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: 
The $d\left(x,y\right)=0$ proof is correct. For the triangle inequality, I would start out with the right hand side. Notice that

\begin{align\*}
\log\left(1+\left|x-z\right|\right)+\log\left(1+\left|z-y\right|\right)&=\log\left(\left(1+\left|x-z\right|\right)\left(1+\left|z-y\right|\right)\right)\\
&=\log\left(1+\left|x-z\right|+\left|z-y\right|+\left|x-z\right|\left|z-y\right|\right).
\end{align\*}

Recall that for all $a,b>0$, $\log\left(a\right)\ge\log\left(b\right)\iff a\ge b$. By the triangle inequality, $\left|x-y\right|\le\left|x-z\right|+\left|z-y\right|$, and so
\begin{align\*}
1+\left|x-y\right|&\le1+\left|x-z\right|+\left|z-y\right|\\
&\le1+\left|x-z\right|+\left|z-y\right|+\left|x-z\right|\left|z-y\right|.
\end{align\*}

Thus,
\begin{align\*}
d\left(x,y\right)=\log\left(1+\left|x-y\right|\right)&\le\log\left(1+\left|x-z\right|+\left|z-y\right|+\left|x-z\right|\left|z-y\right|\right)\\
&=\log\left(1+\left|x-z\right|\right)+\log\left(1+\left|z-y\right|\right)\\
&=d\left(x,z\right)+d\left(z,y\right).
\end{align\*}
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.