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

Question

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.

Answer 1

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.)

Answer 2

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:

• Wikipedia: Summation of Grandi's series#Exponential spacing
• Hardy (1907) "On certain oscillating series", especially page 277
• Keating and Reade (2000) "Summability of alternating gap series" (click on PDF to view article)
• Duren (2012) Invitation to Classical Analysis Ch. 7 "Tauberian Theorems" (PDF preview), especially pages 190-193

Answer 3

$\>$$\>$$\>$$\>$$\>$$\>$A slightly stronger result is true.

$\>$$\>$$\>$$\>$$\>$$\>$If $f:\rightarrow[0,1]$, $$f \text{ is continuous at } 0\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>(1)$$ and $$f(x)+f(x^2)=x \text{, for }x \in (0,1),\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>(2)$$ then $$f(x) = \sum_{\displaystyle i=0}^{\displaystyle \infty} (-1)^{\displaystyle i}x^{2^{\displaystyle i}}, \text{for } x \in [0, 1)\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>(3)$$
and $f$ isn't continuous at $1$.

Proof. Taking the limit in ($2$) as $x\rightarrow 0^{+}$ and using ($1$), we have $f(0)=0.$ So, ($3$) holds if $x = 0$.

$\>\>\>\>\>\>$Suppose $x \in (0,1).$ By ($2$) and induction on $n$,$$f(x) = (-1)^{\displaystyle n+1}f(x^{2^{\displaystyle n+1}})+\sum_{\displaystyle i=0}^{\displaystyle n} (-1)^{\displaystyle i}x^{2^{\displaystyle i}}, \text{for } x \in (0,1).\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>(4)$$

Also, $\displaystyle \lim_{ n\to \infty}x^{2^{\displaystyle n}}=0$. Therefore, $\displaystyle \lim_{n\to \infty}f(x^{2^{\displaystyle n}})=f(0)=0$, by ($1$). Thus, $\displaystyle \lim_{n\to\infty}(-1)^{\displaystyle n}f(x^{2^{\displaystyle n}})=0.$ Since $\displaystyle \sum_{i=0}^{\infty}(-1)^{\displaystyle i}x^{2^{\displaystyle i}}$ is an alternating series whose terms decrease in absolute value to $0$, the series converges and the absolute value of the difference between the series and the sum of its first $n$ terms is at most the absolute value of its $(n + 1)^{\displaystyle st}$ term. Taking the limit in ($4$), as $n \to \infty$, proves ($3$) for $x\in (0,1)$. Further, since $x^2 \in (0, 1),$ $f(x^2) + f(x^4) = x^2$, by ($2$). Therefore, using ($2$) again, $$f(x) = f(x^4) + x - x^2 > f(x^4), \text{ for }x \in (0,1).\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>(5)$$

$\>\>\>\>\>$Let $w = 0.995$ (following the results of Yiorgos S. Smyrlis described above). To avoid round-off error, I obtained the following results using $$www.javascripter.net/math/calculators/100digitbigintcalculator.htm.$$ $w^{2^{\displaystyle 0}} = 0.995,w^{2^{\displaystyle 1}}=0.990025,0.980149<w^{2^{\displaystyle 2}}<0.980150$ and $0.960692<(0.980149)^{2}<w^{2^{\displaystyle 3}}<(0.980150)^{2}<0.960695$. Similarly, $0.922929<w^{2^{\displaystyle 4}}<0.922935, 0.851797<w^{2^{\displaystyle 5}}<0.851810, 0.725558<w^{2^{\displaystyle 6}}<0.725581, 0.526434<w^{2^{\displaystyle 7}}<0.526468, 0.277132<w^{2^{\displaystyle 8}}<0.277169, 0.076802<w^{2^{\displaystyle 9}}<0.076823,0.005898<w^{2^{\displaystyle 10}}<0.005902 \text{ and }w^{2^{\displaystyle 11}}<0.000035.\text{ Hence,}$

$\displaystyle \sum_{i=0}^{10}(-1)^{\displaystyle i}w^{2^{\displaystyle i}}>0.995-0.990025+0.980149-0.960695+0.922929-0.851810+0.725558-0.526468+0.277132-0.076823+0.005898=0.500845.\text{ By (}3\text{), }\displaystyle \sum_{i=0}^{10}(-1)^{\displaystyle i}w^{2^{\displaystyle i}}-f(w)<w^{2^{\displaystyle 11}}<0.000035\text{. Therefore,}$ $$f(w)>0.500845-0.000035=0.50081.\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>(6)$$

$\text{ Let } w(n) = w^{4^{\displaystyle -n}}, \text{ for }n \in \mathbb Z_{ \geq 0}$. Hence, $w(0) = w, w(n) \in (0, 1), (w(n + 1))^4 = w(n)$ and $\displaystyle \lim_{n\to \infty}w(n)=1.$ So, $f(w(n+1)) > f((w(n+1))^4)= f(w(n))$, by ($5$). By induction on $n$ and $\>\>\>\>$ ($6$), $$f(w(n))\geq f(w)>0.50081, \text {for }n \in \mathbb Z_{\geq 0}.\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\> (7)$$
$\>\>\>\>\>\>\text{Suppose }f \text { is continuous at }x = 1\text{. Taking the limit in (}2\text{) as }x \rightarrow 1^{-}\text{, yields }f(1) = 0.5.$ Therefore, $0.5 = f(1) = f(\displaystyle \lim_{n\to\infty}w(n))= \lim_{n\to\infty}f(w(n)),$ contradicting ($7$).