakish Jan 13 '17 at 13:27
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As freakish said in a comment, the key to solution is that the norm on $Y$ is the supremum norm, which implies $\|F\|=\max\|F_j\|$ when a linear operator $F$ is written out in components, $F=(F_1,\dots, F_n)$. So it suffices to get $F_j$ such that $\|F_j\|\le \|f_j\|$ for all $j$, but this this precisely what the Hahn-Banach theorem delivers.

The result would not be true for any other $\ell^p$ norm on $Y$ (with the exception of 2-dimensional case, when $\ell^1$ is isometric to $\ell^\infty$).

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