Prove that a bounded operator $T:X\to Y$ has closed ker, and can have nonclosed image

Question

Related to this question: Kernel of a bounded linear operator on a normed linear space need not be closed or open?

Let $X, Y$ be normed spaces. Define a linear operator $T: X\to Y$. I try to prove that

(1)If T is bounded, then $ker(T)$ is closed.

(2) There are such normed spaces $X$ and $Y$,and $T$ is bounded so that the image of $T$, denoted by $Im(T)$ is not closed.

My proof of (1) is as follows. But I am stuck on (2).

Take a sequence $\{x_n\}\subset Ker(T)$. Suppose that $x_n\to x\in X$ as $n\to \infty$ in $X$. Since $T$ is bounded, then there exists some constants so that $\|T\|\le M$. Then $$ \|Tx_n-Tx\|\le \|T\|\|x_n-x\|\le M\|x_n-x\| $$

As $n\to \infty$, we have $$ 0\le \|Tx\|=\|Tx_n-Tx\|\to 0 $$

Then $x\in Ker(T)$. Thus $ker(T)$ is closed.

For (2), I consider an example of an identical map from $c_{00}$ to $c_0$ equipped with sup norm: $$ T: c_{00}\to c_0 $$

Since $\|T\{x_k\}\|_\infty\le \|\{x_k\}\|_{\infty}$, then for $x=\{x_k\}$, $$ \|T\|=\sup_{\|x\|\le 1} \|Tx\|\le \|x\|\le 1 $$

Also, the image of $c_{00}$ is just $c_{00}$ which is not closed in $c_0$. Because $c_{00}$ is not complete and by the result "Let $X$ be a Banach space. The linear subspace $Y$ of $X$ is complete iff $Y$ is closed in $X$". So $Im(T)=c_{00}$ is not closed in $c_0$.

For (2): Do you know of an example of a subspace $Y_0 \subseteq Y$ such that $Y_0$ is not closed? — Brian Moehring, Oct 09 '23 at 23:51
@BrianMoehring Yes. For example, $c_{00}$ is not a closed subspace of $c_0$. — Hermi, Oct 09 '23 at 23:59
https://math.stackexchange.com/questions/1630556/examples-of-bounded-linear-operators-with-range-not-closed — Jose Avilez, Oct 13 '23 at 05:09

score 1 · Accepted Answer · answered Oct 13 '23 at 05:04

1

Your proof of (1) looks correct.

For (2) let $X$ be any nonclosed subspace of a normed space $Y$. For instanse, we can pick $Y=\ell^2$ and $X=\ell^2_f$. Then the identity embedding $T$ of $X$ into $Y$ is a linear operator with norm $1$, but the image of $X$ is not closed in $Y$.

answered Oct 13 '23 at 05:04

Alex Ravsky

106,166

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Thank you! Can you also check my example in (2)? – Hermi Oct 13 '23 at 05:09
@Hermi Which example do you mean? – Alex Ravsky Oct 13 '23 at 05:10
I just update my answer. – Hermi Oct 13 '23 at 05:11
I consider an example of an identical map from $c_{00}$ to $c_0$ equipped with sup norm: $$ T: c_{00}\to c_0 $$ – Hermi Oct 13 '23 at 05:14
@Hermi As far as I remember, $c_0$ is the space of all sequences convergent to zero endowed the supremum norm and $c_{00}$ is its subspace consisting of sequences which are eventually zero. Then your proof is correct. – Alex Ravsky Oct 13 '23 at 05:19
Yes! That is the definition of these two spaces – Hermi Oct 13 '23 at 05:23
what is $\ell_f^2$? – PNDas Jan 29 '25 at 08:07
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@PNDas The subset of $\ell^2$ consisting of sequences which are eventually zero. – Alex Ravsky Jan 29 '25 at 08:19

Prove that a bounded operator $T:X\to Y$ has closed ker, and can have nonclosed image

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