# Pointwise limit of the sequence of continuously differentiable functions defined inductively.

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

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

• 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 $x = 0$. – Paul Frost Nov 19 '18 at 22:48

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
$$\tag 1 f_{n+1}(x) = x\,\frac{f_1(x/\lambda^n)}{x/\lambda^n} = x\,\frac{f_1(x/\lambda^n)-f_1(0)}{x/\lambda^n}.$$
• @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 $g(0) = f'_1(0)$. – Paul Frost Nov 19 '18 at 22:44