# Extending to closed space

• November 28th 2009, 02:40 PM
putnam120
Extending to closed space
Let $X$ be a normed vector space and $M\subseteq X$ a closed subspace. Show that $M+\mathbb{C}x$ is also closed.

I was told to try using the Hahn-Banach theorem. So I know that there exists a function $f\in X^{*}$ such that $\parallel f\parallel =1$ and $f(x)=\delta$. Where $x\in X\cap M^c$ and $\delta=\inf_{y\in M}\parallel x-y\parallel$

However, I don't see how to use this at all.
• November 29th 2009, 12:17 AM
Opalg
Quote:

Originally Posted by putnam120
Let $X$ be a normed vector space and $M\subseteq X$ a closed subspace. Show that $M+\mathbb{C}x$ is also closed.

I was told to try using the Hahn-Banach theorem. So I know that there exists a function $f\in X^{*}$ such that $\parallel f\parallel =1$ and $f(x)=\delta$. Where $x\in X\cap M^c$ and $\delta=\inf_{y\in M}\parallel x-y\parallel$

However, I don't see how to use this at all.

To show that $M+\mathbb{C}x$ is closed, suppose that you have a convergent sequence $m_k+\lambda_kx\to y$. You need to show that the limit point y is also in $M+\mathbb{C}x$. From the continuity of your function f above, $\lambda_k\delta = f(m_k+\lambda_kx)\to f(y)$ as $k\to\infty$. Deduce that $m_k\to y-\delta^{-1}f(y)x$. Since M is closed, it follows that $z = y-\delta^{-1}f(y)x\in M$, so that $y = z + \delta^{-1}f(y)x\in M+\mathbb{C}x$.
• November 29th 2009, 03:10 PM
putnam120
Thanks. In the time between our post I had figured out that I would have to show that a sequence of that form converged to a point in M. However, I was stuck at that part. (I was thinking about using the fact that the bounded linear functionals separate points somehow)