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We know Hahn-Banach theorem:

"If $V$ is a normed $K$-vector space with linear subspace $U$ (not necessarily closed) and if $f : U \to K$ is continuous and linear, then there exists an extension $g : V \to K$ of $f$ which is also continuous and linear and which has the same norm as $f$."

Does it stay true if we replace $K$ by any $K$-vector space $W$? Or is there a counter-example? To be clear: is there any example of a normed space $V$, a subspace $U \subset V$, a normed space $W$ and a continuous linear map $f : U \to W$ which has no extension $V \to W$?

(I'm trying to see if, given any normed vector spaces $V$ and $W$, it is possible to build a non-trivial continuous linear map $g : V \to W$. Is it always possible? Apparently there are sufficiently many such $g$, see here (and using Hahn-Banach theorem to get $X' \neq \{0\}$)).

Thank you very much!

Alphonse
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  • More generally, what would be the algebraic dimension of $\mathcal L((E, |\cdot|) ; (F, | \cdot |'))$ ? – Alphonse Aug 29 '16 at 14:05
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    The answer of your question in parentheses is clear: given nonzero $V$ and $W$, by the Hahn-Banach theorem you have some nonzero $f : V \to K$, then choose a nonzero vector $w \in W$ and compose $f$ with $K \to W, \lambda \mapsto \lambda w$. This doesn't answer your first question though. – Najib Idrissi Aug 29 '16 at 14:16

2 Answers2

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Hint: Consider $(C^0([0,1],\| \|_{\infty})$ the space of continuous functions on $[0,1]$ and $P([0,1])$ the space of polynomial functions on $[0,1]$. Consider $d:P([0,1]\rightarrow C([0,1])$ the derivative, it is continue. But you cannot extend it to $C([0,1])$

since Stone Weirstrass implies that a continue function is a limit of polynomial. Take the continue $f$ such that $f(x)=2x, x<1/2, f(x)=x+1/2, x\geq 1/2$, it is continue and it is the limit of the sequence of polynomials $p_n$ by Stone Weirstrass, this implies that $lim {p_n}_{\mid [0,1/2]}=2x$ and $lim_n{p_n}_{\mid [1/2,1]}=x+1/2$, if $df$ is defined, $df_{\mid[0,1/2]}=lim_nd{p_n}_{\mid [0,1/2]}=2, df_{\mid[1/2,1]}=1$ impossible since this function is not continue.

Tsemo Aristide
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Whenever a Banach space $V$ (e.g. $V=\ell^\infty$) has a non-complemented subspace $U$ (e.g. $U=c_0$) then the identity $U \to U$ does not have a continuous linear extension $V\to U$ just because this would be a projection onto $U$ (which does not exist by definition of complemented subspaces).

Jochen
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