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Question

Suppose $T:X\rightarrow Y$ is a continuous, injective linear operator between Banach spaces. Suppose, in addition, that $T$ maps norm bounded closed sets in $X$ to closed sets in $Y$. Then the range of $T$ is closed in $Y$.

This is a problem related to one given An Invitation to Operator Theory by Abramovich and Aliprantis and I'd just like to verify my proof.

Attempt

We assume that $T$ is as above, and we shall prove that it has closed range. If $y_n = T x_n$ and $y_n\rightarrow y$, we want to show that $y=Tx$ for some $x\in X$. First, suppose $\{x_n\}_{n\geq 1}$ is unbounded. Then

$$\lim_n\, T(x_n/\|x_n\|) = \lim_n\, y_n/\|x_n\| = 0.$$

But the set $B=\{ x\in X: \| x\|=1\}$ is closed and norm-bounded, so its image under $T$ is closed. In particular, we must have $Tz=0$ for some $z\in B$. This contradicts the fact that $T$ is injective. So the sequence $\{x_n\}_{n\geq 1}$ is bounded in $X$. Since $\{x_n\}_{n\geq 1}$ is bounded, the set $$ A=\mathrm{cl} \{ x_1, x_2, \ldots, x_n, \ldots \}$$ is closed and norm bounded. Hence $T(A)$ is closed in $Y$. In particular, $y=\lim_n\, Tx_n = Tx$ for some $x\in A \subset X$. So the range of $T$ is closed.

Thanks in advance!

  • 2
    Looks good to me.2012-08-16

2 Answers 2