Riesz's Lemma/Proof 1
Jump to navigation
Jump to search
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:
- $\inf \set {\norm {x - y} \colon y \in Y} \ge \epsilon$
For brevity, let:
- $d = \inf \set {\norm {x - y} \colon y \in Y}$
Since $\alpha^{-1} > 1$, there exists $z \in Y$ with:
- $\norm {x - z} < d \alpha^{-1}$
otherwise the infimum would be at least $d \alpha^{-1} > d$, a contradiction.
Since $x \in X \setminus Y$ and $z \in Y$, we clearly have $x \ne z$.
So, we can set:
- $ x_\alpha = \dfrac {x - z} {\norm {x - z} }$
Clearly:
- $\norm {x_\alpha} = \dfrac {\norm {x - z} } {\norm {x - z} } = 1$
Now, for any $y \in Y$ we have:
\(\ds \norm {x_\alpha - y}\) | \(=\) | \(\ds \norm {\frac {x - z} {\norm {x - z} } - y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\norm {x - \paren {z + \norm {x - z} y} } } {\norm {x - z} }\) |
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:
\(\ds \norm {x_\alpha - y}\) | \(=\) | \(\ds \frac {\norm {x - \paren {z + \norm {x - z} y} } } {\norm {x - z} }\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds \frac d {d \alpha^{-1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \alpha\) |
$\blacksquare$