Riesz's Lemma

Theorem
Let $X$ be a normed vector space.

Let $Y$ be a proper closed linear subspace of $X$.

Let $\alpha \in \openint 0 1$.

Then there exists $x_\alpha \in X$ such that:


 * $\norm {x_\alpha} = 1$

with:


 * $\norm {x_\alpha - y} > \alpha$

for all $y \in Y$.

Proof
Since $Y < X$:


 * $X \setminus Y$ is non-empty.

Since $Y$ is closed:


 * $X \setminus Y$ is open.

Let $x \in X \setminus Y$, then there exists $\epsilon > 0$ such that:


 * $\map {B_\epsilon} x \subset X \setminus Y$

So, for all $y \in Y$, we must have:


 * $\norm {x - y} \ge \epsilon$

That is:


 * $\inf \left\{\norm {x - y} \colon y \in Y\right\} \ge \epsilon$

For brevity, let:


 * $d = \inf \left\{\norm {x - y} \colon y \in Y\right\}$

Since $\alpha^{-1} > 1$, there exists $z \in Y$ with:


 * $\norm {x - z} < d \alpha^{-1}$

otherwise the infimum would be at least $d \alpha^{-1}$, a contradiction.

Since $x \in X \setminus Y$ and $z \in Y$, we clearly have $x \ne z$.

So, we can set:


 * $\displaystyle x_\alpha = \frac {x - z} {\norm {x - z} }$

Clearly:


 * $\displaystyle \norm {x_\alpha} = \frac {\norm {x - z} } {\norm {x - z} } = 1$

Now, for any $y \in Y$ we have:

We've already seen that:


 * $\norm {x - z} < d \alpha^{-1}$

Since $Y$ is closed under linear combination, we have:


 * $z + \norm {x - z} y \in Y$

and so:


 * $\norm {x - \paren {z + \norm {x - z} y} } \ge d$

We conclude that: