# Uniform convergence - Length of graph

• MHB
• mathmari
In summary, we have a sequence of functions $f_n$ defined on the interval $[0,1]$, with each function $f_n$ having three different cases depending on the value of $x$. We can use a graphing tool such as Sage to plot the functions $f_1$, $f_2$, and $f_3$. We can also see that as $n$ increases, the functions $f_n$ converge pointwise to a function $f(x)=x$. We can show that the sequence of functions converges uniformly by taking the supremum of $|f_n(x)-f(x)|$ and showing that it approaches $0$ as $n$ increases. Lastly, we can calculate the
mathmari
Gold Member
MHB
Hey! :giggle:

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.Could you give me a hint for (a) ? How do we draw these graphs?

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 $f_{n+1}$ ?

:unsure:

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mathmari said:
Could you give me a hint for (a) ? How do we draw these graphs?

Hey mathmari!

It's a bit awkward to draw them with the usual online graphing tools.
Instead we can use for instance sage to do it.
It is online available on for instance https://cocalc.com.
This page shows how to plot the Cantor set, which is similar to what we want.

mathmari said:
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 $f_{n+1}$ ?

The function doesn't change anymore on $\left[\frac{1}{2n+1},\,1\right]$ does it?

Klaas van Aarsen said:
The function doesn't change anymore on $\left[\frac{1}{2n+1},\,1\right]$ does it?

Do you mean because for $n\rightarrow \infty$ we get the interval $[0,1]$ ? :unsure:

mathmari said:
Do you mean because for $n\rightarrow \infty$ we get the interval $[0,1]$ ?
Yes.
So for any interval $(\varepsilon, 1]$ we can find that the function does not change for sufficiently large $n$.

Btw, I could plot the function with sage and the following code:
Code:
def f(x, n):
if n == 1:
if x <= 1/3:
return 0
if x <= 1:
return abs(2 * x - 1)
return 0
if x <= 1 / (2 * n + 3):
return 0
if x <= 1 / (2 * n + 1):
return abs(2 * (n + 1) * x - 1)
if x <= 1:
return f(x, n - 1)
return 0

plot(lambda x: f(x, 100))

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Klaas van Aarsen said:
Yes.
So for any interval $(\varepsilon, 1]$ we can find that the function does not change for sufficiently large $n$.

Does that mean that the function is then $0$ everywhere, as in the first interval? :unsure:

mathmari said:
Does that mean that the function is then $0$ everywhere, as in the first interval?

No, the function is generally not zero. (Shake)

Klaas van Aarsen said:
No, the function is generally not zero. (Shake)

But when $n$ is large enough what is then the function? I got stuck right now. :unsure:

mathmari said:
But when $n$ is large enough what is then the function? I got stuck right now.

How about we draw $f_1, f_2,$ and $f_3$ as requested in part (a)? (Sweating)
We can use pencil and paper.

Klaas van Aarsen said:
How about we draw $f_1, f_2,$ and $f_3$ as requested in part (a)? (Sweating)
We can use pencil and paper.

We have $$f_1(x)=\begin{cases}0 & \text{ if } x\in \left [0, \frac{1}{3}\right ]\\ |2x-1| & \text{ if } x\in \left [\frac{1}{3}, 1\right ]\end{cases}$$ and the graph is:

$$f_{2}(x)=\begin{cases}0 & \text{ if } x\in \left[ 0, \frac{1}{5}\right ]\\ |4x-1| & \text{ if } x\in \left [\frac{1}{5}, \frac{1}{3}\right ] \\ f_1(x) & \text{ if } x\in \left [\frac{1}{3}, 1\right ] \end{cases}=\begin{cases}0 & \text{ if } x\in \left[ 0, \frac{1}{5}\right ]\\ |4x-1| & \text{ if } x\in \left [\frac{1}{5}, \frac{1}{3}\right ] \\ |2x-1| & \text{ if } x\in \left [\frac{1}{3}, 1\right ] \end{cases}$$
and the graph is:

