Is this type of matrix always positive semidefinite? Let $b\in[0,1]^n$, with $n\in\mathcal{N}^*$. Consider the symmetric matrix $A=(a_{i,j})$, defined by
$$
a_{i,j}=\begin{cases}\min\lbrace b_i,...,b_j\rbrace &\mbox{if } i<j\\
\min \lbrace b_j,...,b_i\rbrace &\mbox{if } j\leq i\end{cases}
$$
Is $A$ always positive semidefinite?
For instance, the matrix generated by the vector $[0. 1, 0.2, 0.3]$ is
$$
    \left(\begin{matrix}
    0.1 & 0.1 & 0.1 \\
    0.1 & 0.2 & 0.2 \\
    0.1 & 0.2 & 0.3 \\
    \end{matrix}\right)
$$
while the one generated by the vector  $[0. 2, 0.1, 0.3]$ is 
$$
    \left(\begin{matrix}
    0.2 & 0.1 & 0.1 \\
    0.1 & 0.1 & 0.1 \\
    0.1 & 0.1 & 0.3 \\
    \end{matrix}\right)\text{.}$$
So far, I have only managed to show that $A$ is positive semidefinite when $b$ is sorted (either in  increasing or decreasing order). The proof is by induction:
The statement clearly holds if $n=1$.
Induction step:
Let's denote $A(b)$ the matrix generated when following the procedure described above with a vector $b$.
Let $n\in\mathcal{N}^*$.  Assume that all the matrices $M\in\lbrace A(b), b\in [0,1]^{i}, i\in\lbrace 1,...n\rbrace, b \mbox{ sorted}\rbrace$ are positive semidefinite.
Consider then a sorted vector $b$ with $n+1$ elements and apply Sylvester's criterion: all the sub-matrices of $A(b)$ corresponding to principal minors belong to $\lbrace A(b), b\in [0,1]^{i}, i\in\lbrace 1,...n\rbrace, b \mbox{ sorted}\rbrace$, so the result holds.
I have also tried to generate counter-examples in the case where $b$ is not sorted, but unsuccessfully.
 A: Yes, it is positive semidefinite. This can be easily proved by mathematical induction on $n$.
To stress its size as well as its dependence on the parameters $b_i$s, let us denote the matrix generated by $(b_1,\ldots,b_n)\in[0,1]^n$ as described in your question as $A(b_1,\ldots,b_n)$. In the inductive step, let $b_i=\min(b_1,\ldots,b_n)$. Then
\begin{aligned}
A(b_1,\ldots,b_n)
&=\left[\begin{array}{c|c|c}
A(b_1,\,\ldots,\,b_{i-1})
&\begin{matrix}b_i\\ \vdots\\ b_i\end{matrix}
&\begin{matrix}b_i&\cdots&b_i\\ \vdots&&\vdots\\ b_i&\cdots&b_i\end{matrix}\\
\hline
\begin{matrix}b_i&\cdots&b_i\end{matrix}&b_i&\begin{matrix}b_i&\cdots&b_i\end{matrix}\\
\hline
\begin{matrix}b_i&\cdots&b_i\\ \vdots&&\vdots\\ b_i&\cdots&b_i\end{matrix}
&\begin{matrix}b_i\\ \vdots\\ b_i\end{matrix}
&A(b_{i+1},\,\ldots,\,b_n)
\end{array}\right]\\
&=\left[\begin{array}{c|c|c}
A(b_1-b_i,\,\ldots,b_{i-1}-b_i)&0&0\\
\hline
0&0&0\\
\hline
0&0&A(b_{i+1}-b_i,\,\ldots,\,b_n-b_i)
\end{array}\right]+b_iE
\end{aligned}
where $E$ denotes the all-one matrix. Since $b_k-b_i\in[0,1]$ for every $k$, we see that $A(b_1,\ldots,b_n)$ is positive semidefinite, by induction assumption.
A: The matrix is indeed positive semi-definite. To see this, we will show there is an alternative way of formulating this matrix as a sum of positive semi-definite matrices.
