[proofplan]
We prove the finiteness and convexity inequalities at the same time. Given two parameters $\lambda_0,\lambda_1\in D_X$, we normalize the positive random variables $e^{\lambda_0X}$ and $e^{\lambda_1X}$ so that each has expectation $1$. A pointwise weighted arithmetic-geometric mean inequality then bounds the normalized exponential at the interpolated parameter, and taking expectations gives the multiplicative moment generating function estimate. Taking logarithms converts that multiplicative estimate into the convexity inequality for $\Lambda_X$.
[/proofplan]
[step:Normalize the endpoint exponentials]
Fix $\lambda_0,\lambda_1\in D_X$ and $\theta\in[0,1]$. Define the interpolated parameter $\lambda_\theta\in\mathbb R$ by
\begin{align*}
\lambda_\theta=(1-\theta)\lambda_0+\theta\lambda_1.
\end{align*}
If $\theta=0$ or $\theta=1$, then $\lambda_\theta$ is one of the endpoint parameters and both desired conclusions follow immediately. Hence assume $0<\theta<1$.
Since $\lambda_0,\lambda_1\in D_X$, the constants
\begin{align*}
m_0=M_X(\lambda_0),\qquad m_1=M_X(\lambda_1)
\end{align*}
are finite. They are also strictly positive because $e^{\lambda_iX}>0$ pointwise for $i\in\{0,1\}$. Define measurable random variables $A:\Omega\to(0,\infty)$ and $B:\Omega\to(0,\infty)$ by
\begin{align*}
A(\omega)=\frac{e^{\lambda_0X(\omega)}}{m_0},
\qquad
B(\omega)=\frac{e^{\lambda_1X(\omega)}}{m_1}.
\end{align*}
Then $A$ and $B$ are integrable and satisfy
\begin{align*}
\mathbb E[A]=1,
\qquad
\mathbb E[B]=1.
\end{align*}
[/step]
[step:Apply the weighted arithmetic-geometric mean inequality pointwise]
[claim:Weighted arithmetic-geometric mean inequality]
For every $a,b>0$ and every $\theta\in(0,1)$, the Weighted Arithmetic-Geometric Mean Inequality states that
\begin{align*}
a^{1-\theta}b^\theta\le (1-\theta)a+\theta b.
\end{align*}
[/claim]
[proof]
Define $\phi:(0,\infty)\to\mathbb R$ by
\begin{align*}
\phi(r)=(1-\theta)+\theta r-r^\theta.
\end{align*}
Then
\begin{align*}
\phi'(r)=\theta-\theta r^{\theta-1}
=\theta(1-r^{\theta-1}).
\end{align*}
Since $\theta-1<0$, we have $\phi'(r)<0$ for $0<r<1$, $\phi'(1)=0$, and $\phi'(r)>0$ for $r>1$. Thus $\phi$ attains its minimum at $r=1$, where $\phi(1)=0$. Hence $\phi(r)\ge 0$ for all $r>0$.
Now set $r=b/a$. Multiplying the inequality
\begin{align*}
\left(\frac{b}{a}\right)^\theta
\le
(1-\theta)+\theta\frac{b}{a}
\end{align*}
by $a>0$ gives
\begin{align*}
a^{1-\theta}b^\theta\le (1-\theta)a+\theta b.
\end{align*}
[/proof]
Applying the claim pointwise with $a=A(\omega)$ and $b=B(\omega)$ gives
\begin{align*}
A(\omega)^{1-\theta}B(\omega)^\theta
\le
(1-\theta)A(\omega)+\theta B(\omega)
\end{align*}
for every $\omega\in\Omega$. Taking expectations and using monotonicity and linearity of expectation for integrable non-negative random variables gives
\begin{align*}
\mathbb E[A^{1-\theta}B^\theta]\le (1-\theta)\mathbb E[A]+\theta\mathbb E[B].
\end{align*}
Using $\mathbb E[A]=1$ and $\mathbb E[B]=1$, the right-hand side is
\begin{align*}
(1-\theta)\mathbb E[A]+\theta\mathbb E[B]=(1-\theta)+\theta=1.
\end{align*}
[guided]
Let $m_0=M_X(\lambda_0)$ and $m_1=M_X(\lambda_1)$. Since $\lambda_0,\lambda_1\in D_X$, both constants are finite, and they are strictly positive because $e^{\lambda_iX}>0$ pointwise for $i\in\{0,1\}$. Define measurable random variables $A:\Omega\to(0,\infty)$ and $B:\Omega\to(0,\infty)$ by
\begin{align*}
A(\omega)=\frac{e^{\lambda_0X(\omega)}}{m_0},
\qquad
B(\omega)=\frac{e^{\lambda_1X(\omega)}}{m_1}.
\end{align*}
Then $A$ and $B$ are integrable and satisfy $\mathbb E[A]=1$ and $\mathbb E[B]=1$. The normalization was chosen so that the endpoint exponentials have expectation exactly $1$, which allows a pointwise inequality to turn directly into a useful expectation estimate.
We first prove the scalar inequality being used. For $a,b>0$ and $0<\theta<1$, define the auxiliary function $\phi:(0,\infty)\to\mathbb R$ by
\begin{align*}
\phi(r)=(1-\theta)+\theta r-r^\theta.