$$f_{3}(x)=\begin{cases}0 & \text{ if } x\in \left[ 0, \frac{1}{7}\right ]\\ |6x-1| & \text{ if } x\in \left [\frac{1}{7}, \frac{1}{5}\right ] \\ f_2(x) & \text{ if } x\in \left [\frac{1}{5}, 1\right ] \end{cases}=\begin{cases}0 & \text{ if } x\in \left[ 0, \frac{1}{7}\right ]\\ |6x-1| & \text{ if } x\in \left [\frac{1}{7}, \frac{1}{5}\right ] \\ |4x-1| & \text{ if } x\in \left [\frac{1}{5}, \frac{1}{3}\right ] \\ |2x-1| & \text{ if } x\in \left [\frac{1}{3}, 1\right ] \end{cases}$$
and the graph is:

So that will be the form of the graph of the functions $f_n$ (as we get at the plot at sage).
So is the functon $f$ of the form $$\begin{cases}|2x-1| \\ |4x-1| \\ |6x-1| \\ \ldots\end{cases}$$ ? :unsure:

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Or do we have for $n\rightarrow \infty$ that $f_n\rightarrow x$ ?
That means that $f_n$ converges pointwise to $f(x)=x$, $x\in [0,1]$, right?
Now we have to check if $\sup |f_n(x)-f(x)|\rightarrow 0$, right?

We have that \begin{equation*}|f_n(x)-f(x)|= \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*}

How do we get the supremum? Is the supremum at the first case because at the other we have $x$ and subtraction but in the first case we have just $x$ ?

:unsure:

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As for (c) are the lengths of $f_n$ equal to \begin{align*}s&=\int_0^1\sqrt{1+\left [f_n'(x)\right ]^2}\, dx\\ & =\int_0^{\frac{1}{2n+1}}\sqrt{1+\left [f_n'(x)\right ]^2}\, dx+\int_{\frac{1}{2n+1}}^{\frac{1}{2n-1}}\sqrt{1+\left [f_n'(x)\right ]^2}\, dx+\int_{\frac{1}{2n-1}}^1\sqrt{1+\left [f_n'(x)\right ]^2}\, dx \\ & =\int_0^{\frac{1}{2n+1}}\sqrt{1+\left [0\right ]^2}\, dx+\int_{\frac{1}{2n+1}}^{\frac{1}{2n-1}}\sqrt{1+\left [2n\right ]^2}\, dx+\int_{\frac{1}{2n-1}}^1\sqrt{1+\left [f_{n-1}'(x)\right ]^2}\, dx \\ & =\int_0^{\frac{1}{2n+1}}1\, dx+\int_{\frac{1}{2n+1}}^{\frac{1}{2n-1}}\sqrt{1+4n^2}\, dx+\int_{\frac{1}{2n-1}}^1\sqrt{1+\left [f_{n-1}'(x)\right ]^2}\, dx \\ & =\left [x\right ]_0^{\frac{1}{2n+1}}+\sqrt{1+4n^2}\cdot \left [x\right ]_{\frac{1}{2n+1}}^{\frac{1}{2n-1}}+\int_{\frac{1}{2n-1}}^1\sqrt{1+\left [f_{n-1}'(x)\right ]^2}\, dx \\ & =\frac{1}{2n+1}+\sqrt{1+4n^2}\cdot \left (\frac{1}{2n-1}-\frac{1}{2n+1}\right )+\int_{\frac{1}{2n-1}}^1\sqrt{1+\left [f_{n-1}'(x)\right ]^2}\, dx \\ & =\frac{1}{2n+1}+\sqrt{1+4n^2}\cdot \frac{1}{4n^2-1}+\int_{\frac{1}{2n-1}}^1\sqrt{1+\left [f_{n-1}'(x)\right ]^2}\, dx \end{align*}
Or do we have to calculate for specific $n$ and then derive a formula? :unsure:

mathmari said:
So that will be the form of the graph of the functions $f_n$ (as we get at the plot at sage).
So is the functon $f$ of the form $$\begin{cases}|2x-1| \\ |4x-1| \\ |6x-1| \\ \ldots\end{cases}$$ ?
Yep. (Nod)