Let $b=(b_1,\ldots,b_n)\in [0,1]^n$ be the given vector. For simplicity, let's suppose all the components are distinct and positive; this is without loss of generality, as the components of $A$ are continuous and the set of positive semi-definite matrices is closed, so we can perturb the entries so they are distinct and positive, then take limits retain positive semi-definiteness. Let's also make the convention that $b_0=b_{n+1}=0$. For each $i=1,\ldots,n$, let $l(i)\leq i\leq r(i)$ denote the largest range of indices containing $i$ such that $b_i$ is minimal in this range, i.e. $$i= \arg\min_j\{b_{l(i)},b_{l(i+1)},\ldots,b_i,\ldots, b_{r(i)}\}\quad\land\quad  b_{l(i)-1},b_{r(i)+1}<b_i.$$
Also, write $\tau(i)$ for $\max\{b_{l(i)-1},b_{r(i)+1}\}$. Finally, for $1\leq i\leq j\leq n$, let $J_{i:j}$ denote the $n\times n$ symmetric matrix with all-$1$ entries in the entries whose row and column indices are in the range $[i,j]$, and $0$ everywhere else.
Then you can check that
\begin{equation}
A=\sum_{i=1}^n (b_{i}-\tau(i))J_{l(i):r(i)}.
\end{equation}
For a sketch of why this is right, suppose $p\leq q$ without loss and consider the $(p,q)$ entry of $A$. Suppose $k= \arg\min\{b_p,\ldots,b_q\}$, and let's think about which terms in the sum have nonzero entries in the $(p,q)$ entry. This happens for every $i$ such that $p,q\in [l(i),r(i)]$. This can't happen for any $i$ such that $b_k<b_i$, as $[l(i),r(i)]$ cannot contain $k$, so cannot contain both $p$ and $q$. Evidently, the term corresponding to $b_k$ is nonzero, as $b_k$ was the minimal entry in the range $[p,q]$, and contributes $b_k-\tau(k)$. Then the unique index $j$ whose value corresponds to $\tau(k)$ is also such that $p,q\in [l(j),r(j)]$, as its interval must contain that of $k$ by construction, so we also get the term $b_j-\tau(j)$, at which point we consider the index corresponding to $\tau(j)$, which in fact must be the other term considered in our construction of $\tau(j)$ i.e. $\min\{b_{l(i)-1},b_{r(i)+1}\}$. Continuing in this fashion, the sum telescopes to just give $b_k$ (I leave it to you to check the details; the way we constructed $\tau$ makes the telescoping work). This shows $A$ is really as you defined above.
The positive semi-definiteness of $A$ is now immediate, as $J_{i:j}$ is positive semi-definite for any $1\leq i\leq j\leq n$, as it can be written as $\mathbf{1}_{i:j}\mathbf{1}^T_{i:j}$, where $\mathbf{1}$ is the vector with $1$s in the entries from $i$ to $j$ and zero elsewhere. These rank-one matrices are multiplied by nonnegative numbers and then summed, so $A$ is positive semi-definite.
Just to give an example, your first matrix is written as
\begin{align}
(0.1-0) J_{1:3}+(0.2-0.1) J_{2:3}+(0.3-0.2)J_{3:3}&=0.1\begin{pmatrix}
1 & 1 & 1\\
1 & 1 & 1\\
1 & 1 & 1
\end{pmatrix}
+.1\begin{pmatrix}
0 & 0 & 0\\
0 & 1 & 1\\
0 & 1 & 1
\end{pmatrix}
+.1\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 1
\end{pmatrix}\\
&=\begin{pmatrix}
.1 & .1 & .1\\
.1 & .2 & .2\\
.1 & .2 & .3
\end{pmatrix},
\end{align}
while your second example can be written as
\begin{align}
(0.2-0.1)J_{1:1}+(0.1-0)J_{1:3}+(0.3-0.1)J_{3:3}&=
.1\begin{pmatrix}
1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix}+0.1\begin{pmatrix}
1 & 1 & 1\\
1 & 1 & 1\\
1 & 1 & 1
\end{pmatrix}
+.2\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 1
\end{pmatrix}\\
&=\begin{pmatrix}
.2 & .1 & .1\\
.1 & .1 & .1\\
.1 & .1 & .3
\end{pmatrix}
\end{align}
