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