We define the sequence of functions $f_n:[0,1]\rightarrow \mathbb{R}$ by $$f_{n+1}(x)=\begin{cases}0 & \text{ if } x\in \left[ 0, \frac{1}{2n+3}\right ]\\ |2(n+1)x-1| & \text{ if } x\in \left [\frac{1}{2n+3}, \frac{1}{2n+1}\right ] \\ f_n(x) & \text{ if } x\in \left [\frac{1}{2n+1}, 1\right ] \end{cases}$$ where $f_1$ is given by $$f_1(x)=\begin{cases}0 & \text{ if } x\in \left [0, \frac{1}{3}\right ]\\ |2x-1| & \text{ if } x\in \left [\frac{1}{3}, 1\right ]\end{cases}$$
(a) Draw the graphs of the functions $f_n$.
(b) Show that the sequence of functions converges uniformly to a continuous function $f:[0,1]\rightarrow \mathbb{R}$.
(c) Calculate the lengths of the graphs of the functions $f_n$.
(d) Show that the graph of $f$ is not a rectifiable curve.
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At (a) we have:
$f_1$ :
$f_2$ :
$f_3$ :
So that will be the form of the graph of $f_n$, right?
At (b) we first show that $f_n$ converges pointwise to a function $f$, right? Taking the limit $n\rightarrow \infty$ then do we look at the intervals that we get at the definition of $f_{n+1}$ ?
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EDIT:
As for the uniform convergence, can we say the following?
From the graph of part (a) we see that for $n\rightarrow \infty$ the $f_n$ converges to a function $f(x)$ with $0\leq f(x)\leq x$, $x\in [0,1]$.
Then we have that: \begin{equation*}|f_n(x)-f(x)|\leq \begin{cases}|0-x| & \text{ if } x\in \left[ 0, \frac{1}{2(n-1)+3}\right ]\\ \left ||2nx-1|-x\right | & \text{ if } x\in \left [\frac{1}{2(n-1)+3}, \frac{1}{2(n-1)+1}\right ] \\ \left |f_{n-1}(x)-x\right | & \text{ if } x\in \left [\frac{1}{2(n-1)+1}, 1\right ] \end{cases}= \begin{cases}x & \text{ if } x\in \left[ 0, \frac{1}{2n+1}\right ]\\ \left ||2nx-1|-x\right | & \text{ if } x\in \left [\frac{1}{2n+1}, \frac{1}{2n-1}\right ] \\ \left |f_{n-1}(x)-x\right | & \text{ if } x\in \left [\frac{1}{2n-1}, 1\right ] \end{cases}\end{equation*} So we get \begin{equation*}\sup_{x\in [0,1]}|f_n(x)-f(x)|\leq \frac{1}{2n+1}\end{equation*} The right side goes to $0$ if $n\rightarrow \infty$.
Therefore $f_n$ converges uniformly to the continuous function $f(x)$.
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At (c) do we just use the formlaof the length and then calculate some terms, for $n=1$, $n=2$, $n=3$ and see that the result is an increasing function and if $n\rightarrow \infty$ then the length goes also to $\infty$ ?
The plotting of the sequence of $\{f_n\}$ gives us this motivation that the limit function $f$ to which $\{f_n\}$ tend is equal to $f_1(x)$ for $x\in[\frac{1}{3},1]$, $f_2(x)$ for $x\in[\frac{1}{5},\frac{1}{3}]$, ... and $f_n(x)$ for $x\in[\frac{1}{2n+1},\frac{1}{2n-1}]$ for $n\in\mathbb N$. Due to the falling maximum of $f_n(x)$ over $[\frac{1}{2n+1},\frac{1}{2n-1}]$, this implies both the pointwise and uniform convergence as constructing the limit function $f(x)$ in interval $[0,\frac{1}{2n+1}]$ does not affect our construction in $[\frac{1}{2n+1},1]$.
(c) and (d) are straightforward by a simple calculation.
P.S.
Plotting the functions was the cleverest part of the solution, I think :)