Let $L, M:U\to V$ be two bounded linear operators where $U$ and $V$ are normed vector spaces. If U has a spanning subset $S$ such that $Lx=Mx \forall x \in S$ , then I need to show that $Lx=Mx \forall x \in X$
Every vector in $ u \in U$ can be expressed as a linear combination of vectors in $S$. i.e.
$u = \sum_{k=1}^{k}a_is_i$ where $a_i$ are scalars and $s_i$ are in $S$
If $S$ has finitely many vectors then the result follows immediately, i.e
$Lu = L\sum_{i=1}^{k}a_is_i =\sum_{i=1}^{k}a_iLs_i=\sum_{i=1}^{k}a_iMs_i=M\sum_{i=1}^{k}a_is_i=Mu$
What if $S$ contains countably infinite number of vectors?
Can I use the above reasoning? i.e.
$Lu = L\sum_{i=1}^{\infty}a_is_i =\sum_{i=1}^{\infty}a_iLs_i=\sum_{i=1}^{\infty}a_iMs_i=M\sum_{i=1}^{\infty}a_is_i=Mu$
I am little unsure if I can switch the infinite sum and the bounded linear operators $L$ and $M$?
The only reason I can think of is that $L,M$ are bounded and linear so therefore must be continuous and the continuity of the operators may permit such switching?
Is my approach for the countable case is correct overall?
Any feedback will be appreciated.
$S $ is a spanning set of $U $ means that every element of $U $ can be expressed as a finite linear combination of elements in $S$. This is true for every spanning set of arbitrary cardinality.