Suppose $f:\Bbb{R}^n \to \Bbb{R}$ be a bounded function such that for any $x\in \Bbb{R}^n$ the following property holds $$f(x) = \frac{1}{m(B(x,r))}\int_{B(x, r)}f(t) ~\mathrm{d}m(t)$$ for all $r<1$. Does it imply that $f$ is constant?
Without that restriction $r<1$, the answer is yes. In that case we could have proceed as shown in the following proof (screenshot attached). But in that proof we actually requires that $r$ can be arbitrarily large which is not possible under the condition $r<1$.
Is it possible to find a counter example to the above problem. Thanks.
There is no counterexample: that function is harmonic. The following works smoothly (pun intended) if $f$ is $\mathcal{C}^2$, but could be adapted to lesser regularity classes, by interpreting the Laplacian in a distributional sense.
There is probably a more direct way, however, fix $x\in\mathbb R^n$ and define $$F(r)=\frac{1}{m(B(x,r))}\int_{B(x,r)}f(t)dm(t)\;,$$ for $r<1$.
Let $m(r)=m(B(x,r))$, then $m'(r)=nm(r)/r$.
Writing the integral in spherical coordinates, you can easily compute the derivative $F'(r)$, obtaining $$F'(r)=\frac{1}{m(r)}\int_{S(x,r)}f(t)d\sigma(t) - \frac{n/r}{m(r)}\int_{B(x,r)}f(t)dm(t)$$ where $\sigma$ is the $n-1$-dimensional measure on $S(x,r)=\{y\in\mathbb R^n\ :\ \|y-x\|=r\}$.
Note that $\sigma(S(x,r))=m'(r)$.
Now, $F(r)$ is constant for $r<1$, so $F'(r)$ should vanish, which implies $$\frac{1}{m(r)}\int_{B(x,r)}f(t)dm(t)=\frac{1}{m'(r)}\int_{S(x,r)}f(t)d\sigma(t)\;.$$
Therefore, it also holds, for $r<1$, that $$f(x)=\frac{1}{\sigma(S(x,r))}\int_{S(x,r)}f(t)d\sigma(t)\;.$$
Now, repeat the trick: define $$G(r)=\frac{1}{\sigma(S(x,r))}\int_{S(x,r)}f(t)d\sigma(t)$$ which is constant for $r<1$ and whose derivative in $r$ is (thanks to the first Green's identity) $$c_r\int_{B(x,r)}\Delta f(t)dm(t)\;.$$ This derivative should vanish identically for $r<1$.
Now, if $\Delta f(x)>0$, then it will be positive in a small ball centered in $x$, and for that ball the previous integral will not vanish, which is a contradiction (same reasoning if $\Delta f(x)<0$). Therefore, $\Delta f(x)=0$; as $x$ was arbitrary, you have your result.
More generally if the mvp holds (ball or sphere version) for a sequence of radii converging to $0$ (and arbitrary center), then the function is harmonic.