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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.

Merkh
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    What have you determined so far about such a function? – Cameron Buie Jun 09 '16 at 17:40
  • I've got a feeling that I should use a fixed point theorem to show uniqueness, but what have I determined about $y(x)?$ Not very much. – Merkh Jun 09 '16 at 17:43
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    Well, first of all you can directly calculate $y(0)$ and $y(1)$ using just algebra. In the interior, notice that $y(x)=e^x+y(x^2)/2=e^x+(e^{x^2}+y(x^4)/2)/2 = \dots$. Then notice that $\lim_{n \to \infty} x^{2^n}=0$... – Ian Jun 09 '16 at 17:44
  • Is $y(x)$ to be assumed continuous, maybe differentiable, or no restrictions on it? – coffeemath Jun 09 '16 at 17:46
  • Yes, what good is this series with a $y(x^{2n})$ term if we are trying to show such a $y$ exists? – Merkh Jun 09 '16 at 17:46
  • No other information is provided, if you can determine $y(x)$ is continuous from the problem statement, it may be used – Merkh Jun 09 '16 at 17:46
  • What I described tells you how to do the problem under pretty mild assumptions (for instance, it's enough to have $y$ bounded in a neighborhood of zero). That's because you get $y(x)$ equal to some explicit function plus $y(x^{2^k})/2^k$, and that $2^k$ blows up. – Ian Jun 09 '16 at 17:48
  • I think if you recursively write down what $y(x)$ is as in an earlier comment, one would get the series $y(x) = \sum_{n=1}^\infty e^{x^{2^n}}/2^n = e^x + e^{x^2}/2 + e^{x^4}/4 + e^{x^8}/8 + \cdots,$ which seems convergent for $x \in [0,1],$ could be wrong though. – Merkh Jun 09 '16 at 18:03
  • You actually have $y(x)=\sum_{n=0}^N \frac{e^{x^{2^n}}}{2^n} + \frac{y \left (x^{2^N} \right )}{2^N}$ for all $N$. Provided that second term goes to zero, we get $y(x)=\sum_{n=0}^\infty \frac{e^{x^{2^n}}}{2^n}$ which is indeed convergent for any $x \in [0,1]$. But I don't see how to deduce that from the problem statement. – Ian Jun 09 '16 at 18:45

1 Answers1

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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 $x \in [0, 1]$.

Alex Wertheim
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  • This is along the lines of what I had done, but my confusion was in the step from $\frac{1}{2} \sup_{x\in [0, 1]} |y_{1}(x^{2}) - y_{2}(x^{2})| \text{; since } x\mapsto x^{2} \text{ is a bijection on } [0, 1],$ $= \frac{1}{2} \sup_{x\in [0, 1]} |y_{1}(x) - y_{2}(x)|.$ Is this truly an equality? It seems weird to me that we could replace the $x^2$ with any bijection on $[0,1].$ Thanks for the answer though, much appreciated! – Merkh Jun 22 '16 at 09:45
  • @Merkh: indeed, I felt a bit guilty posting an answer, since you had already done most of the work. :) I believe that line is correct: if f $f \colon [0, 1] \to [0, 1]$ is a bijection, then the collection of values $A = {y_{1}(x) - y_{2}(x)}{x \in [0, 1]}$ should be the same as $B = {y{1}(f(x)) - y_{2}(f(x))}_{x \in [0, 1]}$, so we should have $\sup A = \sup B$. Is this more convincing? – Alex Wertheim Jun 22 '16 at 17:44
  • Hmm yes that's enough to convince me, sounds good :) – Merkh Jun 22 '16 at 19:14