Solutions to integral equations are equivalence classes of functions, i.e. two functions are in the same class if they are different on a (Lebesgue) measure zero subsetset of their domains. Having said that it is reasonable to look for analytic solutions, i.e. functions which are analytic almost everywhere.
Let's rewrite the integral equation:
$$
I(f;r,r_0)=\frac{1}{2r^{2}_{0}}\int\limits_{r-r_0}^{r+r_0}\left( r^2-r_{0}^{2}-R^2\right)f(R)\,dR=0
$$
This is a functional equation and if $f$ is analytic in the range $(r-r_0,\;r+r_0)$, then also $I(f;\quad,\quad)$ is analitic with respect to its first and second variables.
The integration range is symmetric with respect to the point $R=r$. Let's assume that $f$ is an analytic function, i.e. we assume that there is a range around $R=r$ where $f(R)=\sum_{n=0}^{\infty} a_n (R-r)^n$. Without loss of generality we may assume that the Taylor series is convergent to $f(R)$ in the whole range $(r-r_0,\;r+r_0)$. Now we expand the integral $I(f; r,r_0)$ with respect to $r_0$ obtaining the first three nonvanishing terms:
Collect[1/(2 r0^2) Integrate[ Series[((r - r0) (r + r0) - R^2) f[R], {R, r, 6}],
{R, r - r0, r + r0}, Assumptions -> r > r0 > 0] // Normal,
r0, Simplify] /. r0 -> Subscript[r, 0] // Most // TraditionalForm

and since $I(f; r,r_0)=0$ (with restrictions given by appropriate assumptions) every coefficient of its expansion w.r.t $\;r_0$ has to vanish. The general solution follows readily from the first coefficient DSolve[ 2 f[r] + r f'[r] == 0, f[r], r]
, nonetheless in general there might be no solutions. Thus one should find a function $f(R)$ which makes every coefficient vanish, and since higher order coefficients are not independent we can take the first three nonvanishing (symbolically) coefficients.
coef = DeleteCases[ CoefficientList[
1/(2 r0^2) Integrate[ Series[((r - r0) (r + r0) - R^2) f[R], {R, r, 6}],
{R, r - r0, r + r0}, Assumptions -> r > r0 > 0] //
Normal, r0] // Factor // Most, 0];
sols =
Table[ f[r]/.Flatten @ DSolve[ Thread[# == 0&@coef][[k]], f[r], r,
GeneratedParameters -> (Subscript[c, #, k] &)],
{k, 3}];
Union @@@ (sols /. SolveAlways[Equal @@@ Subsets[sols, {2}], r]) /.
Subscript[c, 1, 3] -> 120 c
{c/r^2}
We have found that the general solution of the above integral equation is $f(R)= {c\over R^2}$ and recalling the introductory remarks $f(R)$ may differ from ${c\over R^2}$ on a measure zero set.
QED.
AsymptoticDSolveValue
orAsymptoticIntegrate
in Mma 10.3 to show that any perturbation from the1/R^2
solution makes the result worse, showing the stationarity of your solution. $\endgroup$DSolve[Integrate[(r^2/(2*r0^2) - 1/2 - R^2/(2*r0^2))*f[R], {R, r - r0, r + r0}, Assumptions -> {r > r0 > 0}] == a, f[x], x] /. a -> 0
gives a warning message:Supplied equations are not differential or integral equations of the given functions
$\endgroup$