My sage graph for $n=100$ is:

Klaas van Aarsen said:
Yep. (Nod)

My sage graph for $n=100$ is:

View attachment 11120
So is it wrong what I wrote in post #10? :unsure:

mathmari said:
Or do we have for $n\rightarrow \infty$ that $f_n\rightarrow x$ ?
That means that $f_n$ converges pointwise to $f(x)=x$, $x\in [0,1]$, right?
Now we have to check if $\sup |f_n(x)-f(x)|\rightarrow 0$, right?

We can see from the graph that $f_n$ does not converge to $x\mapsto x$. (Shake)
Instead it's a sawtooth function that gets more teeth at small values when $n$ increases. The teeth at higher values remain unchanged.
We can also see that $x\mapsto x$ is an upper bound and $x\mapsto 0$ is a lower bound.

Suppose $f$ be the unspecified function that $f_n$ converges to.
Then for any $0 <\delta < 1$ there is an $N$ such that for any $n>N$ and $x$ in $(\delta, 1]$ we have that $f_n(x)=f(x)$.
And for any $x$ in $[0,\delta]$ we have that $0\le f_n(x) \le \delta$, so $|f(x)-f_n(x)|\le \delta$.
Can we find uniform convergence from that? (Wondering)

Klaas van Aarsen said:
We can see from the graph that $f_n$ does not converge to $x\mapsto x$. (Shake)
Instead it's a sawtooth function that gets more teeth at small values when $n$ increases. The teeth at higher values remain unchanged.
We can also see that $x\mapsto x$ is an upper bound and $x\mapsto 0$ is a lower bound.

Suppose $f$ be the unspecified function that $f_n$ converges to.
Then for any $0 <\delta < 1$ there is an $N$ such that for any $n>N$ and $x$ in $(\delta, 1]$ we have that $f_n(x)=f(x)$.
And for any $x$ in $[0,\delta]$ we have that $0\le f_n(x) \le \delta$, so $|f(x)-f_n(x)|\le \delta$.
Can we find uniform convergence from that? (Wondering)

Ah so the supremum of $|f(x)-f_n(x)|$ is the maximum of $\delta$ and $0$, so $\delta$, or not? This $\delta$ is a fraction 1 to a number including n and if n gets large enough the fraction goes to 0, which implies uniform convergence, right? :unsure:

mathmari said:
As for (c) are the lengths of $f_n$ equal to \begin{align*}s&=\int_0^1\sqrt{1+\left [f_n'(x)\right ]^2}\, dx\end{align*}
Or do we have to calculate for specific $n$ and then derive a formula?
We can see from the graph that $f_n$ is not differentiable: it has corners and a jump.
That is, at every interval boundary it is not differentiable.
Instead we should find the lengths within each interval and sum them.
Can we find a series that sums those lengths?

Klaas van Aarsen said:
We can see from the graph that $f_n$ is not differentiable: it has corners and a jump.
That is, at every interval boundary it is not differentiable.
Instead we should find the lengths within each interval and sum them.
Can we find a series that sums those lengths?

Do we not write it as the sum I wrote it in #11 ?
Do you mean an other sum?

