Romanian Mathematical Olympiad

It is readily checked that any affine function satisfies the condition in the statement. To prove the converse, set $y=x$ to write $${\int }_0^{2x}f(t)dt=x(f(2x)+f(0)).(\ast )$$ Since $f$ is integrable, $(\ast )$ shows that the function $g:\mathbb{R}\to \mathbb{R}$, $g(x)=x(f(2x)+f(0))$, is continuous, so $f$ is continuous on ${\mathbb{R}}^{\ast }=\mathbb{R}\setminus \{0\}$. Continuity of $f$ on ${\mathbb{R}}^{\ast }$ and $(\ast )$ then show $g$ differentiable on ${\mathbb{R}}^{\ast }$, so $f$ is differentiable on ${\mathbb{R}}^{\ast }$. Differentiate $(\ast )$ to get $2f(2x)=f(2x)+f(0)+2x{f}^{\prime }(2x)$; that is, $x{f}^{\prime }(x)-f(x)+f(0)=0$ for all $x\ne 0$. Consequently, $${\left(\frac{f(x)-f(0)}{x}\right)}^{\prime }=0\text{for all }x\ne 0,$$ so $$f(x)=\{\begin{array}{ll}ax+f(0),& x<0,\\ bx+f(0),& x>0.\end{array}$$ Set now $x=0$ and $y=1$ in the relation in the statement to write $$\begin{array}{rl}f(1)+f(-1)& ={\int }_{-1}^{1}f(t)dt={\int }_{-1}^0(at+f(0))dt+{\int }_0^1(bt+f(0))dt\\ & =\frac{1}{2}(b-a)+2f(0).\end{array}$$ Plug $f(1)=b+f(0)$ and $f(-1)=-a+f(0)$ into the above relation to get $b-a+2f(0)=\frac{1}{2}(b-a)+2f(0)$, so $a=b$ and $f(x)=ax+f(0)$ for all real numbers $x$.

Alternative Solution: For convenience, consider the integrable function $g:\mathbb{R}\to \mathbb{R}$, $g(x)=f(x)-f(0)$, along with the continuous function $G:\mathbb{R}\to \mathbb{R}$, $G(x)={\int }_0^{x}g(t)dt$. Clearly, $g$ and $G$ both vanish at the origin, and it is readily checked that $$G(x+y)-G(x-y)=y(g(x+y)+g(x-y)),\text{for all real numbers }x\text{ and }y.(1)$$ Let $x=y=t/2$ in (1), and recall that $g$ and $G$ both vanish at the origin, to infer that $$2G(t)=tg(t),\text{for all real numbers }t.\text{\hspace{1em}(2)}$$ Next, let $x=0$ in (1) and use (2) to infer that $G$ is an even function: $G(-y)=G(y)$, for all real numbers $y$. With reference again to (2), it follows that $g$ is odd: $g(-y)=-g(y)$, for all real numbers $y$. It is therefore sufficient to focus on the restrictions $g_{+}$ and $G_{+}$ of $g$ and $G$, respectively, to the ray of positive real numbers. By (2), continuity of $G_{+}$ implies that of $g_{+}$ which in turn implies differentiability of $G_{+}$ and $G_{+}^{\prime }=g_{+}$. Hence $2G_{+}(t)=t{G}_{+}^{\prime }(t)$ for all real $t>0$, showing that the function $t\mapsto t^{-2}{G}_{+}(t)$, $t>0$, is constant; that is, $G_{+}(t)=\alpha t^2$ for some real constant $\alpha$, so $g_{+}(t)=at$, where $a=2\alpha$. Consequently, $g(t)=at$ for all real numbers $t$, and $f(t)=at+f(0)$.