On Wikipedia I found that the following is a corollary of Hahn Banach:
Let $X$ be a normed vector space. $M \subset X$ a linear subspace and $x_0 \in X$ such that $d:=\inf\limits_{y\in M}||x_0-y|| > 0$. Then there exists a continuous functional $f$ such that $f(x_0) = d$, $||f|| = 1$ and $f(y) = 0 \quad \forall y \in M$.
I can consider the projection form $\pi : X \to X/\bar{M}$, so $||\pi||$ with the norm of the quotient space should have all this properties.
So is this really a corollary of Hahn Banach or am I getting something wrong?
Best Answer
This is indeed a corollary of Hahn-Banach, quite a classic one. Here's the proof:
Since $d>0$, we have that $Lin(\{x_0\}) \cap M = \{0\}$, where $Lin(\{x_0\}) =\{tx_0, \, t \in \mathbb{K}\}$. Hence, we can define space $X_1 = Lin(\{x_0\}) \oplus M$ and functional $G: X_1 \rightarrow \mathbb{K}$ with following formula: $$ G(tx_0+y)=td \qquad t\in\mathbb{K}, \ y\in M $$ $G$ is linear and, if $x \in X_1$, $x=tx_0 + y$ we have: $$ |G(x)|=|t|d \le |t| \cdot || x_0 -\big(-\frac{1}{t}\big)y|| = ||tx_0+y|| = ||x|| $$ (we have plugged $(-\frac{1}{t}\big)y$ in since it is in M, and by definition it's "further" from $x_0$ than $d$). This proves that G is also an element of $X_1^*$ and $||G||\le 1$.
Take some $\varepsilon > 0$. There exists $y \in M$ s.t. $||x_0 - y|| < d + \varepsilon $. Let us define: $$ z := \frac{x_0-y}{||x_0-y||} \quad \text{so that }||z||=1 $$ Because $G(x_0 - y)=d$ (by definition) and by linearity of $G$ we get that: $$ |G(z)|=\frac{|g(x_0-y)|}{||x_0-y||} > \frac{d}{d+\varepsilon} $$ Hence $||G||>\frac{d}{d+\varepsilon}$, but $\varepsilon$ has been chosen arbitraly, so going with it to 0 gives us $||G|| \ge 1$, and we get that $||G||=1$.
Now, we do use Hahn-Banach Theorem. It gives us functional $F \in X^*$ s.t. $||F||=1$ and $F(x)=G(x)$ for every $x \in X_1$. In particular, $F(y)=G(y)=0$ for each $y \in M$ and $F(x_0)=G(x_0)=d$.