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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.

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  • $\begingroup$ I am not familiar with topological vector spaces but I suspect that we don't have a sequential criterion on these spaces? Because in any metric space the scalar multiplication is not uniformly continuous. $\endgroup$ – user169373 Dec 28 '14 at 22:01
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You're right, as can be seen in the example you mention, multiplication $\mathbb R\times\mathbb R \to\mathbb R$. And you say this is "clearly" not uniformly continuous, so I don't know if you have a question about proving it. You can take as $V$ the interval $(-1,1)$. Each open $W\subset \mathbb R\times \mathbb R$ containing $(0,0)$ also contains $(\delta,0)$ for some $\delta>0$, but $\left(\frac1\delta+\delta\right)\cdot\frac1\delta - \frac1\delta\cdot\frac1\delta = 1\not\in (-1,1)$ even though $\left(\frac1\delta+\delta,\frac1\delta\right)-\left(\frac1\delta,\frac1\delta\right)\in W$.

You might want to consider for which subsets $A\subset \mathbb R$ the restriction $A\times X\to X$ is uniformly continuous. It is necessary and sufficient that $A$ be bounded.

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