Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X,Y,Z$ Banach spaces, $\text{dom}(S)\subset Y$, let $T:X\rightarrow Y$ be linear and continuous and let $S:\text{dom}(S)\rightarrow Z$ be linear and closed. Show that the composition $ST$ is also closed.

I think the open mapping theorem might be applicable, but I don't know weather $\text{dom}(S)\cap\text{Im}(T)$ is closed.

share|cite|improve this question

it is very straightforward. Let $(x_n) \in \mathrm{dom}(ST) = \{x \in X: Tx \in \mathrm{dom}(S)\}$, with $x_n \to x$ and $STx_n \to z$. Then, as $T$ is continuous, $Tx_n \to Tx$. Now $Tx_n \in \mathrm{dom}(S)$, and hence by closedness of $S$ and $S(Tx_n) \to z$ we have $Tx \in \mathrm{dom}(S)$ and $STx = z$. Hence $x \in \mathrm{dom}(ST)$ and $STx = z$.

So, $ST$ is closed.

share|cite|improve this answer
Why is $\text{dom}(ST)=\text{dom}(S)$? $\text{dom}(ST)\subset X$ and $\text{dom}(S)\subset Y$! – Bolek Jul 15 '12 at 9:56
Upps ... I'll correct it. – martini Jul 15 '12 at 10:03

By definition, a map as $S\circ T:T^{-1}\operatorname{dom}(S)\to Z$ is closed iff its graph $\Gamma(S\circ T)$ is closed in $X\times Z.$

Let be $\{x_n\}$ a sequence in $\operatorname{dom}(S\circ T)$ such that $x_n\to x$ and $S(Tx_n)\to y.$
Being $S$ closed by hypothesis, its graph $\Gamma(S)$ is closed in $Y\times Z$ therefore $Tx\in\operatorname{dom}(S)$ and $y=S(T(x)).$
So we get that $(x,y)=\lim_{n\to\infty}(x_n,(S\circ T)(x_n))\in\Gamma(S\circ T).$

By the arbitrariness of $(x_n)$ this means properly that $\Gamma(S\circ T)$ is closed in $X\times Z.$

share|cite|improve this answer
Doesn't the closed graph theorem state $F:A\rightarrow B$ is continuous iff it's graph is closed? – Bolek Jul 15 '12 at 10:04
Yes, for a linear map $F:A\to B$ with $A$ and $B$ Banach spaces and $\text{dom}(F)=A$ the closedness condition and the continuity condition are equivalent. – Giuseppe Jul 15 '12 at 10:12
Why is this? If $F$ is surjective this follows from the open mapping theorem, but what if $F$ is not surjective? – Bolek Jul 15 '12 at 10:15
One moment please, in your problem the closed graph and the open mapping theorems don't play any role. In my answer what we need is only the definition of closed operators between Banach spaces (cf. – Giuseppe Jul 15 '12 at 10:38
I confused closed map with closed operator! Thank you – Bolek Jul 15 '12 at 10:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.