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A version of the Hahn-Banach Theorem states that:

Let $Y$ be a subspace of a linear normed space $X$ and $\phi$ be a continuous linear functional on $Y$. Then, there exists a continuous linear functional $\Phi$ on $X$ such that $\Phi(y) = \phi(y)$ for all $y\in Y$ and $\|\Phi \| \leq \|\phi \|$

To beter understand this result, I tried to extend $\phi$ naively as:

Let $\Phi(x) = \phi(x)$ if $y \in Y$ and $\Phi(x) = 0$ otherwise.

What is wrong about my $\Phi$?

nan
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1 Answers1

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$\Phi$ is not linear. Take some $y\in Y$ with $\phi(y)\ne 0$ and some $x\in X\setminus Y$. Then $x+y\notin Y$. So $\Phi(x+y)=0$, but $\Phi(x)+\Phi(y)=0+\phi(y)\ne0$.

Harald Hanche-Olsen
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  • Is my $\Phi$ bounded or continuous though – nan Feb 08 '17 at 16:24
  • Not continuous! Let $x\to0$ in my example. Bounded? That is a concept usually reserved for linear functionals or operators in the functional analysis setting. It is certainly bounded as a function on the unit ball, though. – Harald Hanche-Olsen Feb 08 '17 at 16:27
  • If I consider a sequence $x_n \to 0$, how do I know that $x_n$ remains in $X - Y$? I mean, how do I know that I can approach $0$ while staying in $X-Y$? – nan Feb 08 '17 at 16:53
  • Just fix some $x\notin Y$ and take $x_n=\frac1n x$. – Harald Hanche-Olsen Feb 08 '17 at 17:09