Power Function on Strictly Positive Base is Convex

Theorem
Let $a \in \R$ be a real number such that $a > 1$.

Let $f : \R \to \R$ be the real function defined as:
 * $f \left({ x }\right) = a^x$

where $a^x$ denotes $a$ to the power of $x$.

Then $f$ is strictly convex.

Proof
Let $x, y \in \R$.

We first show that $a^x$ is strictly midpoint-convex.

Note that, from Power of Positive Real Number is Positive: Real Number:
 * $\forall t \in \R : a^{t} > 0$.

Hence, the final inequality in the above string of equivalences holds by Cauchy's Mean Theorem.

So $a^{x}$ is midpoint-convex.

Further, from Exponential with Arbitrary Base is Continuous: Real Power, $a^x$ is  continuous.

Thus, from Continuous Midpoint-Convex Functions are Convex: Corollary 1, $a^{x}$ is strictly convex.