Showing bounded linear operator has closed image

Question

I'm trying to show that given a bounded linear operator $T: X \to Y$ with $X$ and $Y$ Banach such that $T$ satisfies:

• For ever $y \in Im(T)$ there is an $x \in X$ with $T(x) = y$ and $||x|| \le M||y||$

Then $T$ has closed image.

Now I think I have a proof of this but I am not sure. If we let $y_n \to y$ be a convergent sequence in $Im(T)$ then by the condition there are $x_n$ with $U(x_n) = y_n$ and from the condition we have that these $x_n$ are Cauchy (as the $y_n$ are Cauchy). Then $X$ Banach means that the $x_n \to x$ and then by continuity of $T$ and uniqueness of limits $T(x) = y$ and hence $y \in Im(T)$

The reason I am unsure is that the the usual proof of this fact has the additional condition that $T$ is injective - but I have not used that here. Does this still apply?

Thanks

Answer 1

You shouldn't say that the $y_n$ are convergent but that they are Cauchy in the image but otherwise I think this is fine. Though you should really fill in the details showing that $T(x_n)\to y = T(x)$. You don't have injectivity but the norm condition supplants it and lets you get away without it.

Answer 2

You might know that a subspace of a Banach space is closed if and only if it is a Banach space. So, it is enough to show that $Im(T)$ is a Banach space. To show that $Im(T)$ is a Banach space I will use the following well known result.

$\textbf{Result}:$ Let $X$ be a normed space. Then absolute convergence of a series implies convergence of the series if and only if $X$ is a Banach space.

So, let $(y_n) \in Im(T)$ be a sequence such that $\sum_\limits{n=1}^{\infty}\|y_n\|<\infty$. In light of the above result it enough to show that $\sum_\limits{n=1}^{\infty}y_n$ converges in $Im(T)$. Let us now show this. By the hypothesis, for each $n$, you can find $x_n \in X$ such that $Tx_n = y_n$ and $\|x_n\|\leq M\|y_n\|$. Since $\sum_\limits{n=1}^{\infty}\|y_n\|<\infty$, it follows that $\sum_\limits{n=1}^{\infty}\|x_n\|<\infty$. Now, by the fact that $X$ is a Banach space and the result, it follows that $\sum_\limits{n=1}^{\infty}x_n$ converges in $X$. Let $\sum_\limits{n=1}^{\infty}x_n = x$. Now, by linearity and continuity of $T$ it follows that $\sum_\limits{n=1}^{\infty}y_n = T(\sum_\limits{n=1}^{\infty}x_n) = Tx \in Im(T).$ This shows that $Im(T)$ is a Banach space and in particular a closed subspace of $Y$.

Answer 3

Let $N$ be the null space of $T$. Then $N$ is closed in $X$ because $T$ is bounded. So $X/N$ is a Banach space. And $T$ becomes $\dot{T}$ on $X/N$, where $\dot{T}(x+N)=T(x)$. Clearly $\dot{T}$ is bounded because $T$ is bounded. $\dot{T}$ has inverse $\dot{T}^{-1}$ from $\mathscr{R}(T)$ to $X/N$ that is also bounded because, by assumption, $$ \|\dot{T}^{-1}y\|_{X/N} = \inf_{\{ x : Tx = y\}} \|x\|_{X} \le M\|y\|_{Y}. $$ So $\dot{T}$ is a linear topological isomorphism between $X/N$ and $\mathscr{R}(T)\subseteq Y$, which reduces this case to one you said you knew about where $\dot{T}$ is injective, and the proof is straightforward.