On the Banach space $(C([0,1]), ||.||_\infty)$, consider the operator given by $$(Tf)(x)=\int_0^1 x^2tf(t) dt+1.$$
a.) Prove that $Tf$ is continuous for all $f\in C([0,1])$.
b.) Show that $T$ is a contraction mapping.
c.) Let $f_n(x)=\frac{2}{3}\left(1-\frac{1}{4^{n-1}}\right)x^2+1$, where $n\geq 1$. What is the solution of $$f(x)=\int_0^1 x^2tf(x) dt +1?$$
For a.) I say that we need to show that $T$ maps $(C([0,1]), ||.||_\infty)$ to itself.
So, $x\longrightarrow 1$ is constant and enables us to write,
$$|1|.\left|\int_0^1x^2tf(t)dt-\int_0^1x'^2tf(t)dt\right|=|x^2-x'^2|.\left|\int_0^1tf(t)dt\right|.$$
However, I think that instead of saying $|x^2-x'^2|$, we should say $|x^2|$? Is this correct? And then as $f$ is continuous, $tf(t)$ is continuous and integrable and so $Tf$ is continuous.
b.) To show $T$ is a contraction, we say $$\left|Tf(x)-Tg(x)\right| = \left|\left|x^2\int_0^1t(f(t)-g(t))\right|\right|_\infty\leq ||f-g||_\infty.$$
I am uncertain if I am missing a coefficient in front of $||f-g||_\infty$?
I would then say that $||Tf-Tg||_\infty\leq ||f-g||_\infty$.
Lastly, for c.), I think we should use Banach's fixed point theorem? And this would allow us to say $f(x)=?$ Or use Picard iteration?
Please let me know if you notice any flaws and help me rectify them.
Thank you
Hints: Note that $$ (Tf)(x) = \left(\int_0^1 tf(t)\,dt \right)x^2 + 1 $$ For b), note that $$ \left|Tf(x)-Tg(x)\right| = \left\|x^2\int_0^1t(f(t)-g(t))dt\right\|_\infty\leq \left\| x^2 \int_0^1 t(\|f-g\|_\infty)\,dt \right\|_\infty = \cdots ? $$
For c): in order to find a fixed point, solve the equation $$ \int_0^1 t(at^2 + 1)\,dt = a $$ for $a$. Use some theorem to show that the corresponding fixed point is unique.