# Closed Linear Subspace

• Nov 28th 2010, 11:45 AM
featherbox
Closed Linear Subspace
Show that $\displaystyle W=\{ f \in C[0,1]: f(0)=0\}$ is a closed linear subspace of $\displaystyle (C[0,1] , \lVert \cdot \rVert_\infty)$. Where $\displaystyle \lVert f \rVert_\infty=sup_{t\in[0,1]}\abs{f(t)}$.

Please could you explain how to show something is a closed linear subspace in a more general case too?
• Nov 28th 2010, 07:21 PM
Drexel28
Quote:

Originally Posted by featherbox
Show that $\displaystyle W=\{ f \in C[0,1]: f(0)=0\}$ is a closed linear subspace of $\displaystyle (C[0,1] , \lVert \cdot \rVert_\infty)$. Where $\displaystyle \lVert f \rVert_\infty=sup_{t\in[0,1]}\abs{f(t)}$.

Please could you explain how to show something is a closed linear subspace in a more general case too?

You show that it's a linear subspace and you show it's closed with the given topology.

Spoiler:
It's evidently a subspace since if $\displaystyle f,g\in W$ then

$\displaystyle (\alpha f+\beta g)(0)=\alpha f(0)+\beta g(0)=\alpha 0+\beta 0=0$

so that $\displaystyle \alpha f+\beta g\in W$. To prove it's closed first recall that the evaluation linear functional

$\displaystyle \varphi:C[0,1]\to\mathbb{R}:f\mapsto f(0)$

is continuous since

$\displaystyle \|f-g\|_{\infty}<\varepsilon\implies |\varphi(f)-\varphi(g)|=|f(0)-g(0)|<\varepsilon$

Then, note that since $\displaystyle \mathbb{R}$ is a metric space singletons are closed and since the preimage of closed sets under continuous maps is closed it follows that $\displaystyle \ker \varphi=\varphi^{-1}(\{0\})$ is closed. Note though that $\displaystyle W=\ker\varphi$.

For more general cases it can be more difficult. Showing that it's a subspace is usually simple, and showing it's closed is just usual topology/analysis.

Ask if you have more questions.