Uniform convergence defined recursively Define a sequence of functions $y_n(x)$ by
$y_0(x)=0$, $y_n(x) = Ty_{n-1}(x)$, where 
$$Tf(x) = x^3 + \int_0^x t^3 f(t)dt$$
Where I've shown that $T:B \rightarrow B$ is a contraction in the Banach Space
$ B = \{ f \in C([0,1]) : \|f\| := \sup_{[0,1]}\{|f(t)|e^{-x^2}\} < \infty\}$
I need to determine whether or not the sequence of functions defined recursively above converges uniformly or not. And if they do converge uniformly, to what?
I am unsure how to approach this problem, however I did try to compute a few terms of $y_n$ to see if anything familiar would pop up, and that was a little futile.
 A: The limit point of $y_n(x)$ is the fixed point of the contracting operator $T$. See the proof of the Banach fixed point theorem: http://en.wikipedia.org/wiki/Banach_fixed-point_theorem.
Indeed in the proof you find the fixed point of the operator considering exactly your sequence and proving that it is a Cauchy sequence in your Banach space, which, by force, must converge to the fixed point of your operator.
A: Since you have shown that $T$ is a contraction, $y_n$ will converge to some $y$ in the norm you defined. That is, 
$$\sup_{x\in [0,1]} e^{-x^2} |y_n(x) - y(x)| \to 0$$
But $e^{-x^2}\ge  e^{-1}$ on $[0,1]$, so you actually have
$$\sup_{x\in [0,1]} |y_n(x) - y(x)| \to 0$$
thus $y_n$ converges to $y$ uniformly. To find out $y$, note that $y$ is the fixed point of $T$, so 
\begin{equation}
\begin{split}
y(x) = x^3 + \int_0^x t^3 y(t) dt &\Rightarrow y'(x) = 3x^2 + x^3 y(x)\\
&\Rightarrow y'(x) - x^3 y(x) = 3x^2
\end{split}
\end{equation}
If you solve this, you got 
$$y(x) = 3e^{\frac{x^4}{4}} \int_0^x s^2 e^{-\frac{s^4}{4}}ds$$
It seems that you do not have a closed form to this $y$. 
