Crit. for a Pt. to be in the Closure of a Subspaces of Normed Lin. Sps.

# Criterion for a Point to be in the Closure of a Subspaces of Normed Linear Spaces

Theorem 1: Let $X$ be a normed linear space and let $Y \subseteq X$ be a subspace. Then a point $x_0 \in X$ is in the closure $\overline{Y}$ if and only if whenever $\varphi \in X^*$ is such that $\varphi(y) = 0$ for all $y \in Y$ then $\varphi(x_0) = 0$. |

**Proof:**$\Rightarrow$ Suppose that $x_0 \in X$ is such that $x_0 \in \overline{Y}$. Let $\varphi \in X^*$ be such that $\varphi(y) = 0$ for all $y \in Y$. Since $x_0 \in \overline{Y}$ we have that $x_0$ is an accumulation point of $Y$. So there exists a sequence of points $(y_n)_{n=1}^{\infty} \subseteq Y$ such that $\displaystyle{\lim_{n \to \infty} y_n = x_0}$. Since $\varphi$ is continuous we have that:

\begin{align} \quad \varphi (x_0) = \varphi \left ( \lim_{n \to \infty} y_n \right ) = \varphi \left ( \lim_{n \to \infty} 0 \right ) = \varphi (0) = 0 \end{align}

- $\Leftarrow$ Now suppose that whenever $\varphi \in X^*$ is such that $\varphi(y) = 0$ for all $y \in Y$ then $\varphi(x_0) = 0$.

- Assume that $x_0 \not \in \overline{Y}$. Define a function $\varphi : \overline{Y} \oplus \mathrm{span} (x_0) \to \mathbb{C}$ for all $y + \lambda x_0 \in \overline{Y} \oplus \mathrm{span} (x_0)$ by:

\begin{align} \quad \varphi (y + \lambda x_0) = \lambda \quad (*) \end{align}

- It is easy to show that $\varphi$ is a linear functional. We now show that $\varphi$ is continuous. Observe that since $\overline{Y}$ by definition is closed that then $X \setminus \overline{Y}$ is open and by assumption, $x_0 \in X \setminus \overline{Y}$. So there exists an open ball centered at $x$ that is fully contained in $\overline{Y}$, i.e., there exists an $r > 0$ such that:

\begin{align} \quad B(x_0, r) \cap \overline{Y} = \emptyset \end{align}

- This implies that:

\begin{align} \quad \| x_0 - y \| \geq r, \quad \forall y \in \overline{Y} \end{align}

- Now for all $\lambda \in \mathbb{C}$ with $\lambda \neq 0$ we have that:

\begin{align} \quad \| y + \lambda x_0 \| = | \lambda | \biggr \| \underbrace{\frac{1}{|\lambda|} y}_{\in \overline{Y}} - x_0 \biggr \| \geq | \lambda | r \quad \Leftrightarrow \quad \frac{1}{r} \| y + \lambda x_0 \| \geq | \lambda| \quad (**) \end{align}

- From $(*)$ and $(**)$ we have that:

\begin{align} \quad | \varphi(y + \lambda x_0) | = |\lambda| \leq \frac{1}{r} \| y + \lambda x_0 \| \end{align}

- So indeed, $\varphi : Y \oplus \mathrm{span} (x_0) \to \mathbb{C}$ is continuous. By the Hahn-Banach theorem there exists a linear function $\Phi : X \to \mathbb{C}$ such that:

\begin{align} \quad \Phi(y + \lambda x_0) = \varphi (y + \lambda x_0), \quad \forall y + \lambda x_0 \in \overline{Y} \oplus \mathrm{span} (x_0) \\ \end{align}

- So $\Phi(y) = 0$ for all $y \in \overline{Y}$. However, $\Phi(x_0) = 1$. This is a contradiction. So the assumption that $x_0 \not \in \overline{Y}$ was false. Hence:

\begin{align} \quad x_0 \in \overline{Y} \quad \blacksquare \end{align}