Prove there is a unique $y:[0,1] \to \mathbb{R}$ solving $y(x) = e^x + \frac{y(x^2)}{2}$ for $x \in [0,1].$ The title is the problem statement, but to reiterate,
Prove there is a unique $y:[0,1] \to \mathbb{R}$ solving $y(x) = e^x + \frac{y(x^2)}{2}$ for $x \in [0,1].$
Looking for hints/solutions, thanks in advance.

Edit 6/10/16: Progress?
If we define the operator $Ty(x) = e^x + \frac{y(x^2)}{2},$ then $|Ty(x) - Tz(x)| = 1/2|y(x^2) - z(x^2)|,$ which might be a contraction map (not exactly sure how)... then we could use a fixed point theorem to conclude that for some guess $y_0,$ the series  $T^N y_0(x) = \frac{y_0(x^{2^N})}{2} + \sum_{n=0}^N \frac{\exp(x^{2^n})}{2^n}$ converges to $y(x)$?

Could anybody confirm if one is allowed to claim this
$$ |Ty(x) - Tz(x)| = 1/2|y(x^2) - z(x^2)|
$$
 is a contraction map?  On one hand, it look like
$$ |Ty - Tz| = 1/2|y - z|,
$$
but on the other, the arguments are not the same for $y, z$ on both sides of the equation.  
 A: As you suspect, the tool you want to use here is the Banach fixed point theorem. Consider the space $C[0, 1]$ of real valued continuous functions on $[0, 1]$ with the sup norm, which makes $C[0, 1]$ into a complete metric space with metric $d$. Define
$$f\colon C[0, 1] \to C[0, 1] \text{ by } f(y(x)) = e^{x} + \frac{y(x^{2})}{2}$$
Then for any $y_{1}, y_{2} \in C[0, 1]$,
\begin{align*}
d(f(y_{1}), f(y_{2})) = & \bigg\lvert\bigg\lvert \left(e^{x} + \frac{y_{1}(x^{2})}{2}\right) - \left(e^{x} + \frac{y_{2}(x^{2})}{2}\right) \bigg\rvert\bigg\rvert \\
= & \bigg\lvert\bigg\lvert \frac{y_{1}(x^{2})}{2} - \frac{y_{2}(x^{2})}{2} \bigg\rvert\bigg\rvert \\  
= & \frac{1}{2} \sup_{x\in [0, 1]} |y_{1}(x^{2}) - y_{2}(x^{2})| \text{; since } x\mapsto x^{2} \text{ is a bijection on } [0, 1], \\
= & \frac{1}{2} \sup_{x\in [0, 1]} |y_{1}(x) - y_{2}(x)| \\
= & \frac{1}{2} d(y_{1}(x), y_{2}(x))
\end{align*}
Hence, $f$ is a contraction with constant (say) $L = 2/3 < 1$. Thus, since $C[0, 1]$ is complete, by the Banach fixed point theorem, $f$ has a unique fixed point, i.e. there exists unique $y \in C[0, 1]$ such that $y(x) = f(y(x)) = e^{x} + y(x^{2})/2$ for all $x \in [0, 1]$.  
