Let $(M,g)$ be an $m$-dimensional Riemannian symmetric space and $d(\cdot,\cdot)$ be the geodesic distance function. Given any fixed $\alpha\in M$ and $v\in T_\alpha M$ with $\|v\|=1$. Define $$ g(t)=\int_M f(t,x)e^{-f(t,x)}dx, $$ where $f(t,x)=d^2(Exp_\alpha(tv),x).$ Question: Is $g$ second-order differentiable at $t=0$?

Answer: Here $g$ is a constant function so the answer is yes.

There was a typo in the above question. To fix it, we define $$ h(t)=\int_M f(t,x)e^{-f(0,x)}dx, $$ where $f(t,x)=d^2(Exp_\alpha(tv),x).$ Question: Is $h$ second-order differentiable at $t=0$?

Let $(M,g)$ be an $m$-dimensional Riemannian symmetric space and $d(\cdot,\cdot)$ be the geodesic distance function. Fix any $\alpha\in M$ and $v\in T_\alpha M$ with $\|v\|=1$.

Q1: Define $$ g(t)=\int_M f(t,x)e^{-f(t,x)}dx, $$ where $f(t,x)=d^2(Exp_\alpha(tv),x).$ Question: Is $g$ second-order differentiable at $t=0$?

Answer: Here $g$ is a constant function so the answer is yes. This was given by Willie Wong.

Q2: There was a typo in the above question. To fix it, we define $$ h(t)=\int_M f(t,x)e^{-f(0,x)}dx, $$ where $f(t,x)=d^2(Exp_\alpha(tv),x).$ Question: Is $h$ second-order differentiable at $t=0$?

Let $(M,g)$ be an $m$-dimensional Riemannian symmetric space and $d(\cdot,\cdot)$ be the geodesic distance function. Given any fixed $\alpha\in M$ and $v\in T_\alpha M$ with $\|v\|=1$. Define $$ g(t)=\int_M f(t,x)e^{-f(t,x)}dx, $$ where $$ f(t,x)=d^2(Exp_\alpha(tv),x). $$ Question: is it possible to show that $g$ is second-order differentiable at $t=0$?

Let $(M,g)$ be an $m$-dimensional Riemannian symmetric space and $d(\cdot,\cdot)$ be the geodesic distance function. Given any fixed $\alpha\in M$ and $v\in T_\alpha M$ with $\|v\|=1$. Define $$ g(t)=\int_M f(t,x)e^{-f(t,x)}dx, $$ where $f(t,x)=d^2(Exp_\alpha(tv),x).$ Question: Is $g$ second-order differentiable at $t=0$?

Answer: Here $g$ is a constant function so the answer is yes.

There was a typo in the above question. To fix it, we define $$ h(t)=\int_M f(t,x)e^{-f(0,x)}dx, $$ where $f(t,x)=d^2(Exp_\alpha(tv),x).$ Question: Is $h$ second-order differentiable at $t=0$?

Let $(M,g)$ be an $m$-dimensional complete Riemannian manifold with bounded sectional curvatures and $d(\cdot,\cdot)$ be the geodesic distance function. Suppose $\alpha\in M$ and $v\in T_\alpha M$ with $\|v\|=1$. Define $$ g(t)=\int_M f(t,x)e^{-f(t,x)}dx, $$ where $$ f(t,x)=d^2(Exp_\alpha(tv),x). $$ Question: is it possible to show that $g$ is second-order differentiable at $t=0$?

Let $(M,g)$ be an $m$-dimensional Riemannian symmetric space and $d(\cdot,\cdot)$ be the geodesic distance function. Given any fixed $\alpha\in M$ and $v\in T_\alpha M$ with $\|v\|=1$. Define $$ g(t)=\int_M f(t,x)e^{-f(t,x)}dx, $$ where $$ f(t,x)=d^2(Exp_\alpha(tv),x). $$ Question: is it possible to show that $g$ is second-order differentiable at $t=0$?