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Let $R = [0,1] \times [0,1]$, $M = \{ f(x,y) \in C(R), 0\leq |f(x,y)|\leq 1\}$, and $0<\alpha< \frac{4}{3}$. I need to show that $T$ is a contraction on $M$ where $T$ is given by $$ Tf(x,y) = \frac{xy}{3} + \alpha\int_{R}(x+y) st(f(s,t))^2 dsdt.$$

I have succeeded in showing that $||Tf - Tg||_{\infty} \leq \frac{4}{3} ||f-g||_{\infty}$ for $f,g \in M$...I can't seem to get the stronger $< 1$ needed...

Here is my work: Choose $f, g \in M$. Then $$|Tf(x,y) - Tg(x,y)| = |\alpha \int_{R} (x+y) st (f(s,t)^2-g(s,t)^2)dsdt|$$ $$= |\alpha(x+y)\int_{R} st (f(s,t) - g(s,t))(f(s,t)+g(s,t))dsdt| $$ $$\leq \alpha|x+y|\int_R||f-g||_\infty |st(f(s,t)+g(s,t))|dsdt$$ $$\leq \alpha|x+y|||f-g||_\infty \int_R 2stdsdt$$ $$=\alpha |x+y|||f-g||_\infty 2 \cdot \frac{1}{4}$$ $$\leq \frac{4}{3}||f-g||_\infty$$

So I have $||Tf-Tg||_\infty \leq \frac{4}{3} ||f-g||_\infty$....

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    Could you explain what are $x$ and $y$ in the definition of $T$ ? What is $\frac{xy}{3}$ ? – 2017-01-05
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    Does my edit clarify? Functions $f(x,y)$ in the domain of $T$ are from $M$, which is the set of continuous functions on the "unit box". I.e., the set of all functions such that $f(x,y)$ continuous, $x \in [0,1], y \in [0,1]$. – 2017-01-05
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    OK. How did you bound $|f(s,t)+g(s,t)|$ by $2$ ? When defining $T$, did you mean to write $$ (Tf)(x,y):=\frac{xy}{3}-\alpha\int_{R}(x+y)st(f(s,t))^2\,ds\,dt\quad? $$ If so, unless I'm mistaken, in the case $\alpha=1$, taking $f\equiv10$ and $g=0$ leads to $$ \|Tf-Tg\|_{\infty}\geq\left.\frac{100(x+y)}{4}\right\vert_{(x,y)=(1,1)}=50>10=\|f-g\|_{\infty} $$ so that $T$ is *not* a contraction on $M$. – 2017-01-05
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    Completely my fault--missed copying a key piece of information on $M$! Editing now! I'm sorry for the inconvenience--don't know where my brain was! – 2017-01-05
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    The equation $\displaystyle Tf(x,y) = \frac{xy}{3} = \alpha\int_{R}(x+y) st(f(s,t))^2 dsdt$, doesn't quite make sense. Note that there are two equal signs. – 2017-01-05
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    If you let $f(x,y) = 1$, $g(x,y) = 0.9$, and $\alpha = 1.2$, then $(Tf-Tg)(x,y) = -0.057(x+y)$, and so $\|Tf-Tg\|_{\infty} = 0.114 > 0.1 = \|f-g\|_{\infty}$. – 2017-01-05

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