Solving Equations Involving integration Suppose $f:R \to R$ is a continuous function such that $$f(x)=\frac{1}{t}\int^t _0f(x+y)-f(y)dy$$ for every $x$ and for all $t>0$. Prove that there exists a constant $c$ such that $f(x)=cx$ for every $x$
 A: First note that
$$f(0) = \frac{1}{t}\int_0^t f(0+y) - f(y) \: dy = 0$$
Take the derivative of $f$ w.r.t. $x$
$$f'(x) = \frac{1}{t}\int_0^t f'(x+y)\:dy = \frac{f(x+t)-f(x)}{t}$$
From here we can deduce that
$$f'(0) = \frac{f(0+t)-f(0)}{t} = \frac{f(t)}{t}$$
$$\implies f(t) = f'(0)\cdot t$$
for all $t>0$. If we want to prove that it holds for negative values as well, take $g(x) = f(-x)$ then consider
$$g'(x) = \frac{f(-x)-f(t-x)}{t}$$
and plug in $x=t$ to get the differential equation
$$t\cdot g'(t) = g(t)$$
which only has $ct$ as a solution. Thus $f(-t)$ is also a linear function for $t>0$ and we are done.
A: Multiply both sides by $t$: $$t f(x) = \int_0^t (f(x+y) - f(y))\,dy$$ and apply the partial derivative with respect to $t$ to both sides: $$f(x) = f(x+t) - f(t)\qquad \qquad  (1)$$
which holds for all $x$ and for all $t > 0$.  

Now we need a few facts that come from $(1)$:


*

*Setting $x=0$ in $(1)$ gives $f(0) = 0.$

*Setting $x=-t$ in $(1)$ gives $f(-t) = f(0) - f(t) = -f(t).$

*If $t < 0,$ then replacing $t$ with $-t>0$ and $x$ with $x+t$ gives $$f(x+t) = f(x) - f(-t) = f(x)+f(t)$$ which shows that $(1)$ is true when $t<0.$

*Since $f(x) = f(x+0) - 0 = f(x+0) - f(0),$ we also see that $(1)$ is true when $t=0.$
In summary, we have shown that $f(x+t) = f(x) + f(t)$ holds for all real $x,t,$ so $f$ is additive over the reals.
Since $f$ is additive, it's an easy exercise to show that $f(x) = f(1)\cdot x$ for all $x\in \mathbb{Q}.$
Since $f$ is continuous and $\mathbb{Q}$ is dense in $\mathbb{R}$, it follows that $$f(x) = f(1)\cdot x, \qquad \forall x\in \mathbb{R}$$ which is exactly what we wanted to prove (where $c=f(1)$)
A: The equation can be written as $\int_0^{t} [f(x+y)-f(y)-f(x)]dt=0$. Since this holds for all $t >0$ we get $f(x+y)-f(y)-f(x)=0$ whenever $y>0$. Since $f(0)=0$ (letting $t \to 0$ in the given equation we get $f(x)=f(x)-f(0)$ so $f(0)=0$)  this holds for $y \geq 0$. Now  a standard argument shows that $f(x)=cx$ for all $x \geq 0$ where $c=f(1)$. Now let $x<0$. For $y>0$ large enough we get $f(x+y)=c(x+y)=f(x)+f(y)=f(x)+cy$ which gives $f(x)=cx$. 
