2
$\begingroup$

Let $X,Y$ be two Banach Spaces and let $T:X\to Y$ be a bounded linear operator. Show that either $T$ is onto or else $T(X)$ is a meager set.

I assumed that $T$ is not onto .Then we need to show that $T(X)$ is meager i.e. it is a set of first category.

Though my attempt is useless,I am still giving it.Will someone please help me how to proceed here.

  • 0
    Theorem 7.15, along with proof in lemmas 7.16-7.18, of this handout: http://www.ucl.ac.uk/~ucahad0/3103_handout_7.pdf will be helpful to you. – 2017-01-07
  • 0
    Why is a linear open map surjective?@астонвіллаолофмэллбэрг – 2017-01-07
  • 0
    The answer to that question is extremely easy, but I will still refer you to a page: http://math.stackexchange.com/questions/1749555/show-that-an-open-linear-mapping-between-normed-spaces-is-surjective – 2017-01-07

1 Answers 1

2

To have a self-contained answer, I'll summarize the proof following the handout pointed out in a comment. Let $B_X$ and $B_Y$ be the open unit balls in $X$ and $Y$, respectively.

Case 1: $\overline{T(B_X)}$ has empty interior. Then it's a nowhere dense closed set, which allows us to cover $T(X)$ with countably many such sets: $T(X)\subset \bigcup_{n=1}^\infty \overline{T(nB_X)}$. Thus $T(X)$ is meager (of the 1st category).

Case 2: $\overline{T(B_X)}$ has nonempty interior. Since it's convex and symmetric about $0$, it follows that $\overline{T(B_X)}\supset rB_Y$ for some $r>0$. We will prove that $T(B_X)$ itself contains a neighborhood of $0$ in $Y$, which will imply $T$ is onto. Given $y\in B_Y$ with $\|y\|

The last part of the proof is what we do in the proof of the Open Mapping Theorem; and indeed, in the lecture notes the result about $TX$ is meager is incorporated into a stronger form of the OMT: if a bounded linear operator between Banach spaces has a nonmeager image, then it is an open map.