TITLE: Very basic question on continuous representations
QUESTION [5 upvotes]: Suppose $G$ is a topological group, i.e. 
$$G \times G \to G, (g,h) \mapsto gh ~~~~ G \to G, g \mapsto g^{-1}$$
are continuous. Suppose $K$ is a field with norm $|\cdot|$ on it and $V$ is a normed vector space over $K$. Then, the continuous, linear maps from $V$ to $V$ can be endowed with the norm topology. Let $\rho : G \to GL(V)$ be a group homomorphism (i.e. a representation). I wonder whether the following are equivalent:
(a$_0$) For each $v \in V$ fixed, $g \mapsto \rho(g)v$ is continuous
(a) $G \times V \to V, (g,v) \mapsto \rho(g)v$ is continuous
(b) $G \to GL(V), g \mapsto \rho(g)$ is continuous
(c) For every $v \in V$ fixed, the map $g \mapsto \rho(g)v$ is continuous at $g=1_G$ (the neutral element) and for every $\epsilon > 0$ there exists a neighbourhood $U$ around $1_G$ such that for every pair $g, h \in G$, $g^{-1}h \in U$ implies $|||\rho(g^{-1}h)||| < \epsilon$ where $|||\cdot|||$ is the operator norm.
This much I know/suspect:
The implication (b) $\Rightarrow$ (a) is straightforward, and (c) $\Rightarrow$ (b) one can find in Knapp, Representation Theory of Semisimple Groups. Now the question is: 
does (a) imply (c)?
Obviously, (a) implies that $g \mapsto \rho(g)v$ is completely continuous for all $v \in V$.
On $a_0$: I am pretty sure that it is not equivalent. Tried though I have, I was unable to construct a counterexample. This might be the reason: It is equivalent to (a) if $G$ is either finite or compact and $V$ is a finite-dimensional vector space over $\mathbb{R}$ or $\mathbb{C}$: Then one can endow $V$ with a scalar product such that the rep. becomes unitary, i.e. $(\rho(g)v, \rho(g)w) = (v,w)$. From this one can deduce ($a_0$) $\Rightarrow$ (a). So here, the question is:
Can someone provide a counter example s.t. ($a_0$) is satisfied but (a) is not?
best wishes,
Fabian Werner

REPLY [2 votes]: (a) does not imply (c): The second condition in (c) may be written in the following equivalent and simpler way:
$$\forall\epsilon>0 \quad \exists U\in\mathcal{N}_1 \quad \forall g\in U \quad |||\rho(g)|||<\epsilon.$$
(Here $\mathcal{N}_1$ denotes the collection of all neighborhoods of $1$ in $G$.) However, one cannot expect above to hold true since in any case $U$ will contain $1$ and $|||\rho(1)|||=|||\mathrm{id}_V|||=1.$
For almost all cases of interest, (a$_0$) and (a) are equivalent: If $G$ is a locally compact Hausdorff group and $V$ a Banach space, then (a$_0$) and (a) are indeed equivalent (see Corollary 2-2 on p. 49 of the book Fourier Analysis on Number Fields by Ramakrishnan and Valenza). This means that if there really is an example of the sort you ask, it should be quite unusual.