📂Banach Space

Proof of Lissajous's Auxiliary Lemma

Theorem ¹

In the normed space $(X , \| \cdot \| )$, let’s say $Y$ is a closed set for the subspace $Y \subsetneq X$. For all $\theta \in (0,1)$ and $y \in Y$, there exists $x_{\theta} \in X$ that satisfies $\| x_{ \theta } \| = 1$ and $\| x_{ \theta } - y \| > \theta$.

Proof

Strategy: Show that a concrete $x_{\theta}$ exists and then $\| x_{ \theta } - y \| > \theta$ holds.

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Let’s say $x_{0}$ is $x_{0 } \notin Y$ and $x_{0 } \in X$. Assuming $d=0$, there exists a sequence $\left\{ y_{n} \right\}_{n \in \mathbb{N} }$ of $Y$ that satisfies $\displaystyle \lim_{ n \to \infty } \| x_{0} - y_{n} \| = 0$. This means $x_{0} \in \overline{Y}$, but since $\overline{Y} = Y$, it leads to a contradiction, hence it must be $d> 0$.

Now, there exists $y_{0} \in Y$ that satisfies $\displaystyle 0 < \| x_{0} - y_{0} \| < {{d} \over {\theta }}$, further away than the distance $d$ between the edge of $x_{0}$ and $\overline{Y}$. Let’s call this $\theta$. If $y \in Y$, then

$$\| x_{ \theta} - y \| = \left\| {{ x_{0} - y_{0} } \over { \| x_{0} - y_{0} \| }} - y \right\| = {{1 } \over { \| x_{0} - y_{0} \| }} \left\| x_{0} - y_{0} - \| x_{0} - y_{0} \| y \right\|$$

Since $(y_{0} + \| x_{0} - y_{0} \| y) \in Y$ and $\left\| x_{0} - y_{0} - \| x_{0} - y_{0} \| y \right\| \ge d$,

$$\| x_{ \theta} - y \| \ge {{1 } \over { \| x_{0} - y_{0} \| }} d$$

Because of $\displaystyle \| x_{0} - y_{0} \| < {{d} \over {\theta }}$, the following holds.

$$\| x_{ \theta} - y \| > {{ \theta } \over {d }} d = \theta$$

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