Proving a real valued function does not exist

[_N3WTON_]

Homework Statement

Show that if $\lambda > \frac{1}{2}$ there does not exist a real-valued function 'u' such that for all x in the closed interval $0 <= x <= 1$, $u(x) = 1 + \lambda\int_{x}^{1} u(y)u(y-x)\,dy$

Homework Equations

The Attempt at a Solution

I started off by assuming that such a function does in fact exist, so:
$u(x) = 1 + \lambda\int_{x}^{1}u(y)u(y-x)\,dy$
$\lambda > \frac{1}{2}$
$0 <= x <= 1$
$\mu = \int_{0}^{1} u(x)\,dx$
$= \int_{0}^{1}\,dx + \int_{0}^{1}\int_{x}^{1} u(y)u(y-x) dy \hspace{1 mm} dx$
Here, I changed the order of integration:
$= 1 + \lambda\int_{0}^{1}\int_{0}^{y} u(y)u(y-x)dx \hspace{1 mm} dy$
$= 1 + \lambda\int_{0}^{1}u(y)\int_{0}^{y}u(y-x)dx \hspace{1 mm} dy$
At this point I am a bit confused about where to go, so I am starting to think that perhaps I am not going in the right direction; I was hoping someone could give me some advice. Thanks :)

 

[_N3WTON_]

Ok, I worked on this for a quite a while last night before bed and I think I have a solution, hopefully someone can confirm/deny...I'll start where I left off above, where I take Y constant:
$z = y - x$
$dz = -dx$
$\mu = 1 + \lambda\int_{0}^{1}u(y)\int_{y}^{0} u(z)(-dz)\,dy$
$= 1 + \lambda\int_{0}^{1}u(y)\int_{0}^{y}u(z)dz\,dy$
${Let} \hspace{1 mm} f(y) = \int_{0}^{y} u(z)\,dz \hspace{2 mm} {then,} \hspace{2 mm} f'(y) = u(y), \hspace{1 mm} f(0)=0, \hspace{1 mm} f(1) = \mu$
$\mu = 1 + \lambda\int_{0}^{1}f'(y)f(y)\,dy$
$= 1 + \frac{\lambda {\mu}^2}{2}$
$\lambda {\mu}^2 - 2\mu + 2 = 0$
So now, if mu is going to exist, the discriminant of that quadratic cannot be negative:
$4 - 8\lambda >= 0$
$\lambda <= \frac{1}{2}$
-This contradicts the problem statement that $\lambda > \frac{1}{2}$. Hence, it is confirmed that for $\lambda > \frac{1}{2}$ there does not exist $u = 1 + \lambda\int_{x}^{1}u(y)u(y-x)\,dy$.

 

[pasmith]

Looks correct.

 

[_N3WTON_]

thanks :)