Suppose $E=\{x=(x_1,\dots,x_n)\in \mathbb R^n_{+}:\sum x_i=1\}$, and $E^{o}=\{x\in E : x_i>0 \forall i\}$, define $d(x,y)=\max\limits_{i} \ln\frac{x_i}{y_i}-\min\limits_{j} \ln\frac{x_j}{y_j}\forall x, y\in E^o$. Could any one tell me $(E^o, d)$ is complete metric space or not? I have not seen such metric before. Thanks for helping.
I feel like I can write $e^{d(x,y)}=\frac{\max\limits_{i}\frac{x_i}{y_i}}{{\min\limits_{j}\frac{x_j}{y_j}}}$
Yes, $(E^o, d)$ is complete. To see this, consider the function
$$\phi : E^o \rightarrow \mathbb R^n, \text { with } \phi((x_1,x_2,\ldots,x_n)):=(\ln x_1,\ln x_2,\ldots,\ln x_n).$$
Assume for $x,y \in E^o$ that $d(x,y)=\delta$. Note that due to the sum condition on elements of $E$, in your distance definition the term $\max\limits_{i} \ln\frac{x_i}{y_i}$ will always be non-negative and the term $\min\limits_{j} \ln\frac{x_j}{y_j}$ will always be non-positive. That means
\begin{align*} \left\lvert\max\limits_{i} \ln\frac{x_i}{y_i}\right\rvert \le \delta \text{ and } \left\lvert\min\limits_{j} \ln\frac{x_j}{y_j}\right\rvert \le \delta \end{align*}
But that means that for any index $k$, if $x_k \ge y_k$, then
$$ 0 \le \ln x_k- \ln y_k = \ln \frac{x_k}{y_k} \le \max\limits_{i} \ln\frac{x_i}{y_i} \le \delta$$
and hence
$$\lvert \ln x_k- \ln y_k \rvert \le \delta.$$
On the other hand, if $x_k < y_k$, then
$$ 0 > \ln x_k- \ln y_k = \ln \frac{x_k}{y_k} \ge \min\limits_{j} \ln\frac{x_j}{y_j} \ge -\delta$$
and hence again.
$$\lvert \ln x_k- \ln y_k \rvert \le \delta.$$
That means
$$\lVert \phi(x,y) \rVert_\infty \le \delta.$$
So if you have a Cauchy sequence $x^{(1)},x^{(2)},\ldots$ in $(E^o, d)$, then $\phi(x^{(1)}),\phi(x^{(2)}), \ldots$ is a Cauchy sequence in $(\mathbb R^n,\lVert\cdot\rVert_\infty)$.
Since the latter space is complete, that sequence $\phi(x^{(1)}),\phi(x^{(2)}), \ldots$ converges to a limit $r=(r_1,r_2,\ldots, r_n)$.
Define
$$\psi: \mathbb R^n \rightarrow \mathbb R \text{ as } \psi((x_1,x_2,\ldots,x_n)) :=\sum_{i=1}^n e^{x_i}$$
By the sum condition on $E$, we find that
$$\forall x \in E^o: \psi(\phi(x))=1.$$
So we have in particular $\forall i=1,2,\ldots: \psi(\phi(x^{(i)}))=1$. Now $\psi$ is a continuous function (using $\lVert\cdot\rVert_\infty$ in the domain). Since we know that $\lim_{i \to \infty} \phi(x^{(i)}) = r$ in $(\mathbb R^n,\lVert\cdot\rVert_\infty)$, we can conclude that $\psi(r)=1.$
Now set $s=(e^{r_1},e^{r_2},\ldots,e^{r_n})$. By what we just proved, we know that $s \in E^o$.
It remains to show that our original Cauchy sequence $x^{(1)},x^{(2)},\ldots$ converges to $s$. Be the definition of $\phi$ and $r$, we have componentwise convergence for the logarithms:
$$\forall k=1,2,\ldots,n: \lim_{i \to \infty} \ln x^{(i)}_k = r_k.$$
Again, by $e^x$ being a continuous function, we can conclude
$$\forall k=1,2,\ldots,n: \lim_{i \to \infty} x^{(i)}_k = e^{r_k}=s_k.$$
So the considered Cauchy sequence converges componentwise to $s$.
That means for a given $\epsilon > 0$ there is for each index $k$ a number $N_k$ such that $$\forall i > N_k: \lvert x^{(i)}_k - s_k \rvert < \min (s_ke^{\frac{\epsilon}2}-s_k, s_k - s_ke^{\frac{-\epsilon}2}) \Rightarrow e^{\frac{-\epsilon}2} < \frac{x^{(i)}_k}{s_k} < e^{\frac{\epsilon}2}$$
From that we get
$$ \forall i > N_k: -\frac{\epsilon}2 < \ln \frac{x^{(i)}_k}{s_k} < \frac{\epsilon}2$$
Combining all the componentwise results we get
$$ \forall k,\, \forall i > \max(N_1,\ldots,N_n): -\frac{\epsilon}2 < \ln \frac{x^{(i)}_k}{s_k} < \frac{\epsilon}2$$
This means
$$ \forall i > \max(N_1,\ldots,N_n): \max\limits_{k}\ln \frac{x^{(i)}_k}{s_k} < \frac{\epsilon}2$$
and
$$ \forall i > \max(N_1,\ldots,N_n): \min\limits_{k}\ln \frac{x^{(i)}_k}{s_k} > -\frac{\epsilon}2$$
and finally
$$\forall i > \max(N_1,\ldots,N_n): d(x^{(i)}, s) < \epsilon.$$
This concludes the prof, as $s$ has been established as the limit of the Cauchy series $x^{(1)},x^{(2)},\ldots$ in $(E^o, d)$
