Mathematics Prelims

July 17, 2008

Riesz’s Lemma

Filed under: Analysis,Functional Analysis — cjohnson @ 1:35 pm

If X is a normed space (of any dimension), Z is a subspace of X and Y is a closed proper subspace of Z, then for every \theta \in [0, 1] there exists a z \in Z such that \|z\| = 1 and \|z - y\| \geq \theta for every y \in Y.

Proof: Let v \in Z \setminus Y and let a = \inf_{y \in Y} \| v - y\|.  As Y is closed and v \notin Y, we have a > 0.  Now let \theta \in (0, 1) and note that there exists a y_0 \in Y such that a\leq \| v - y_0 \| \leq \frac{a}{\theta} (as \theta < 1, we have \frac{a}{\theta} > a).  Let

\displaystyle z = \frac{v - y_0}{\| v - y_0 \|}

Obviously z \in Z and \| z \| = 1.  Let y be any element of Y.  We have the following.

\displaystyle \|z-y\| = \left\| \frac{1}{\|v - y_0\|} (v - y_0) - y\right\|

\displaystyle \qquad = \frac{1}{\|v - y_0\|} \| v - y_0 - (\|v - y_0\|) y\|

\displaystyle \qquad \geq \frac{a}{\| v - y_0 \|}

\displaystyle \qquad \geq \frac{a}{a/\theta}

\displaystyle \qquad = \theta


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