Metrizable topological group : if $f$ is a continuous homomorphism, then $f$ is uniformly continuous 
We say that a metrizable group $G$ (topological group) satisfies condition $(L)$ when exists $c>0$ such that $d(x\cdot u, y \cdot u)\leq c \cdot d(x,y)$ for any $x,y,u\in G$. Prove that if $G,H$  satisfies condition $(L)$, then every continuous homomorphism $f:G\rightarrow H$ is uniformly continuous.

I'm actually taking a course on metric spaces and i  haven't taken any course on groups yet. So i'm not very familiar with the notion of group and of homorphism and i dunno exactly how to proceed on this question.
All i've done was to apply the definition of a continuous map at $x_0\in G$: taking any $\epsilon>0$, one can find $\delta>0$ such that $d(x_0,x)<\delta$ implies $d(f(x_0),f(x)) <\epsilon.$
I think that i must write $x_0=a\cdot u$ and $x = b \cdot u$ for some $u\in G$ and then 
$d(x_0\cdot u, x\cdot u) \leq c \cdot d(x_0,x)<c\delta$  $\implies d(f(x_0)\cdot f(u),f(x)\cdot f(u))\leq k\cdot d(f(x_0),f(x))<k \epsilon$ 
but i cannot see how to choose a suitable $u$ and to properly manipulate those constants in order to prove the result.
Any help would be much appreciated.
 A: Just two comments before we get to the answer. In the yellow box right at the start of your question, you forgot to say that the homomorphism considered in the proposition is continuous, even though you said it in the title, I think that in general more attention is given to the body of the question than to the title. The second thing is that I believe you have mistakengly written homorphism instead of homomorphism in your question.
Now here is my attempt at the proof. I think you are on the right track. A hint would be to work with the fixed element being $x_0=1$. Now, using the fact that $G$ satisfies condition $(L)$ we deduce that $d(x,y) = d(1 \cdot x, (yx^{-1}) \cdot x) \leq c d(1,yx^{-1})$, but we may as well write $d(1,yx^{-1}) = d(x\cdot x^{-1},y\cdot x^{-1}) \leq cd(x,y)$ so by making $d(x,y)$ sufficiently small, $d(1,yx^{-1})$ also becomes small.
Now, if you make the above distance small enough and use the continuity of $f$ to obtain
\begin{equation}
d(f(1),f(yx^{-1}))<\frac{\varepsilon}{k}
\end{equation}
(where $k$ denotes the constant associated with the fact that $H$ satisfies the $(L)$ condition) then by using the same trick from before:
\begin{align} d(f(x),f(y)) &= d(1 \cdot f(x),(f(y) \cdot f(x)^{-1})\cdot f(x))\\ &= d(1 \cdot f(x),f(yx^{-1}) \cdot f(x))\\&\leq kd(1,f(yx^{-1}))\\ &< k \frac{\varepsilon}{k} = \varepsilon,\end{align}
whic proves the uniform continuity. The essence of the result seems to be that the $(L)$ condition alows us to relate the distance between arbitrary points with some specific distance to a point that is fixed, and this relation is in both directions.
Just some final notes. You said you are not very familiar yet with the notion of groups and homomorphisms of groups. I did use some facts about homomorphisms in the proof above, namely:


*

*If $G,H$ are groups with identity elements $1_G,1_H$ respectively and $f:G \to H$ is a group homomorphism, then $f(1_G) = f(1_H)$;

*If $f:G \to H$ is a group homomorphism, then for every $x \in G$ we have $f(x^{-1}) = f(x)^{-1}$.
You can take a look at these results in the literature, but I do recommend trying to prove them by yourself.
