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 $a, b$,
$$
\log(1 + a + b) \leq \log(1 + a + 1 + b).
$$
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.