a little question about the definition of a group representation My textbook give the following definition.

Let $G$ be any topological group. A representation of $G$ on a nonzero complex Hilbert space $V$ is a group homomorphism $\phi$ of $G$ into the group of bounded linear operators on $V$ with bounded inverses, such that the resulting map $ G\times V\to V$ is continuous. 

And it says that to have the continuity property it is enough to have that the map $g\mapsto \phi(g)v$ from $G$ to $V$ is continuous at $g=1$ for all $v\in V$ and a uniform bound for $\|\phi(g)\|$ in some neighbourhood of $1$.
I want to show that the map $G\times V\to V$ given by $(g,v)\mapsto \phi(g)v$ is continuous under these assumptions.
My attempt
Let $(g_0,v_0)\in G\times V$. Let $\epsilon>0$. We need to find an open neighbourhood $U\subseteq G\times V$ of $(g_0,v_0)$ such that $\|\phi(g)v-\phi(g_0)v_0\|<\epsilon$ for all $(g,v)\in U$. By the continuity of the map $G\to V$ by $g\mapsto \phi(g)v_0$ at $g=1$, there is an open neighbourhood $N$ of $1$ such that $\|\phi(g)v_0-v_0\|<\epsilon$ for all $g\in N$. On the other hand, for any $g\in N$ we have
\begin{align*}
\|\phi(g)v-\phi(g_0)v_0\|&=\|\phi(g_0^{-1}g)v-v_0\| \qquad \text{by the unitarity} \\&\leq \|\phi(g_0^{-1}g)v-\phi(g_0^{-1}g)v_0\|+\|\phi(g_0^{-1}g)v_0-v_0\| \\& \leq \|\phi(g_0^{-1}g)\|_{op} \|v-v_0\|+\underbrace{\|\phi(g_0^{-1}g)v_0-v_0\|}_\text{$(*)$}
\end{align*}
Here I could not find an upper bound for the term $(*)$. Could anyone help me? Thanks.
Update: I have realized that we have no unitarity assumption. So I will think on this. 
 A: You are almost there, but have confused yourself a bit with the neighbourhoods of $1$ I think.  Let me redo it using continuity along nets:
Let $(g_\alpha,v_\alpha)\in G\times V$ converge to $(g,v)$, then:
\begin{align}
\|\phi(g_\alpha)v_\alpha-\phi(g)v\|&=\|\phi(g)\left(\phi(g^{-1}g_\alpha)v_\alpha-v\right)\|\\
&≤\|\phi(g)\|\,\|\phi(g^{-1}g_\alpha)\|\,\|v_\alpha-v\|+\|\phi(g)\|\,\|\phi(g^{-1}g_\alpha)v-v\|\\
&≤\|\phi(g)\|\,\left(\|\phi(g^{-1}g_\alpha)\|\,\|v_\alpha-v\|+\|\phi(g^{-1}g_\alpha)-1\|\,\|v\|\right)
\end{align}
Now $g^{-1}g_\alpha\to1$ and $\phi$ is continous at $1$, so the term on the right vanishes and also $\|\phi(g^{-1}g_\alpha)\|\to1$. Since $v_\alpha\to v$ so $\|v_\alpha-v\|\to0$ the term on the left also vanishes.
This means $(g,v)\mapsto\phi(g)v$ is continuous.
A: While s.harp's answer is fine, we can obtain a similar proof that fits the neighborhood-style proof you're pursuing.  
Fix $(g_0,v_0)\in G\times V$ and $\varepsilon>0$.  By hypothesis, there is some $M>0$ and some neighborhood $U'$ of $1\in G$ such that $\|\phi(h)\|\leq M$ for all $h\in U'$.  Additionally, there is some neighobrhood $U''$ of $1\in G$ such that $\|\phi(h)v_0-v_0\|<\varepsilon$ for all $h\in U''$.  Put $U=U'\cap U''$.  Now if $\|v-v_0\|<\varepsilon$ and $g\in g_0U$, we have 
\begin{align}
\|\phi(g_0)v_0-\phi(g)v\|&\leq\|\phi(g_0)\|\|\phi(g_0^{-1}g)v-v_0\|\\
&\leq\|\phi(g_0)\|\left[\|\phi(g_0^{-1}g)\|\|v-v_0\|+\|\phi(g_0^{-1}g)v_0-v_0\|\right]\\
&\leq\|\phi(g_0)\|\left(M+1\right)\varepsilon.
\end{align}
So (after some rescaling), $g_0U\times B_\varepsilon(v_0)$  is our desired neighborhood of $(g_0,v_0)$.
