[proofplan]
The proof is the supporting-line property of a differentiable concave function, written directly so no external result is needed. We first show that each tangent line at a support point $s_j$ lies above $\ell$ on all of $I$. Since this upper bound holds for every support point, taking the minimum over the finitely many tangent lines preserves the upper bound. Finally, the monotonicity of the exponential function transfers the logarithmic envelope inequality to the unnormalised density.
[/proofplan]
[step:Show that each tangent line at a support point dominates $\ell$]
Fix an index $j \in \{1,\dots,m\}$ and define the affine map $a_j: I \to \mathbb{R}$ by
\begin{align*}
a_j(x) := \ell(s_j)+\ell'(s_j)(x-s_j).
\end{align*}
We prove that $\ell(x) \leq a_j(x)$ for every $x \in I$.
If $x=s_j$, then $\ell(x)=a_j(x)$. Suppose first that $x>s_j$. For every $t \in (0,1)$, define $y_t := (1-t)s_j+t x$. Since $I$ is an interval, $y_t \in I$. By concavity of $\ell$,
\begin{align*}
\ell(y_t) \geq (1-t)\ell(s_j)+t\ell(x).
\end{align*}
Subtracting $\ell(s_j)$ and using $y_t-s_j=t(x-s_j)>0$ gives
\begin{align*}
\frac{\ell(y_t)-\ell(s_j)}{y_t-s_j} \geq \frac{\ell(x)-\ell(s_j)}{x-s_j}.
\end{align*}
As $t \downarrow 0$, we have $y_t \to s_j$ from the right through points of $I$, and differentiability of $\ell$ at $s_j$ relative to $I$ implies that the left-hand side converges to $\ell'(s_j)$. Hence
\begin{align*}
\ell'(s_j) \geq \frac{\ell(x)-\ell(s_j)}{x-s_j}.
\end{align*}
Multiplying by the positive number $x-s_j$ gives $\ell(x) \leq a_j(x)$.
Suppose now that $x<s_j$. For every $t \in (0,1)$, define $y_t := (1-t)s_j+t x$. Again $y_t \in I$, and concavity gives
\begin{align*}
\ell(y_t) \geq (1-t)\ell(s_j)+t\ell(x).
\end{align*}
Subtracting $\ell(s_j)$ and using $y_t-s_j=t(x-s_j)<0$ reverses the inequality when dividing:
\begin{align*}
\frac{\ell(y_t)-\ell(s_j)}{y_t-s_j} \leq \frac{\ell(x)-\ell(s_j)}{x-s_j}.
\end{align*}
Letting $t \downarrow 0$ and using differentiability at $s_j$ relative to $I$ yields
\begin{align*}
\ell'(s_j) \leq \frac{\ell(x)-\ell(s_j)}{x-s_j}.
\end{align*}
Multiplying by the negative number $x-s_j$ reverses the inequality and gives $\ell(x) \leq a_j(x)$. Therefore $\ell(x) \leq a_j(x)$ for every $x \in I$.
[guided]
Fix an index $j \in \{1,\dots,m\}$. The tangent line at $s_j$ is the affine map $a_j: I \to \mathbb{R}$ defined by
\begin{align*}
a_j(x) := \ell(s_j)+\ell'(s_j)(x-s_j).
\end{align*}
We want to prove that this affine map lies above the graph of $\ell$ on the whole interval $I$.
The point $x=s_j$ is immediate because
\begin{align*}
a_j(s_j)=\ell(s_j)+\ell'(s_j)(s_j-s_j)=\ell(s_j).
\end{align*}
Now take $x>s_j$. The idea is to compare the secant slope from $s_j$ to $x$ with slopes from $s_j$ to nearby points between $s_j$ and $x$. For $t \in (0,1)$, define
\begin{align*}
y_t := (1-t)s_j+t x.
\end{align*}
Because $I$ is an interval and both $s_j$ and $x$ lie in $I$, the convex combination $y_t$ lies in $I$. Concavity of $\ell$ gives
\begin{align*}
\ell(y_t) \geq (1-t)\ell(s_j)+t\ell(x).
