If $X$ and $Y$ are topological vector spaces over $\mathbb R$, then a map $f:X\to Y$ is called uniformly continuous if for each neighborhood $V\subseteq Y$ of $0\in Y$, there exists a neighborhood $W\subseteq X$ of $0\in X$ such that for any $x,y\in X$ satisfying $x-y\in W$, it follows that $f(x)-f(y)\in V$.
It is not difficult to show that addition $+:X\times X\to X$ is a uniformly continuous function when $X\times X$ is endowed with the product topology. Also, for a given $\alpha\in\mathbb R$, the map $x\mapsto \alpha x$ is a uniformly continuous function from $X$ to $X$.
However, I suspect that scalar multiplication $\cdot:\mathbb R\times X\to X$ is not necessarily uniformly continuous when $\mathbb R$ is endowed with the usual topology and $\mathbb R\times X$ is endowed with the product tvs structure. Indeed, if $X=\mathbb R$ under the usual topology, then the function $(\alpha,x)\mapsto \alpha x$ from $\mathbb R^2$ to $\mathbb R$ is clearly not uniformly continuous.
Can anybody confirm this? I'm asking because several textbook references on topological vector spaces claim that both addition and multiplication are uniformly continuous functions, but I suspect uniform continuity of scalar multiplication can only be established in a restricted sense, i.e., when it is interpreted for a given scalar and not when $\mathbb R$ is considered as a tvs in its own right to be multiplied with $X$ to form a product tvs. In particular, I'm confused as to which of these two notions of “scalar multiplication” is appropriate when one is talking about the uniform continuity of such an operation.