I want to prove the following corollary of the Hahn-Banach theorem.
Let $X$ be a normed space. For every closed linear subspace $Y\subseteq X$ and $x\in X-Y$, there exists $x'\in X $ such that $x'|Y=0$ and $x'(x)\neq 0$. My plan was as follows. Consider the linear subspace $M:=\{Y+\lambda x\,\big|\,\lambda\in\mathbb{K}\}$ and define $f:M\to\mathbb{K}$ by $f(y+\lambda x_0)=\lambda$. Then, $f$ is clearly linear and $f|Y=0$. I could then use the normed version of the Hahn-Banach theorem to extend $f$ to $x'\in X'$ such that $x'|M=f$. However, I need boundedness of $f$ and that is where I come into problems, I want to calculate
$$
|f|_{op}=\sup\{|f(y+\lambda x_0)|\,\big|\,||y+\lambda x_0||\leq 1\}
$$
but I don't know how to show this is finite, I guess it has to do with closedness of $Y$. Any help would be appreciated.
Another option I had was projecting down to the quotient space, which is a normed space since $Y$ is closed, but I get stuck in this way as well.
Best Answer
Here's a minor modification of your first approach. Let $d:= dist(x,Y)=\inf_{y \in Y}\|x-y\|$. Since $Y$ is closed and $x \not\in Y$, we have $d>0$. Now define $M$ as you did, but let $f:M \to \Bbb{K}$ by given by $f(y + \lambda x)=\lambda d$. It's clear that $f$ is linear. To check that $f$ is bounded, suppose first that $\lambda=0$. Here $$ |f(y)|=0 \leq \|y\| $$ Now, if $\lambda \neq 0$, we have $$ |f(y+\lambda x)| = |\lambda|d=|\lambda|d \frac{\|y+\lambda x\|}{\|y+\lambda x\|} = \frac{d\|y+\lambda x\|}{\|(y/\lambda)+x\|} \leq \|y+\lambda x\| $$ where the inequality follows from definition of $d$. Thus, $f$ is indeed bounded, so you can proceed as usual to get $F: X \to \Bbb{K}$ so that $F|_{M}=f$. This gives of course that $F(y)=0$ for any $y$ in $Y$ and that $F(x)=d \neq 0$.