# How prove there is no continuous functions $f:[0,1]\to \mathbb R$, such that $f(x)+f(x^2)=x$.

Prove that there is no continuous functions $f:[0,1]\to R$, such that $$f(x)+f(x^2)=x.$$

My try. Assume that there is a continuous function with this property. Thus, for any $n\ge 1$ and all $x\in [0,1]$, \begin{align*} f(x)&=x-f(x^2)=x-\big(x^2-f(x^4)\big)=x-x^2+\big(x^4-f(x^8)\big)=\cdots\\ &=x-x^2+x^4-\cdots+(-1)^n\left(x^{2^n}-f\big(x^{2^{n+1}}\big)\right) \end{align*}

since $f(0)=0$ and $\displaystyle\lim_{n\to \infty}x^{2^{n+1}}=0$ for any $x\in(0,1)$,it follows by the continuity of $f$ that $\displaystyle\lim_{n\to \infty}f\big(x^{2^{n+1}}\big)=0$, hence $$f(x)=x-x^2+x^4-x^8+\cdots+(-1)^nx^{2^n}+\cdots$$ for any $x\in (0,1)$

Why do I have prove that there is exists such a continuous functions $f$? Maybe my example is wrong? Why?

if my method is wrong,then How prove this problem ? Thank you.

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Why is there an $x^6$ in the infinite sum ? – Amr Jan 26 '14 at 9:20
sorry,I have edit.Thank you – math110 Jan 26 '14 at 9:23
I think the problem is extending your power series to the endpoint $x=1$, in a continuous manner. – user1337 Jan 26 '14 at 9:27
It is actually a famous example, starting with your power series, that said power series does not extend continuously to $x=1$. – alex.jordan Jan 26 '14 at 9:37
See the following elementary proof of non-existence of $f$: [1]: math.harvard.edu/~elkies/Misc/sol8.html – Yiorgos S. Smyrlis Jan 26 '14 at 11:03

Clearly, for every $x\in[0,1)$,
$$f(x)=\sum_{n=0}^\infty (-1)^nx^{2^n}.$$ G. H. Hardy (“On certain oscillating series”, Quart. J. Math. 38 (1907), 269–288) proved that $f$ has inﬁnitely many, very small oscillations as $x\to 1^−$, see attached Figure,

and the limit of $f$, as $x\to 1^-$, does not exist. See Duren's book.

Update. For an elementary proof, it can be easily calculated that $f(0.995)>.5$ (in fact $f(.995)=.500881586206$). Let $x_0=0.995$, then $$f(x_0^{1/4})=x_0^{1/4}-f(x_0^{1/2})=x_0^{1/4}-x_0^{1/2}+f(x_0)>f(x_0).$$ In this way we obtain a strictly increasing sequence $x_n=x_0^{1/4^n}$, such that $x_n\to 1^-$ and $f(x_n)>f(x_0)>\frac{1}{2}$. At the same time $$f(x_n^2)=x_n-f(x_n)<1-\frac{1}{2}=\frac{1}{2},%\quad\text{and}\quad %f(x_{n+1}^2)=x_{n+1}-f(x_{n+1})=x_{n+1}-x_{n+1}^{1/2}+f(x_{n+1}^{1/2})$$ and hence the sequence $y_n=x_n^{2}\to 1^-$, and $f(y_n)<1/2$. Thus the limit $\lim_{x\to 1^-}f(x)$ does not exist. (See also Puzzle 8.)

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Hello,I can't see this book? and I have google this book,But I can't see this book PDF,so I can't read this book,can you post this solution ? Thank you – math110 Jan 26 '14 at 10:01
Wow, it readily extends to a special case of Ostrowski-Hadamard gap theorem! A nice argument. – Sangchul Lee Jan 27 '14 at 18:54
It will be great (and very helpful) if you can write the answer of Duren's book (if you have the time of course) – user119228 Mar 7 '14 at 17:15

The easy approach is to note that the convergence of $f$ at $1$ is equivalent to the Abel summability of a certain series, whose partial sums are bounded, so it's also equivalent to Cesaro summability, but the averaged partial sums clearly oscillate between $1/3$ and $2/3$, so it isn't Cesaro summable.

There are also harder approaches. For details, try these free resources:

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Thanks for the links! I like the Keating-article very much, it gives me a fresh view into things – Gottfried Helms Jan 29 '14 at 10:58