I am trying to see that if I have $X$ a banach space and $S:X\rightarrow X$ a closed operator then it's image is closed in $X$?
We know since $X$ is closed that by the closed-graph theorem we will have that $S$ is bounded, but still I can't quite see why $S(X)$ would be closed, and I can't find a counterexample.
Any help is aprecciated.
Best Answer
Let $X=\{f\in C[0,1]: f(0)=0\}$ with the sup norm. Define $T: X \to X$ by $Tf(x)=\int_0^{x} f(t)dt$. Then $T$ is continuous and hence it is a closed operator. Its range is dense (by Weierstrass approximation). If the range is closed then it would be equal to $X$. However every function in the range is continuously differentiable so the range cannot be $X$.