The set of continuous linear operators is a vector subspace of the space of all linear operators.

Question

I am trying to proof that the set $L(X,Y)$ of continuous linear operators between to topological vector spaces is a vector subspace of the space of all linear operators $Hom(X,Y)$. I proved that linear operator $$T:X \longrightarrow Y$$ is continuous iff it's continuous at 0. Using this i am trying to prove that $f + h$ is continuous, where $f, h$ are continuous linear operators, but i can't find for an neighborhood of zero $W(0) \subset Y$ any open set $U(0) \subset X$ such that $$(f+h)(U(0)) \subset W(0)$$ I know that sum and scalar multiplications are continuous, but i don't understand how to prove that the topology is shift invariant and using this proof the statement.

Answer 1

If $V$ is an open set in $Y$ containing $0$ then there exist open sets $V_1$ and $V_2$ containing $0$ such that $V_1+V_2 \subseteq V$. [This is by continuity of addition]. Choose open sets $U_1$ and $U_2$ in $X$ containing $0$ such that $f(U_1) \subseteq V_1$ and $h(U_2) \subseteq V_2$. If $U=U_1\cap U_2$ then $(f+h)(U)\subseteq V$.

Answer 2

You just need to use the properties of topological vector spaces. The sum operator

$+: X^2 \rightarrow X$

is continuous, and since $f$ and $g$ are continous, their cartisian product

$(f,g): X^2 \rightarrow X^2$

is continuous, where $(f,g)(x,y) = (f(x),g(y))$. We also have that the function

$i: X \rightarrow X^2$

defined by $i(x)=(x,x)$ is continuous, so since composition of functions is continuous we get

$f+g = + \circ (f,g) \circ i$

is continous.