:unsure:

mathmari said:
Ah so the supremum of $|f(x)-f_n(x)|$ is the maximum of $\delta$ and $0$, so $\delta$, or not?
It's not the supremum. (Shake)
Instead it's an upper bound. The supremum must be somewhere between $0$ and $\delta$, and we don't have to know what it is exactly. It suffices that it is $\le \delta$.

mathmari said:
This $\delta$ is a fraction 1 to a number including n and if n gets large enough the fraction goes to 0, which implies uniform convergence, right?
True. Perhaps we should be a bit more formal though.
Something like: for any $\varepsilon > 0$ there is an $N$ such that...
... so that for any $x$ in $[0,1]$ we have that $\sup|f_n(x) -f(x)| < \varepsilon$.

mathmari said:
Do we not write it as the sum I wrote it in #11 ?
Do you mean an other sum?
Ah right.
I triggered on the use of $f_n'$ and the use of an integral on a domain with missing points...
I guess we can do that if we implicitly leave out the points of the domain where $f_n'$ is undefined, and if we use a more general integral than the Riemann integral. :unsure:

Then it looks as if your $s$ is actually an $s_n$, and we have to find what it converges to.
Oh, and I didn't check your calculations yet, but the approach seems correct.

Klaas van Aarsen said:
Perhaps we should be a bit more formal though.
Something like: for any $\varepsilon > 0$ there is an $N$ such that...
... so that for any $x$ in $[0,1]$ we have that $\sup|f_n(x) -f(x)| < \varepsilon$.

I got stuck right now. So we don't need to find a specific $f$ that $f_n$ converges pointwise?
We just describe what $f$ has to look like from the graph of $f_n$ ?

We have that the graph of $f_n$ is between $y=x$ and $y=0$. So does $f$ be also there? :unsure:

Klaas van Aarsen said:
Ah right.
I triggered on the use of $f_n'$ and the use of an integral on a domain with missing points...
I guess we can do that if we implicitly leave out the points of the domain where $f_n'$ is undefined, and if we use a more general integral than the Riemann integral. :unsure:

Then it looks as if your $s$ is actually an $s_n$, and we have to find what it converges to.
Oh, and I didn't check your calculations yet, but the approach seems correct.

I tried some examples for $n$.

For $n=1$ :
\begin{align*}s_1&=\int_0^1\sqrt{1+\left [f_n'(x)\right ]^2}\, dx\\ & = \int_0^{\frac{1}{3}}\sqrt{1+\left [f_n'(x)\right ]^2}\, dx+\int_{\frac{1}{3}}^1\sqrt{1+\left [f_n'(x)\right ]^2}\, dx\\ & = \int_0^{\frac{1}{3}}\sqrt{1+\left [0\right ]^2}\, dx+\int_{\frac{1}{3}}^1\sqrt{1+\left [2\right ]^2}\, dx \\ & = \int_0^{\frac{1}{3}}\sqrt{1}\, dx+\int_{\frac{1}{3}}^1\sqrt{1+4}\, dx\\ & = \int_0^{\frac{1}{3}}1\, dx+\int_{\frac{1}{3}}^1\sqrt{5}\, dx\\ & = \frac{1}{3}+\sqrt{5}\left (1-\frac{1}{3}\right ) \\ & = \frac{1}{3}+\sqrt{5}\cdot \frac{2}{3} \end{align*}

