Is there a function of bounded variation on $[0,1]$, say $f$, such that $f(x)-2x \sqrt{V^x_0(f)}= \frac{5}{4}x^2$ for all $x \in [0,1]$? Is there a function of bounded variation on $[0,1]$, say $f$, such that $f(x)-2x \sqrt{V^x_0(f)}= \frac{5}{4}x^2$ for all $x \in [0,1]$?
Here, $V^b_a(f) = \sup_{\mathcal{P}}\{ \sum_{j=1}^{N}|f(x_j) - f(x_{j-1})|: \mathcal{P}=\{a=x_0<x_1<...<x_N =b\} \}$
And how do we find such a function $f$? Would it be unique?
I am working on the above question for my qual prep but I am stuck. As a hint, I was told that I would start with showing if $f$ is monotonic but I have no clue how it helps. Any helps would be highly appreciated!
 A: Observing that $f$ is monotonic is indeed the key to this problem. We will now prove the existence and uniqueness of such $f$.
We can rewrite the main equation as follows
$$
\tag{1} f(x) = 2x \sqrt{V_0^x (f)} + \frac 54 x^2.
$$
The function $V_0^x(f)$ is non-decreasing in $x$ for any $f$ of bounded variation, hence the right-hand side of $(1)$, as a sum of two monotonic functions is itself monotonic. Thus $f$ is increasing on $[0,1]$. Observe also, that substituting $x=0$ in $(1)$ implies that $f(0) = 0$.
Now, from the definition of variation and the fact that $f$ is increasing we have $V_0^x(f) = f(x) - f(0) = f(x)$ as $f(0) = 0$. Substituting this in $(1)$ we arrive at
$$
f(x) = 2x \sqrt{f(x)} + \frac 54 x^2.
$$
Completing the square in the last equation implies
$$
\left(\sqrt{f(x)} - x\right)^2 = \frac 94 x^2.
$$
From the last expression we have $\sqrt{f(x)} -x = \pm \frac 32 x$ for any $x\in [0,1]$ and since $f$ is increasing with $f(0) = 0$ we get that $\sqrt{f(x)} - x = \frac 32 x$ for all $x \in [0,1]$, arriving thus at
$$
f(x) = \frac {25}{4} x^2,
$$
which is the unique solution to the given functional equation.
