Let $f_1:[0,1]\to\mathbb{R}$, $f_1(0)=0$ be continuously differentiable function and $\lambda&gt;1$. Consider the sequence of function defined by  $f_k(x):=\lambda f_{k-1}(x/\lambda)$, $k≥2,$ , $x\in [0,1]$. Find the pointwise limit of the sequence of function.  My Approach:  By the recurring relation, we get,  $f_n(x)=\lambda^{n-1} f_1(x/\lambda)$. If we take the limit and using the continuity of $f_1$, we come across a situation of $\infty •0$. Please help me further.
Let $f_1 : [-1, 1] \rightarrow \mathbb{R};\: f_1(0) = 0 $ be a continuously differentiable function and $\lambda &gt; 1$. 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.  $\rule{17cm}{1pt}$  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  $$f_n\rightarrow\begin{cases}0&amp;x=0\\x\:f'_1(0) &amp;x\neq0\end{cases}=f $$
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