Proving $f(x)=cx$ for constant $c$ where $f(x)=\frac{1}{t}\int_0^t (f(x+y)-f(y))\,dy \quad \forall x\in \mathbb{R}$ 
If $f:\mathbb{R} \to \mathbb{R}$ be continuous such that $f(x)=\frac{1}{t}\displaystyle\int_0^t (f(x+y)-f(y))\,dy \quad \forall x\in \mathbb{R}$ and $\forall t>0$, then show that there exists a constant $c$ such that $f(x)=cx \forall x$

Does this have anything to do with Cauchy's functional equation $f(x)=f(x+y)-f(y)$ which also gives the same solution for $f$?
 A: As suggested in a comment: since this holds for all $t>0$, you have the equivalent formulation
$$
t f(x)= \int_0^t (f(x+y)-f(y))\,dy \quad \forall x\in \mathbb{R}, \forall t > 0 \tag{1}
$$
which implies, by differentiating with regard to $t$:
$$
f(x)= f(x+t)-f(t) \quad \forall x\in \mathbb{R}, \forall t >0 \tag{2}
$$
which is exactly Cauchy's functional equation (recall that $f$ is also assumed continuous in our case). Hence, any solution to (1) must be a solution to (2), and we know the latter: $f\colon x\mapsto cx$ for some $c\in\mathbb{R}$. It only remains to check that indeed these solutions satisfy (1); which is immediate.
A: Assuming f is infinitely differentiable. (so it is a bit out of scope)
$$f'(x)=\frac{1}{t}\int_0^t \frac{d(f(x+y)-f(y))}{dx}\,dy$$
$$=\frac{1}{t}\int_0^t f'(x+y)\,dy$$
$$=\frac{1}{t} (f(x+t)-f(x))$$
$$=f'(x)+\frac{f''(x)}{2}t+...$$
$$\rightarrow f''(x)=0$$
$$\rightarrow f'(x)=constant$$
$$\rightarrow f(x)=cx + constant$$
A: Assume f is differentiable
$$f'(x)=\frac{1}{t}\int_0^t \frac{d(f(x+y)-f(y))}{dx}\,dy$$
$$=\frac{1}{t}\int_0^t f'(x+y)\,dy$$
$$=\frac{1}{t} (f(x+t)-f(x))$$
$$\to f'(x)=\frac{1}{t} (f(x+t)-f(x))$$
$$\to \exists c \ \in \ \mathbb R \ \text{s.t.} \ f'(x)=c$$
$$\to f(x)=cx+d$$
Find d:
$$cx+d = \frac{1}{t} \int_0^t (c(x+y)+d - cy - d) dy $$
$$\to cx+d = \frac{1}{t} \int_0^t (cx) dy$$
$$\to cx+d = (cx) \frac{1}{t} \int_0^t  dy$$
$$\to cx+d = (cx)$$
$$\to d=0$$

Otherwise:
See Clement C.'s answer:
https://math.stackexchange.com/a/1859672/140308
