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If $X$ is an infinite dimensional hilbert space and $X_n$ be its $n$ dimensional subspace. Let $P$ denotes the orthogonal projection on $X_n$ from $X$. Also $T$ is a bounded linear operator from $X$ to hilbert space $Y$.

Now How to prove that range of this map $TP:X \to Y$ is closed.

I know the following results

  1. For an orthogonal projection range$(P)$ is closed.

  2. the range of $T$ is closed in $Y$ if $T$ is open.

But how to prove this i don't know.

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    $\begingroup$ A finite dimensional subspace is closed. $\endgroup$
    – copper.hat
    Aug 10, 2016 at 14:54

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The range of $TP$ is a finite dimensional submanifold, so it is closed.

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  • $\begingroup$ can you please explain how? I have Range $P$ is finite dimensional but then after that? $\endgroup$
    – kapil
    Aug 10, 2016 at 15:11
  • $\begingroup$ A finite dimensional vector subspace of a Banach space is closed $\endgroup$
    – Tsemo Aristide
    Aug 10, 2016 at 15:12
  • $\begingroup$ yes i know this result. But my doubt is which property of $T$ we are using here? Thanks in advance $\endgroup$
    – kapil
    Aug 10, 2016 at 15:15
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    $\begingroup$ To see that remark that every norm on a finite dimension space are equivalent, so if you restrict the norm to the finite dimensional subspace, it is equivalent to the norm sup, so if a sequence converges, its coordinates relatively to a base are Cauchy sequence which converges. $\endgroup$
    – Tsemo Aristide
    Aug 10, 2016 at 15:15
  • $\begingroup$ If $S$ is a finite dimensional subspace, then so is $TS$. $\endgroup$
    – copper.hat
    Aug 10, 2016 at 15:47

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