\end{align*}
Subtracting $\ell(s_j)$ from both sides gives
\begin{align*}
\ell(y_t)-\ell(s_j) \geq t(\ell(x)-\ell(s_j)).
\end{align*}
Since $y_t-s_j=t(x-s_j)$ and $x-s_j>0$, division by $y_t-s_j$ preserves the inequality:
\begin{align*}
\frac{\ell(y_t)-\ell(s_j)}{y_t-s_j} \geq \frac{\ell(x)-\ell(s_j)}{x-s_j}.
\end{align*}
As $t \downarrow 0$, the point $y_t$ approaches $s_j$ from the right through points of $I$, and differentiability of $\ell$ at $s_j$ relative to $I$ implies that these difference quotients converge to $\ell'(s_j)$. Therefore
\begin{align*}
\ell'(s_j) \geq \frac{\ell(x)-\ell(s_j)}{x-s_j}.
\end{align*}
Multiplying by $x-s_j>0$ gives
\begin{align*}
\ell(s_j)+\ell'(s_j)(x-s_j) \geq \ell(x).
\end{align*}
Thus $\ell(x) \leq a_j(x)$ for all $x>s_j$.
It remains to treat $x<s_j$. The same interpolation point
\begin{align*}
y_t := (1-t)s_j+t x
\end{align*}
still lies in $I$, and concavity again gives
\begin{align*}
\ell(y_t) \geq (1-t)\ell(s_j)+t\ell(x).
\end{align*}
After subtracting $\ell(s_j)$, we have
\begin{align*}
\ell(y_t)-\ell(s_j) \geq t(\ell(x)-\ell(s_j)).
\end{align*}
Now $y_t-s_j=t(x-s_j)<0$, so dividing by $y_t-s_j$ reverses the inequality:
\begin{align*}
\frac{\ell(y_t)-\ell(s_j)}{y_t-s_j} \leq \frac{\ell(x)-\ell(s_j)}{x-s_j}.
\end{align*}
Letting $t \downarrow 0$ and using differentiability at $s_j$ relative to $I$ gives
\begin{align*}
\ell'(s_j) \leq \frac{\ell(x)-\ell(s_j)}{x-s_j}.
\end{align*}
Multiplying by $x-s_j<0$ reverses the inequality one more time:
\begin{align*}
\ell(s_j)+\ell'(s_j)(x-s_j) \geq \ell(x).
\end{align*}
Hence $\ell(x) \leq a_j(x)$ for all $x<s_j$ as well. Combining the three cases $x=s_j$, $x>s_j$, and $x<s_j$, the tangent line $a_j$ dominates $\ell$ on all of $I$.
[/guided]
[/step]
[step:Take the minimum over the tangent upper bounds]
Let $x \in I$ be arbitrary. From the previous step, for every $j \in \{1,\dots,m\}$,
\begin{align*}
\ell(x) \leq \ell(s_j)+\ell'(s_j)(x-s_j).
\end{align*}
Since the index set $\{1,\dots,m\}$ is nonempty and finite, the minimum defining $u(x)$ exists. A real number bounded above by every member of a finite family is bounded above by the minimum of that family, so
\begin{align*}
\ell(x) \leq \min_{1 \leq j \leq m}\{\ell(s_j)+\ell'(s_j)(x-s_j)\}=u(x).
\end{align*}
Because $x \in I$ was arbitrary, $\ell(x) \leq u(x)$ for every $x \in I$.
[/step]
[step:Exponentiate the logarithmic envelope inequality]
Define the functions $f: I \to (0,\infty)$ and $g: I \to (0,\infty)$ by
\begin{align*}
f(x) := \exp(\ell(x)).
\end{align*}
and
\begin{align*}
g(x) := \exp(u(x)).
\end{align*}
The exponential function $\exp: \mathbb{R} \to (0,\infty)$ is increasing. Applying it to the pointwise inequality $\ell(x) \leq u(x)$ gives
\begin{align*}
f(x)=\exp(\ell(x)) \leq \exp(u(x))=g(x)
\end{align*}
for every $x \in I$. Hence $\exp \circ u$ is a pointwise envelope for the unnormalised density $\exp \circ \ell$.
[/step]
