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 :)