\end{align*}
Its derivative is
\begin{align*}
\phi'(r)=\theta-\theta r^{\theta-1}
=\theta(1-r^{\theta-1}).
\end{align*}
Because $\theta-1<0$, the factor $r^{\theta-1}$ is greater than $1$ when $0<r<1$, equal to $1$ when $r=1$, and less than $1$ when $r>1$. Therefore $\phi$ decreases on $(0,1)$, increases on $(1,\infty)$, and has its global minimum at $r=1$. Since
\begin{align*}
\phi(1)=(1-\theta)+\theta-1=0,
\end{align*}
we obtain $\phi(r)\ge 0$ for every $r>0$. Substituting $r=b/a$ and multiplying by $a>0$ gives
\begin{align*}
a^{1-\theta}b^\theta\le (1-\theta)a+\theta b.
\end{align*}
Now apply this inequality at each point $\omega\in\Omega$ with
\begin{align*}
a=A(\omega),
\qquad
b=B(\omega).
\end{align*}
The random variables $A$ and $B$ are strictly positive, so the scalar inequality applies pointwise and gives
\begin{align*}
A(\omega)^{1-\theta}B(\omega)^\theta
\le
(1-\theta)A(\omega)+\theta B(\omega).
\end{align*}
The right-hand side is integrable because $A$ and $B$ are integrable. Taking expectations and using monotonicity of expectation gives
\begin{align*}
\mathbb E[A^{1-\theta}B^\theta]\le \mathbb E[(1-\theta)A+\theta B].
\end{align*}
Linearity of expectation applies because $A$ and $B$ are integrable, so
\begin{align*}
\mathbb E[(1-\theta)A+\theta B]=(1-\theta)\mathbb E[A]+\theta\mathbb E[B].
\end{align*}
Finally, the normalization gives
\begin{align*}
(1-\theta)\mathbb E[A]+\theta\mathbb E[B]=(1-\theta)+\theta=1.
\end{align*}
This is the key estimate: after normalization, the exponential corresponding to the interpolated parameter has expectation at most $1$.
[/guided]
[/step]
[step:Convert the normalized estimate into the moment generating function inequality]
By the definitions of $A$, $B$, $m_0$, and $m_1$, for every $\omega\in\Omega$,
\begin{align*}
A(\omega)^{1-\theta}B(\omega)^\theta
=
\left(\frac{e^{\lambda_0X(\omega)}}{m_0}\right)^{1-\theta}
\left(\frac{e^{\lambda_1X(\omega)}}{m_1}\right)^\theta.
\end{align*}
Combining the powers in the numerator and denominator gives
\begin{align*}
A(\omega)^{1-\theta}B(\omega)^\theta
=
\frac{e^{((1-\theta)\lambda_0+\theta\lambda_1)X(\omega)}}{m_0^{1-\theta}m_1^\theta}.
\end{align*}
Using the definition of $\lambda_\theta$, this becomes
\begin{align*}
A(\omega)^{1-\theta}B(\omega)^\theta
=
\frac{e^{\lambda_\theta X(\omega)}}{m_0^{1-\theta}m_1^\theta}.
\end{align*}
Taking expectations and using the estimate from the previous step,
\begin{align*}
\frac{M_X(\lambda_\theta)}{m_0^{1-\theta}m_1^\theta}
=
\mathbb E[A^{1-\theta}B^\theta]
\le 1.
\end{align*}
Therefore
\begin{align*}
M_X(\lambda_\theta)
\le
m_0^{1-\theta}m_1^\theta
=
M_X(\lambda_0)^{1-\theta}M_X(\lambda_1)^\theta.
\end{align*}
The right-hand side is finite, so $M_X(\lambda_\theta)<\infty$. Hence $\lambda_\theta\in D_X$.
[/step]
[step:Take logarithms to prove convexity of $\Lambda_X$]
Since $M_X(\lambda_0)$ and $M_X(\lambda_1)$ are strictly positive and finite, and since $M_X(\lambda_\theta)$ is also strictly positive and finite, the logarithm is defined at all three values. The logarithm is increasing on $(0,\infty)$, so the moment generating function estimate gives
\begin{align*}
\Lambda_X(\lambda_\theta)=\log M_X(\lambda_\theta)\le \log\left(M_X(\lambda_0)^{1-\theta}M_X(\lambda_1)^\theta\right).
\end{align*}
Using the logarithm product and power identities for positive [real numbers](/page/Real%20Numbers),
\begin{align*}
\log\left(M_X(\lambda_0)^{1-\theta}M_X(\lambda_1)^\theta\right)=(1-\theta)\log M_X(\lambda_0)+\theta\log M_X(\lambda_1).
\end{align*}
By the definition $\Lambda_X(\lambda)=\log M_X(\lambda)$ on $D_X$, this becomes
\begin{align*}
\Lambda_X(\lambda_\theta)\le (1-\theta)\Lambda_X(\lambda_0)+\theta\Lambda_X(\lambda_1).
\end{align*}
Because $\lambda_0,\lambda_1\in D_X$ and $\theta\in[0,1]$ were arbitrary, $D_X$ is convex and $\Lambda_X$ is convex on $D_X$.
[/step]
