Existence of function such that $f(x)=1+\alpha\int_x^1f(t)f(t-x)dt.$ Let $\alpha>1/2$ be a real number. Prove that there does not a real function $f$ such that
$$f(x)=1+\alpha\int_x^1f(t)f(t-x)dt.$$
for all $t\in [0,1].$
I trid to differentiate both sides but I ended up with a even more nasty looking expression:
$$f'(x)=-\alpha \left(f(0)f(x)+\int_x^1f(t)f'(t-x)dt \right).$$
But I couldn't integrate the integral on the right hand side.
 A: Assume that the function $f$ is continuous with real values. Put $I = \int_{0}^{1}f(x)\, \mathrm{d}x$. If we integrate the integral equation
\begin{equation*}
f(x) = 1+\alpha\int_{x}^{1}f(t)f(t-x)\, \mathrm{d}t 
\end{equation*}
and use Fubini's theorem we get
\begin{gather*}
I = 1+\alpha\int_{0}^{1}\left(\int_{x}^{1}f(t)f(t-x)\, \mathrm{d}t\right)\,\mathrm{d}x = 1+ \alpha\int_{0}^{1}\left(\int_{0}^{t}f(t)f(t-x)\, \mathrm{d}x\right)\,\mathrm{d}t = \\[2ex]1+ \alpha\int_{0}^{1}f(t)\left(\int_{0}^{t}f(t-x)\, \mathrm{d}x\right)\,\mathrm{d}t \tag{1}
\end{gather*}
But
\begin{equation*}
\int_{0}^{t}f(t-x)\, \mathrm{d}x =[t-x=y] = \int_{0}^{t}f(y)\, \mathrm{d}y 
\end{equation*}
which is a primitive function of $f(t)$. 
Then we return to (1).
\begin{equation*}
I = 1+\alpha\int_{0}^{1}f(t)\left(\int_{0}^{t}f(y)\, \mathrm{d}y\right)\,\mathrm{d}t =1+\dfrac{\alpha}{2}\left[\left(\int_{0}^{t}f(y)\, \mathrm{d}y\right)^{2}\right]_{0}^{1} = 1 + \dfrac{\alpha}{2}I^{2}.
\end{equation*}
But since $\alpha>\frac{1}{2}$ this a contradiction.
\begin{gather*}
I=1+\dfrac{\alpha}{2}I^2 \Leftrightarrow I^2 -\dfrac{2}{\alpha}I +\dfrac{2}{\alpha}=0 \Leftrightarrow \\[2ex]
0\le\left(I-\dfrac{1}{\alpha}\right)^{2}=\dfrac{1-2\alpha}{\alpha^{2}}<0.
\end{gather*}
