I was wondering whether the following handwaving line of thought can be used to show that the IV of an Asian option is less than the IV of a vanilla option with the same strike and time to maturity:

For simplicity I'll take $r=q=0$. Furthermore I am going to assume (as pointed out in the comment by Kevin) that the asset $S\_u$ is a diffusion (no jumps).

The price of an Asian option is
$$
E\_0\left[ \left( \frac1T \int\_0^T S\_u\, du - K\right)\_+ \right].
$$
According to the intermediate/mean value property for integrals, there exists *at least one* $t \in[0,T] $ such that
$$
S\_t = \frac1T \int\_0^T S\_u\, du.
$$
Let $t^\*$ be the first such $t$. It's clear that $t^\*$ will be a random variable which is always less than or equal to $T$.

We can therefore write
\begin{align}
E\_0\left[ \left( \frac1T \int\_0^T S\_u\, du - K\right)\_+ \right] &= E\_0 \left[ \left( S\_{t^\*} - K \right)\_+ \right] \\ &\leq E\_0 \left[ \left( S\_T - K \right)\_+ \right].
\end{align}

I think this is OK, but I still have some lingering doubts as $t^\*$ is a random time.

Does anyone spot a blatant error above? Better yet, would someone be able to make make the 'proof' more rigorous (if it is correct) or point out where it is incorrect?