Deducing a property for any function $f$ using the wave equation I am given the wave equation in spherical coordinates for a wave function only depending on $r$:
$$\frac{1}{v^2} \frac{\partial^2 \xi(r,t)}{\partial t^2} = \left( \frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r} \right) \xi(r,t).$$
Also, I know that this equation can be solved using the ansatz $\xi(r,t) = A(r) f(kr \pm \omega t)$. Now I have to show that for any (twice differentiable) function $f$,
$$A(r) = \frac{C}{r}$$
with some constant $C$ must hold.
To prove this, I used the given ansatz and computed the partial derivatives of the wave function and plugged them into the wave equation. Doing this, I end up with the condition
$$A''(r) f(kr\pm\omega t) + A'(r)[2(f'(kr\pm \omega t)k + \frac{1}{r} f(kr\pm \omega t))] + A(r) \frac{2k}{r} f'(kr \pm \omega t) = 0,$$
where $A', A''$ and $f'$ denote the derivatives of those functions.
My question is: How do I proceed? 
Obviously, this equation holds for $f = 0$. If we have $f = m$ for some $m \in \mathbb{R} \setminus \{0\}$, the condition will simplify to the second order differential equation 
$$A''(r) + \frac{2 A'(r)}{r} = 0.$$
Using substitution ($B(r) := A'(r)$), we find $A(r) = - \frac{c_1}{r} + c_2$ and since $A(r) \to 0$ as $r \to \infty$, we can conclude $c_2 = 0$.
But how can I similarly (?) show $A(r) = \frac{C}{r}$ for any function $f$?
Thank you very much in advance for any help.
 A: There is a difference between the statement 

for any $f$, $A(r)$ must equal $C/r$

and 

were $A(r)$ to hold for any $f$, $A(r)$ must equal $C/r$

The first statement $A(r)$ is allowed to depend on $f$. The second statement $A(r)$ cannot. 
The first statement is false. Take $\omega = 0$ and set $A(r) f(kr) = 1$, then for any non-vanishing $f$ this particular $A(r)$ will work. 
To show the second statement is true, you are mostly done. From the equation you derived by plugging in the ansatz, you have that, grouping terms by $f$, 
$$ \left(A''(r) + A'(r)\frac{2}{r}\right) f(kr\pm\omega t) +\left( A'(r)  +  A(r) \frac{1}{r}\right)2k f'(kr \pm \omega t) = 0 $$
For this to be true for any $f$, the "coefficients" inside the parentheses must vanish identically (consider (a) the constant function $f\equiv 1$ and (b) the function $f$ defined so that on an interval $[a,b]$, $f(x) = x$). So you get
$$ A'' + 2A'/r = 0 \qquad A' + A/r = 0 $$
or equivalently 
$$ (r^2 A')' = 0 \qquad (rA)' = 0 $$
which gives you precisely that $rA = C$. 
