Cut locus, conjugate points and smoothness of distance function

I had a couple of related questions on the cut locus, conjugate points and smoothness of distance function. Let $(M,g)$ be a smooth complete Riemannian manifold and $r(x) = d(p,x)$ the distance function to a fixed point $p\in M$. Recall that the cut locus $Cut(p)$ consists of two kinds of points - points that are conjugate to $p$ along some geodesic, or points that have multiple minimal geodesics connecting them to $p$.

1) I know that $r$ is smooth on $M\backslash (Cut(p)\cup \{p\}$. Is it true that $r$ is in fact smooth at $x$ as long as there is a unique minimal geodesic connecting it to $p$? The only proof of smoothness involves the exponential map, but the exponential map is not invertible at conjugate points, so that approach wont work.

2) Is there a simple example of $(M,g,p)$ with points conjugate to $p$ but with unique minimal geodesics to $p$.

1. The distance function has to be differentiable at $$x$$ if there is a unique minimizing geodesic $$[p,x]$$. It means that the differential $$d_xr$$ is well defined linear function on the tangent space $$\mathrm{T}_x$$. The proof is straightforward. But, it does not mean that $$r$$ is smooth in a neighborhood of $$x$$ --- evidently it does not hold if $$x$$ is conjugate to $$p$$ along the geodesic.
2. Take a surface of revolution for an even function $$f$$; say $$f(0)=1$$, $$f''<0$$. You can make it so that the curvature on the equator $$(f(0)\cdot \sin t,f(0)\cdot\cos t,0)$$ is positive and it takes maximal value on the surface. Then the maximal minimizing arc on the equator is unique minimal geodesics between its ends.
• I'm sorry to disturb six years later, but I really don't see why 1) is true. That there is a unique minimizing geodesic from $p$ to $x$ does not mean the same is true in a neighborhood of $x$, so we can't apply the first variation formula. So how do we prove the distance function is differentiable? Feb 9, 2020 at 10:32
• @Colescu If the function is not differentiable at $x$, then there is a sequence $x_n\to x$ such that $\measuredangle[x_p^{x_n}]+\measuredangle[{x_n}_p^{x}]>\pi+\varepsilon$ for some fixed $\varepsilon>0$. Observe that the geodesics $[p, x_n]$ do not converge to $[p,x]$, whence its limit is another geodesic from $p$ to $x$. Feb 10, 2020 at 0:04
• @Mathy angles.     Feb 13, 2020 at 6:30