# Discontinuous linear function

Consider the set of all real-valued functions defined on the integers. This is a vector space over the reals with the obvious definition of addition and scalar multiplication. We can consider this a topological space as the set of doubly infinite real sequences with the product topology.

Is there a linear transformation from this set to itself which is not continuous?

• Yes. We can give the space a complete norm, and it's infinite-dimensional, so there exists an unbounded linear map from the space to itself. – AJY Apr 20 '17 at 4:24
• @AJY: Is it possible to construct an explicit example? I tried searching, but I am probably not using the right keywords. – Arin Chaudhuri Apr 20 '17 at 14:05
• @AJY: I fear that the space is not normable since no neighbourhood of the origin is bounded... – haemi Apr 20 '17 at 19:13
• @haemi You're right. I was going to come back to this and see if I could cook up such a map explicitly. – AJY Apr 20 '17 at 19:15
• Any linear transformation mapping $e_n$ to $n\cdot e_0$ where $e_n=(\delta_{mn})_{m\in\mathbb{Z}}$ (using the Kronecker delta notation) should do the trick, doesn't it? – haemi Apr 20 '17 at 19:27

Consider the sequences $f_n:=(\delta_{mn})_{m\in\mathbb{Z}}$. They are linearly independent, hence - by basic linear algebra - there is a (algebraic, or "Hamel-") basis $B$ of our space (calling it $X$ from now on) which includes all the $f_n$. Therefore there exists a (unique) linear mapping $T$ from $X$ in itself with $T(f_n)=|n|\cdot f_0$ and $T(g)=0$ for each $g\in B\setminus\{f_n:n\in\mathbb{Z}\}$. Now, arguing as the OP in a comment, $T$ cannot be continuous: the sequence of sequences $(y_m)_m=((z_{mn})_{n\in\mathbb{Z}})_{m\in\mathbb{N}}$ where $z_{mn}=1$ iff $|n|\leq m$, else 0 converges pointwise, hence wrt the product topology, to $(1)_{n\in\mathbb{Z}}$. Hence $(T(y_m))_{m\in\mathbb{N}}=((\sum_{|n|\leq m}|n|)\cdot e_0)_{m\in\mathbb{N}}$ must converge wrt the product topology, especially pointwise. Consider the 0-component, $(\sum_{|n|\leq m}|n|)_{m\in\mathbb{N}}$, which evidently doesn't converge.