Convex Pi

Claim: $\pi_L > 2$

Proof: We represent the convex polygon as $z_0, z_1 , \ldots, z_{n-1}$ where $z_i \in \mathbb{C}$ represent the vertices in that order.

Note that $D_L = \max_{i, j} |z_i - z_j|$ and that $P_L = \sum_{k = 0}^{n - 1} |z_{k+1} - z_k|$ where the sub-scripts are modulo $n$ .

Now, if $D_L = |z_r - z_l|$ , $r < l$ , the following holds : $D_L \leq \sum_{i = r}^{l -1} |z_{i+1} - z_i| \\ D_L \leq \sum_{i=l}^{r-1}|z_{i+1}-z_i|\\\implies 2D_L \leq P_L$

by the triangle inequality. Note that equality can't hold as the polygon is convex.

Claim: $\pi_L < \pi$

Proof: Now, we construct a circle with the diameter as the end points $z_r$ and $z_l$ (using the previous notations).

Note that $P_L$ is bounded above (stricly) by the perimeter of this circle, i.e., $P_L < \pi D_L$