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
Best Answer
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.