For $n=2$ :
\begin{align*}s_2&=\int_0^1\sqrt{1+\left [f_n'(x)\right ]^2}\, dx\\ & = \int_0^{\frac{1}{5}}\sqrt{1+\left [f_n'(x)\right ]^2}\, dx+\int_{\frac{1}{5}}^{\frac{1}{3}}\sqrt{1+\left [f_n'(x)\right ]^2}\, dx+\int_{\frac{1}{3}}^1\sqrt{1+\left [f_n'(x)\right ]^2}\, dx\\ & = \int_0^{\frac{1}{5}}\sqrt{1+\left [0\right ]^2}\, dx+\int_{\frac{1}{5}}^{\frac{1}{3}}\sqrt{1+\left [4\right ]^2}\, dx+\int_{\frac{1}{3}}^1\sqrt{1+\left [2\right ]^2}\, dx \\ & = \int_0^{\frac{1}{5}}\sqrt{1+0}\, dx+\int_{\frac{1}{5}}^{\frac{1}{3}}\sqrt{1+16}\, dx+\int_{\frac{1}{3}}^1\sqrt{1+4}\, dx \\ & = \int_0^{\frac{1}{5}}1\, dx+\int_{\frac{1}{5}}^{\frac{1}{3}}\sqrt{17}\, dx+\int_{\frac{1}{3}}^1\sqrt{5}\, dx \\ & = \frac{1}{5}+\sqrt{17}\left (\frac{1}{3}-\frac{1}{5}\right )+\sqrt{5}\left (1-\frac{1}{3}\right ) \\ & = \frac{1}{5}+\sqrt{17}\cdot \frac{2}{15}+\sqrt{5}\cdot \frac{2}{3} \end{align*} For $n=3$ :
\begin{align*}s_3&=\int_0^1\sqrt{1+\left [f_n'(x)\right ]^2}\, dx\\ & = \int_0^{\frac{1}{7}}\sqrt{1+\left [f_n'(x)\right ]^2}\, dx+\int_{\frac{1}{7}}^{\frac{1}{5}}\sqrt{1+\left [f_n'(x)\right ]^2}\, dx+\int_{\frac{1}{5}}^{\frac{1}{3}}\sqrt{1+\left [f_n'(x)\right ]^2}\, dx+\int_{\frac{1}{3}}^1\sqrt{1+\left [f_n'(x)\right ]^2}\, dx\\ & = \int_0^{\frac{1}{7}}\sqrt{1+\left [0\right ]^2}\, dx+\int_{\frac{1}{7}}^{\frac{1}{5}}\sqrt{1+\left [6\right ]^2}\, dx+\int_{\frac{1}{5}}^{\frac{1}{3}}\sqrt{1+\left [4\right ]^2}\, dx+\int_{\frac{1}{3}}^1\sqrt{1+\left [2\right ]^2}\, dx \\ & = \int_0^{\frac{1}{7}}\sqrt{1}\, dx+\int_{\frac{1}{7}}^{\frac{1}{5}}\sqrt{37}\, dx+\int_{\frac{1}{5}}^{\frac{1}{3}}\sqrt{17}\, dx+\int_{\frac{1}{3}}^1\sqrt{5}\, dx \\ & = \frac{1}{7}+\sqrt{37}\cdot \frac{2}{35}+\sqrt{17}\cdot \frac{2}{15}+\sqrt{5}\cdot \frac{2}{3}\end{align*}

Does this help us to check where $s_n$ converges? :unsure:

As for the uniform convergence, can we not 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 $f(x)$.

:unsure:

## 1. What is uniform convergence?

Uniform convergence is a type of convergence in which the limit of a sequence of functions is independent of the choice of point in the domain. This means that the rate of convergence is the same for all points in the domain, and the functions are said to converge uniformly to their limit.

## 2. What is the significance of uniform convergence?

Uniform convergence is important because it guarantees that the limit function will also be continuous, which is a desirable property in many mathematical and scientific applications. It also allows for easier manipulation of limits and integrals.

## 3. How is uniform convergence different from pointwise convergence?

Pointwise convergence only requires that the limit of a sequence of functions exists at each point in the domain, while uniform convergence requires that the limit function is the same for all points in the domain. This means that uniform convergence is a stronger form of convergence.

## 4. How is the length of a graph related to uniform convergence?

The length of a graph is a measure of the distance between the points on the graph. In uniform convergence, the distance between the points on the graph approaches zero as the sequence of functions converges to its limit. This means that the length of the graph also approaches zero, indicating that the functions are getting closer and closer to their limit.

## 5. How is uniform convergence used in real-world applications?

Uniform convergence is used in many areas of mathematics and science, such as in the study of differential equations, Fourier series, and numerical analysis. It is also used in physics and engineering to model and analyze real-world phenomena.

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