Question: > 
> Show that the sequence of functions $\{f\_n(z)=e^{-(z-n)^2}\}$ converges uniformly on the set $E=D(0,1)$.
> 
> 
> 

$|e^{-(z-n)^2}-0|=|e^{-z^2}.e^{-n^2}.e^{2nz}|\leqslant |e^{-n^2}.e^{2nz}| $

However I do not know how to proceed.

**Questions:**

Is the proof right till the last step? How should I finish the proof?

Thanks in advance!

Answer: 
You have the right idea about how to show that $d(x,y)=0$ implies $x=y$.

Hints for the triangle inequality:

* \begin{aligned}d(x,z)+d(z,y)&=\log(1+|x-z|)+\log(1+|z-y|) \\&= \log(1+|x-z|+|z-y|+|x-z||z-y|)
\end{aligned}
* The function $d': \mathbb R \times \mathbb R \to \mathbb R$ given by $d'(x,y)=|x-y|$ is a metric on $\mathbb R$.
* $\log(x)$ is nondecreasing on $[1, \infty)$.
It is obvious that $1+a+b \leq (1+a)(1+b)$ for $a, b \geq 0$. Hence $\log(1+a+b) \leq \log(1+a)+\log(1+b)$. This and the fact that $\log$ is increasing show that for any metric $d$, $d'(x,y)=\log(1+d(x,y))$ is also a metric.