Let $M$ be a complete Riemannian manifold. Now for a fixed $p\in M$, is there any non-constant smooth function $u:M\rightarrow\mathbb{R}$ such that
$$\int\_{B\_r}udV=0\ \forall 0\leq r<\infty,$$
where $B\_r$ is a ball with center at $p$ and radius $r$. I can not find such an example. I know that in $\mathbb{R}$, the above equation implies that $u$ is zero, but for general case I can not find any example. Please help!



Example of a smooth function in a manifold whose integration vanishes
On the circle $x^2+y^2=1$, the function $x$ has this property for $p=(0,1)$ or $p=(0,-1)$.
On the sphere $|x|^2=1$ in $\mathbb{R}^{n+1}$, every nonzero linear function restricts to a function with this property, around any point $p$ at which it vanishes. On hyperbolic space, in the ball model, any linear function on the ball has this property about $p$ the origin. Similarly for complex or quaternionic hyperbolic space. 
Suppose that $M$ is a Riemannian manifold and take a point $p$ of $M$. Suppose that $r$ is the injectivity radius of $M$ at $p$. Denote the exponential map as $e \colon B\_r(0) \to B\_r(p)$ between the balls in $T\_p M$ and in $M$. Take the Euclidean volume form $\Omega\_p$ on $T\_pM$ and the volume form $\Omega\_M$ of $M$. Let $F>0$ be the function on $B\_r(p)$ so that $e^{-1\*}\Omega\_p=F\Omega\_M$ on $B\_r M$. Let $h$ be a linear function on $T\_p M$. Let $k$ be a smooth rotationally symmetric function on $T\_p M$ with compact support contained in the open ball $B\_r(0)$. Then the function $f$ equal to $(e^{-1\*}hk)F$ in $B\_r(p)$ and equal to zero outside $B\_r(p)$ has zero integral on any ball around $p$ in $M$, as the integral is just the Euclidean integral of $hk$ around $0$ in $B\_r(0)$.


