Riesz's Lemma

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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:

$\norm {x_\alpha} = 1$


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

for all $y \in Y$.


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:

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


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

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:

\(\ds \norm {x_\alpha - y}\) \(=\) \(\ds \frac {\norm {x - \paren {z + \norm {x - z} y} } } {\norm {x - z} }\)
\(\ds \) \(>\) \(\ds \frac d {d \alpha^{-1} }\)
\(\ds \) \(=\) \(\ds \alpha\)


Source of Name

This entry was named for Frigyes Riesz.