As Matt F. says, his answer is not optimal, but the optimal solution, for most polygons (see (!!) below) comes from a similar construction using just arcs of circumference and segments. This answer gives a finite dimensional family of curves containing the optimal curve, thus reducing the problem to minimizing some functions (with explicit expressions) from $[0,1]^n$ to $\mathbb{R}$.
We will begin by studying which "convex" curves under certain boundary conditions (the initial/final positions and speeds) minimize curvature.
Fix $(a,b)\in\mathbb{R}^2\setminus\{0\}$ with $a,b\geq0$ and some unit vector $v=(v_1,v_2)\in\mathbb{S}^1$ with $v_2>0$, and let $\Gamma$ be the set of $C^1$ curves $\gamma:[0,T_\gamma]\to\mathbb{R}^2$ parametrized by arc length such that:
$\gamma(0)=(0,0)$, $\gamma'(0)=v$, $\gamma(T_\gamma)=(a,b)$ and $\gamma'(T_\gamma)=(1,0)$.
Let $\alpha_{\gamma}(t)$ be the angle that $\gamma'(t)$ forms with the positive $x$-axis. Then $\alpha_\gamma(t)\in[0,\pi]\forall t$ and $\alpha_\gamma(t)$ is decreasing.
An example of such a curve is given by $c:[0,T_c]\to\mathbb{R}^2$ formed by a circumference arc and a horizontal segment, as in the figure below. Such a curve $c$ is unique in $\Gamma$, and exists if $\alpha(0)\geq\arccos(\frac{a}{\sqrt{a^2+b^2}})$ (we will assume that is the case).
Claim: $c$ is the only curve of $\Gamma$ with least maximal curvature.
(If curvature is not defined everywhere, we can just interpret maximum curvature of $\gamma$ as the minimum Lipschitz constant of the map $t\mapsto\gamma'(t)$)
$\textit{Proof of the claim:}$ We will suppose that $\gamma\in\Gamma$ has less maximum curvature than $c$ and prove that $\gamma=c$. Both $\alpha_\gamma(t)$ and $\alpha_c(t)$ are decreasing, and the maximum curvature of $\gamma$ being less than that of $c$ implies that $\alpha_c(t)\leq\alpha_\gamma(t)\forall t$. Note that $\gamma(t)=\left(\int_0^t\cos(\alpha_\gamma(t))dt,
\int_0^t\sin(\alpha_\gamma(t))dt\right)$ for any $\gamma\in\Gamma$.
So as $\cos(\alpha_\gamma(t))\leq\cos(\alpha_c(t))\forall t$ and $a=\int_0^{T_\gamma}\cos(\alpha_\gamma(t))dt=\int_0^{T_c}\cos(\alpha_c(t))dt$, we have $T_\gamma\geq T_c$. Now first suppose that $\alpha_c(0)<\frac{\pi}{2}$, so that $\sin(\alpha_c)(t),\sin(\alpha_\gamma)(t)>0$ for all $t$.
Then $b=\int_0^{T_c}\sin(\alpha_c(t))dt\leq\int_0^{T_c}\sin(\alpha_\gamma(t))dt
\leq\int_0^{T_\gamma}\sin(\alpha_\gamma(t))dt=b$. So all the inequalities are equalities, meaning that $\alpha(t)=\beta(t)\forall t$, that is, $c$ and $\gamma$ are the same curve.
If $\alpha(0)>\frac{\pi}{2}$, the proof I found is more complicated, here is a sketch. let $t_\gamma=\inf\{t;\alpha_\gamma(t)=\frac{\pi}{2}\},t_c=(\alpha_c)^{-1}(\frac{\pi}{2})$. Then we can prove that $\gamma(t_\gamma)-c(t_c)$ has negative $x$-coordinate and, as above, deduce that $T_c-t_c\leq T_\gamma-t_\gamma$, so, letting $y(p)$ be the second coordinate of $p\in\mathbb{R}^2$, we get $b-y(c(t_c))\leq\int_{t_c}^{T_c}\sin(\alpha_c(t))dt\leq\int_{t_\gamma}^{T_\gamma}\sin(\alpha_\gamma(t))dt=b-y(\gamma(t_\gamma))$, with equality iff $T_c-t_c=T_\gamma-t_\gamma$ $\alpha_\gamma(t_\gamma+t)=\alpha_c(t_c+t)$ for all $t>0$.
