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: 
Yes, your proof that $d(x,y)=0$ implies $x=y$ makes sense. All you need is the fact that the $\log(s)$ function on $s > 0$ has only one zero, and that zero is $s=1$. Therefore, if $\log(s) = 0$, then $s$ can only be equal to $1$, hence if $1 = s = 1 + |x-y|$, then $|x-y|=0$, which implies $x=0$.

As for the triangle inequality, ask, for a general function $f(s), s > 0$, how the condition
$$
f(a+b) \leq f(a) + f(b)
$$
relates to the concavity of that function. Examine a few examples: e.g., the convex function $f(s) = s^2$, the concave function $f(s)=s^{1/3}$, the linear function $f(s) = 7s$.

And note that, for nonnegative $a, b$,
$$
\log(1 + a + b) \leq \log(1 + a + 1 + b).
$$
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\*}