A question involving implied differentiability I was working on the following problem:
Given:
$$ g(x) = \int_0^x f(y) (y - x) dy $$ 
And $g(x)$ is exactly 3 times continuously differentiable. What is the greatest integer n for which $f$ must be $n$ Times continuously differentiable? (source ETS GRE practice exam).
So an idea was to apply integration by parts:
$$ \frac{1}{2} f(y) (y - x)^2 dy |_0^x  - \frac{1}{2} \int_0^{x} f'(y) (y - x)^2  = \int_0^x f(y) (y - x) dy$$
Now application of bounds and the left side becomes:
$$ -\frac{1}{2} f(0) x^2 - \frac{1}{2} \int_0^{x} f'(y) (y - x)^2 = g(x)$$
But this didn't seem fruitful, and my 1 minute block of thinking that i can reasonably allocate for this ran out.
How does one resolve this problem?
 A: We can write
$$g(x) = \int_0^x f(y)y\, dy - x\int_0^xf(y)\,dy.$$
Since $f$ is continuous, we can differentiate the right side using the FTC. After simplifying we have
$$g'(x) = -\int_0^x f(y)\, dy.$$
Applying FTC again, we get $g''(x) = - f(x).$ Since $g$ is given to be $C^3,$ we get $g'''(x) =-f'(x).$ Hence $f$ must be at least $C^1.$
A: There is a slightly shorter path reducing computation but necessitating to use (causal) convolution, Dirac's $\delta$ distribution and some of their properties. 
It is based upon the observation that
$$\tag{1}g=\tilde f \ast h$$
with $\ast$ the operation of convolution (https://en.wikipedia.org/wiki/Convolution), with $h$ defined in the following way :
$$h(x)=\begin{cases}0 & if \ x<0\\
x & if \ x>0\end{cases}$$ 
and where $\tilde f=f.H$ is the "causal" function obtained by multiplying $f$ by Heaviside function ($1$ for $x \geq 0$, $0$ for $x<0$), which is the derivative of $h$. 
Using classical properties of convolution and $\delta$ distribution (see Remark below), and denoting by $(u)^{(n)}$ the $n$th derivative of $u$,
$$\tag{2} (f \ast h)^{(n)}=f^{(n-2)}\ast h''=f^{(n-2)}\ast\delta=f^{(n-2)}$$
Conclusion: Taking $n=3$, it is necessary and sufficient that $f^{(3-2)}=f'$ exists.
Remark: Properties being used in (2):


*

*the fact that $F^{(m)}\ast G^{(p)}=(F \ast G)^{(m+p)}$, 

*the fact that the second derivative of $h$ is the first derivative of $H$, itself equal to $\delta$, 

*and finally the fact that $\delta$ is neutral for convolution.
Final comment: The main interest of this presentation is that it sheds some light on the apparently intriguing fact that for a function being $C^3$, one has only to achieve that another one is $C^1$. We are at the heart of one of the main properties of convolution, i.e., its regularizing ability.
Had we taken $g(x) := \int_0^x f(y) (y - x)^5 dy$, for example, we would have "jumped" from $f \in C^1$ to $g \in C^7$... 
