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Let $f_1 : [-1, 1] \rightarrow \mathbb{R};\: f_1(0) = 0 $ be a continuously differentiable function and $\lambda > 1$. Consider the sequence of functions defined inductively by $f_k(x) := \lambda f_{k-1}\left(\dfrac{x}{\lambda}\right);\: k \geq 2;\: x\in [-1, 1]$. Find the pointwise limit of the sequence of functions $(f_n)$.

I figured out that $f_k(0)=0 \:\forall\: k\implies f_k\rightarrow 0 $ when $ x=0$. Also $f_k$ is continuously differentiable $\forall \:k$.

$\Bigg($Since $\quad\displaystyle\lim_{h\rightarrow0}\dfrac{f_2(x+h)-f_2(x)}{h}=\lambda\lim_{h\rightarrow0}\dfrac{f_1\left(\frac{x+h}{\lambda}\right)-f_1\left(\frac{x}{\lambda}\right)}{h}.\: $Since$ f_1 $is continuously differentiable we see $ f_2 $ is differentiable and its derivative is continuous and hence $f_k$. $\Bigg)$.

Further since $f_1(0)=0$. So I assumed $f_1(x)=x\cdot g(x)$, where $g(x)$ is continuously differentiable and $g(0)\neq0$. Assume $x\neq0,\:f_2(x)=\lambda\cdot\dfrac{x}{\lambda}g\left(\dfrac{x}{\lambda}\right)=x\cdot g\left(\dfrac{x}{\lambda}\right)\implies f_3(x)=x\cdot g\left(\dfrac{x}{\lambda^2}\right)$ $\implies f_k(x)=x\cdot g\left(\dfrac{x}{\lambda^{k-1}}\right)$.

So taking limit as $k\rightarrow \infty$, we get $\displaystyle \lim_{k\rightarrow\infty}f_k(x)=x\cdot g(0)$. Now I am not sure how to proceed or whether there is something wrong here. Please provide hints or suggestions.

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To sum up (using the arguments made by $\textbf{zhw}$ and $\textbf{Paul Frost}$)

When $\displaystyle x\neq0, \lim_{n\rightarrow\infty}f_{n+1}(x) = \lim_{n\rightarrow\infty}x\,\frac{f_1(x/\lambda^n)}{x/\lambda^n} = \lim_{n\rightarrow\infty}x\,\frac{f_1(x/\lambda^n)-f_1(0)}{x/\lambda^n}=x\:f'(0).$ Thus

$$f_n\rightarrow\begin{cases}0&x=0\\x\:f'_1(0) &x\neq0\end{cases}=f $$

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  • $\begingroup$ You do not need $g$ continuously differentiable. This in general not true (also $g(0) \ne 0$ is in general not true). Your approach shows that it suffices to know that $g$ is continuous at $x = 0$. $\endgroup$ – Paul Frost Nov 19 '18 at 22:48
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It looks to me like you have the right answer, but I don't think defining $g$ is necessary. Additionally, I think we need only assume $f_1'(0)$ exists.

Hint: First verify $f_{n+1}(x) = \lambda^n f_1(x/\lambda^n),$ which you can do by induction. Suppose $x\in [-1,1]\setminus\{0\}.$ Then

$$\tag 1 f_{n+1}(x) = x\,\frac{f_1(x/\lambda^n)}{x/\lambda^n} = x\,\frac{f_1(x/\lambda^n)-f_1(0)}{x/\lambda^n}.$$

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    $\begingroup$ @YadatiKiran Your approach is very simliar to zhw's. For $x \ne 0$ define $g(x) = \frac{f_1(x)}{x} = \frac{f_1(x) - f_1(0)}{x - 0}$. This is a continuous function on $[-1,1] \setminus \{ 0 \}$. But $f_1$ is differentiable at $x = 0$, hence $g$ has a unique continuous extension to $[-1,1]$ by taking $g(0) = f'_1(0)$. $\endgroup$ – Paul Frost Nov 19 '18 at 22:44

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