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Let $f \in C[0,1]$ be a function satisfying the a Lipschitz condition, so that $|f(x,y_2)-f(x,y_1)\le L|y_2-y_1|$ for all $(x,y_2), (x,y_1) \in [0,1]\times \mathbb{R}$, where $L<8$. Show that the integral operator

$$T(u) = \int_0^1 g(x,\xi) f(\xi, u(\xi))d\xi$$

is a contraction mapping on the normed space of continuous functions $\left(C[0,1], \|\cdot\|_\infty\right)$.

Where

$$g(x,\xi)=\cases{\xi(x-1) & if $0\le \xi\le x \le 1$\\ x(\xi-1) & if $0\le x \le \xi \le 1$}$$

$T(u)$ is actually related to the boundary value problem $u''(x)=f(x,u), u(0)=u(1)=0$.

Here's how I approached this problem:

$$\left|T(u_1)-T(u_2)\right|=\left| \int_0^1 \left( g(x,\xi)f(\xi, u_1(\xi)) - g(x,\xi)f(\xi, u_2(\xi)) \right) d\xi \right|$$ $$ \le \int_0^1 | g(x,\xi) | |f(\xi, u_1(\xi)) - f(\xi, u_2(\xi)) | d\xi \le L \int_0^1 | g(x,\xi) | | u_1(\xi) - u_2(\xi) | d\xi$$

Thus

$$\left\|T(u_1)-T(u_2)\right\|_\infty \le L \| g(x,\xi) \|_\infty \int_0^1 \| u_1(\xi) - u_2(\xi) \|_\infty d\xi=L \| g(x,\xi) \|_\infty \| u_1(\xi) - u_2(\xi) \|_\infty$$ $$<8 \| g(x,\xi) \|_\infty \| u_1(\xi) - u_2(\xi) \|_\infty$$

So, as far as I understand, one needs to prove here that $\| g(x,\xi) \|_\infty < \frac18$.

Would appreciate some help.

sequence
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1 Answers1

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By definition $g(1/2,1/2)=-1/4$, then $\|g\|_\infty\ge 1/4>1/8$. But we can do a better estimate in your calculation. When you arrive to:

$$L\int_0^1|g(x,\xi)||u_1(\xi)-u_2(\xi)|$$

it is convenient to estimate as follows:

$$\begin{align}L\int_0^1|g(x,\xi)||u_1(\xi)-u_2(\xi)|&\le L\|u_1-u_2\|_\infty\int_o^1|g(x,\xi)|=\\&=L\|u_1-u_2\|_\infty\bigg[\int_0^x|\xi(x-1)|d\xi+\int_x^1|x(\xi-1)|d\xi\bigg]=\\&= L\|u_1-u_2\|_\infty\frac{1}{2}(-x^2+x)\le\\&\le L\|u_1-u_2\|_\infty\frac{1}{8} \end{align}$$

just by doing the explicit calculation of the integrals and considering $x\in[0,1]$.

PozzPlot
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  • Can you please clarify how you integrated the absolute values? Also, how did you get $(-x^2+x)\le 1/4$? – sequence Oct 24 '17 at 23:21
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    @sequence I give you some hints, you can understand by yourself. First, recalling that $x,\xi\in[0,1]$, you have that $|x-1|=1-x$ and $|\xi-1|=1-\xi$. Second, $-x^2+x$ is just a parabola on $[0,1]$, you can find its maximum. – PozzPlot Oct 25 '17 at 06:48