Riesz's Lemma
Riesz's Lemma
Theorem 1 (Riesz's Lemma): Let $X$ be a normed linear space and let $Y \subseteq X$ be a proper and closed linear subspace of $X$. Then for all $\epsilon$ such that $0 < \epsilon < 1$ there exists an element $x_0 \in X$ with $\| x_0 \| = 1$ such that $\| x_0 - y \| > 1 - \epsilon$ for every $y \in Y$. |
- Proof: Let $X$ be a normed linear space and let $Y$ be a proper and closed linear subspace of $X$ and let $\epsilon$ be such that $0 < \epsilon < 1$.
- Since $Y$ is a proper subset of $X$ we have that $X \setminus Y$ is nonempty. So take any $x \in X \setminus Y$. Since $Y$ is closed, $X \setminus Y$ is open. So there exists an $r > 0$ such that the open ball centered at $x$ with radius $r$ is fully contained in $X \setminus Y$, that is:
\begin{align} \quad B(x, r) \subseteq X \setminus Y \end{align}
- In other words:
\begin{align} \quad \| x - y \| \geq r, \quad \forall y \in Y \end{align}
- Now let $d$ be the infimum of $\| x - y \|$ such that $y \in Y$. Then:
\begin{align} \quad d = \inf \{ \| x - y \| : y \in Y \} \geq r > 0 \end{align}
- Observe that since $0 < \epsilon < 1$ we have that $\displaystyle{\frac{d}{1 - \epsilon} > d}$. Choose any $y^* \in Y$ such that:
\begin{align} \quad \| x - y^* \| < \frac{d}{1 - \epsilon} \quad (*) \end{align}
- We now define $x_0$ as:
\begin{align} \quad x_0 = \frac{x - y^*}{\| x - y^* \|} \end{align}
- Then clearly we have that $\displaystyle{\| x_0 \| = \frac{\| x - y^* \|}{\| x - y^* \|} = 1}$. Furthermore, for any $y \in Y$ we have that:
\begin{align} \quad x_0 - y = \frac{x - y^*}{\| x - y^* \|} - y = \frac{1}{\| x - y^* \|} (x - y^* - \| x - y^*\|y) = \frac{1}{\| x - y^*\|} [x - (y^* + \| x - y^* \|y)] \end{align}
- Observe that $y^* + \| x - y \|^* y \in Y$ since $Y$ is a linear subspace. Therefore:
\begin{align} \quad \| x - (y^* + \| x - y^* \|y \| \geq d \end{align}
- Now from $(*)$ we have that $\displaystyle{\frac{d}{\| x - y^* \|} > 1 - \epsilon}$, and so:
\begin{align} \quad \| x_0 - y \| = \frac{\| x - (y^* + \| x - y^* \|y)\|}{\| x - y^* \|} \geq \frac{d}{\| x - y^* \|} > 1 - \epsilon \end{align}
- So $x_0 \in X$ is such that $\| x_0 \| = 1$ and for all $y \in Y$, $\| x_0 - y \| > 1 - \epsilon$. $\blacksquare$