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Thread: Closed Linear Subspace

  1. #1
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    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?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by featherbox View Post
    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.

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