We also have that $t_c\leq t_\gamma$, so $y(c(t_c))
=\int_0^{t_c}\sin(\alpha_c(t))dt=
\int_{\frac{\pi}{2}}^{\alpha_c(0)}\min\{t;\alpha_c(t)>\alpha\}d\alpha
\leq
\int_{\frac{\pi}{2}}^{\alpha_\gamma(0)}\min\{t;\alpha_\gamma(t)>\alpha\}d\alpha$
$=\int_0^{t_\gamma}\sin(\alpha_\gamma(t))dt=y(c(t_\gamma))$, where the non trivial equalities come from Fubini (we calculate the area under the graph of the function horizontally instead of vertically). So finally, we have $b=y(c(t_c))+(b-y(c(t_c)))\leq y(\gamma(t_\gamma))+(b-y(\gamma(t_\gamma)))=b$, with equality iff $\alpha_c(t)=\alpha_\gamma(t)$ for all $t$, so $c=\gamma$. $\square$
Now back to the question: suppose that we have a convex curve $\gamma$ passing through all the points of $X$ (the points of $X$ are the vertices of a polygon $\mathcal{P}$, and maybe some additional points in edges of $\mathcal{P}$). Let $x,y$ be points of $X$ that appear consecutively in $\gamma$, with $\gamma(a)=x$, $\gamma(b)=y$ for some $a,b$. We can find some curve $f:[a,b]\to\mathbb{R}^2$ going from $x$ to $y$, with $f'(a)=\gamma'(a)$, $f'(b)=\gamma'(b)$ and such that $f$ has less curvature that $\gamma$ in [a,b].
(!!) We can only apply the claim when the sum of the two angles $\gamma$ forms with each side is $<\pi$ (indeed, in the claim the angle $\alpha(t)$ has a total variation $<\pi$). However if the sum of the angles in the claim was $\pi$, we can extend the curve by segments without increasing the curvature, and if it is $>\pi$, we can achieve arbitrarily low curvature using the same trick. From now on we will assume that these sums of angles are $<\pi$. This is always true for polygons in which the sum of any two consecutive exterior angles is $<\pi$ (this is not a very restrictive condition, as all exterior angles add up to $2\pi$), but it may not work with triangles for example.
We can summarize this in the following result:
For any convex convex curve $\gamma$ passing through the points of $X$, there is another convex curve $f$ passing through the points of $X$ such that:
- $f'(x)=\gamma'(x)\;\forall x\in X$.
- The arc of $f$ between any two consecutive points $x,y\in X$ is formed by a segment and/or an arc of circumference.
- $f$ has less maximal curvature than $\gamma$.
In fact the curve $f$ is uniquely determined by the values of $\gamma'(x)$ for each $x$ in $X$. So what we have to do is look for the values of $\gamma'(x)$ for each $x$ in $X$ which make the maximal curvature of $f$ be minimal.
We can also use this to prove that the curve with min-max curvature exists: first of all, let $\mathcal{E}$ be the set of edges of $\mathcal{P}$, and let $\mathcal{A}\subseteq\mathcal{E}$ be the subset of edges of $\mathcal{P}$ contained in $\gamma$; then $\mathcal{A}$ can only take finitely many values, and for each one we can prove that there is a curve of min-max curvature using a compactness argument (curvature depends continuously on the values $f'(x)$ for $x\in X$, except when $f'(x)$ is parallel to a segment of $\mathcal{P}$ containing $x$, and when $f'(x)$ approaches that segment the curvature gets large so we don't care).
So to minimize the maximal curvature you can, for each value of $\mathcal{A}$, write the maximal curvature of $f$ in terms of the angles $f'(x)$ forms with the sides of $\mathcal{P}$, and then the problem essentially becomes minimizing some functions with domain $[0,1]^n$. Probably dividing in cases depending on $\mathcal{A}$ of edges contained in $\gamma$ will be unavoidable, but I'm sure there is some way to make this more efficient than just checking all the subsets $\mathcal{A}$.
Edit: I decided to add the explicit expression of the radius of curvature.
Suppose that $l$ is a side of $\mathcal{P}$ with length $x$ and the convex curve $c$ formed by segments and arcs of circumference is like in the figure below, forming angles $\alpha\geq\beta$ (with $\alpha+\beta<\pi$) with the side $l$.
Then it's not difficult to check that the radius of curvature of the circumference arc is $x\frac{\sin(\beta)(1+\cos(\alpha+\beta))}{\sin(\alpha+\beta)^2}$. Note that for fixed $\alpha+\beta$, the minimum curvature is achieved for $\alpha=\beta$, and in the case $\alpha=\beta\geq0$, the curvature increases with $\alpha$.
Claim: If the points of $X$ are in a circumference and the condition in (!!) is satisfied, then $c$ achieves min-max curvature.
$\textit{Proof:}$ Let $\gamma$ be a $C^1$ convex curve containing all points of $X$ which minimizes maximal curvature. For each side $l$ let $\alpha_l=\beta_l$ be the angles that $c$ forms with $l$, and let $\alpha_l'\geq\beta_l'$ be the angles that $\gamma$ forms with $l$. Then $\sum_l\alpha_l+\beta_l=2\pi=\sum_l\alpha_l'+\beta_l'$. But for each $l$ we have $\alpha_l+\beta_l\geq\alpha_l'+\beta_l'$, because $\gamma$ minimizes curvature, so $\alpha_l+\beta_l=\alpha_l'+\beta_l'$ for all $l$. Again, as $\gamma$ minimizes curvature, we need to have $\alpha_l'=\beta_l'$ for all $l$, so $\gamma=c$ as we wanted. $\square$
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