Trace of operators in Dirac's bra-ket notation Let $H$ be a Hilbert space.
Let $p,s$ be two density operators.

How do I show that the trace of their product lies between $0$ and $1$, i.e. $0\leq$tr$(ps)\leq1$?

I know that density operators are operators $A$ where $A\geq 0$ and tr$(A)=1$.
Furthermore I know that a density operator can be written as $A=\sum_{i=1}^ne_i|f_i\rangle\langle f_i|$, where $e_i\geq 0$, $\sum_i e_i=1$, and $|f_i\rangle$ form an orthonormal basis for $H$.
My attempt was this: we write $p=\sum_i e_i|f_i\rangle\langle f_i|$ and $s=\sum_i a_i|f_i\rangle\langle f_i|$. Then
$$ps=(\sum_i e_i|f_i\rangle\langle f_i|)(\sum_i a_i|f_i\rangle\langle f_i|)=\sum_ie_ia_i|f_i\rangle\langle f_i|.$$
Then we find tr$(ps)=\sum_ie_ia_i$ tr ($|f_i\rangle\langle f_i|)=\sum_i e_ia_i\langle f_i|f_i\rangle=\sum_i e_ia_i$.
Is this a correct way of proving the question? Why must $\sum_i e_ia_i$ be between $0$ and $1$?
 A: $\newcommand{\bra}[1]{\langle{#1}\rvert}
\newcommand{\ket}[1]{\lvert{#1}\rangle}
\newcommand{\braket}[2]{\langle{#1}|{#2}\rangle}
\newcommand{\N}{\mathbb{N}}
\DeclareMathOperator{\tr}{tr}$For the sake of brevity I'll write the basis as $\{\ket{i}\}_{i\in\N}$.
If $p=\sum_{i\in\N}a_i\ket{i}\bra{i}$ and $s=\sum_{i\in\N}b_i\ket{i}\bra{i}$
then their product is (you have to use two different indices!)
\begin{equation}
ps=\sum_{i\in\N}a_i\ket{i}\bra{i}\sum_{j\in\N}b_j\ket{j}\bra{j}=
\sum_{i\in\N}\sum_{j\in\N}a_ib_j\ket{i}\braket{i}{j}\bra{j}=
\sum_{i\in\N}\sum_{j\in\N}a_ib_j\delta_{ij}\ket{i}\bra{j}=
\sum_{i\in\N}a_ib_i\ket{i}\bra{i}
\end{equation}
so its trace is
\begin{equation}
\tr(ps)=\sum_{k\in\N}\bra{k}ps\ket{k}=
\sum_{k\in\N}\sum_{i\in\N}a_ib_i\braket{k}{i}\braket{i}{k}=
\sum_{k\in\N}\sum_{i\in\N}a_ib_i\delta_{ki}\delta_{ik}=
\sum_{k\in\N}a_kb_k
\end{equation}
then, since $a_i\le 1$ and $b_i\le 1$ you have $a_ib_i\le a_i$ $\forall i\in\N$, therefore
\begin{equation}
\tr(ps)=\sum_{k\in\N}a_kb_k\le\sum_{k\in\N}a_k=\tr(p)=1.
\end{equation}
A: Very late to the party, but I'd like to point out that there is no reason for $p,s$ to be diagonal with respect to the same orthonormal basis. Assuming so makes things a bit more easy (but not too much, admittedly). The following general proof holds for infinite-dimensional (even non-separable) Hilbert spaces $H$.
We can diagonalize the density operators $p=\sum_i p_i\langle f_i,\cdot\rangle f_i$ and $s=\sum_j s_j\langle g_j,\cdot\rangle g_j$ with $p_i,s_j\in(0,1]$, $\sum_i p_i=\sum_j s_j=1$ and orthonormal systems $(f_i)_{i\in I}, (g_j)_{j\in J}$, $I,J\subseteq\mathbb N$ in $H$. Thus, we obtain
$$
\operatorname{tr}(ps)=\sum_{j}s_j\langle g_j,pg_j\rangle=\sum_{j,i}s_j p_i|\langle f_i,g_j\rangle|^2
$$
which is obviously a non-negative expression, i.e., $\operatorname{tr}(ps)\geq 0$. Furthermore, by the Cauchy-Schwarz inequality one gets $|\langle f_i,g_j\rangle|^2\leq \|f_i\|^2\|g_j\|^2=1$ for all $i\in I,j\in J$ and
$$
0\leq\operatorname{tr}(ps)\leq\sum_{j,i}s_j p_i\leq \Big(\sum_j s_j^2\Big)^{1/2}\Big(\sum_i p_i^2\Big)^{1/2}\leq \Big(\sum_j s_j\Big)^{1/2}\Big(\sum_i p_i\Big)^{1/2}=1\,.
$$
Note that $s_j^2\leq s_j$ and $p_i^2\leq p_i$ are obvious due to $s_j,p_i\in[0,1]$.
