# Weierstrass function

#### mathmari

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Hey!! I am looking at the following example of a continuous function $f:\mathbb{R}\rightarrow \mathbb{R}$ that is not differentiable at any $x\in \mathbb{R}$.

For $x\in [-1,1]$ we define $\phi (x)=|x|$ and then we extend $\phi$ to the whole $\mathbb{R}$ such that $\phi (x+2)=\phi (x)$.
Then the desired $f:\mathbb{R}\rightarrow \mathbb{R}$ is $$f(x)=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )$$

Then $f$ is continuous and bounded on $\mathbb{R}$.
For $x_0\in \mathbb{R}$ and $k\in \mathbb{N}^{\star}$ take $\delta_k=\pm \frac{1}{2}4^k$, where the sign is chosen such that there is no integer between $4^kx_0$ and $4^k(x_0+\delta_k)$ (it is $4^k\mid \delta_k$).
Define $\gamma_n=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]$.
For $n>k$ it is $\gamma_n=0$ and for $0\leq n\leq k$ it is $|\gamma_n|\leq 4^n$.
Since $|\gamma_k|=4^k$ we have that $$\left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\sum_{n=0}^k\left (\frac{3}{4}\right )^n\gamma_n\right |\geq 3^k-\sum_{n=0}^{k-1}3^n=\frac{1}{2}\left (3^k+1\right )$$ so for $k\rightarrow +\infty$, $\delta_k\rightarrow 0$ and $3^k\rightarrow +\infty$, it follows that....

This is not very clear for me.

Let's start from the beginning.

Why do we define in that they the function $\phi$ ? Then $f$ is continuous and bounded. Could you give me a hint how we can see that? #### Klaas van Aarsen

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Hey mathmari !!

The function $\phi$ is just a helper function to define the actual function $f isn't it? Let's start with bounded. The function$\phi$has a range of$[0,1]$doesn't it? Isn't the series bounded above by a geometric series then? Will the series be convergent for any$x$? #### mathmari ##### Well-known member MHB Site Helper The function$\phi$is just a helper function to define the actual function$f isn't it? Ahh ok!

The function $\phi$ has a range of $[0,1]$ doesn't it?
I got stuck right now. We have at the definition that the range is $[0,1]$, yes. But at the definition we have that $x\in [-1,1]$, is $x$ from the same interval at $\phi \left (4^nx\right )$ ? Isn't the series bounded above by a geometric series then? Will the series be convergent for any $x$? So taking that the range of $\phi$ is in general $[0,1]$, then we have the following:
\begin{align*}0\leq \phi \left (4^nx\right ) \leq 1 & \Rightarrow 0\leq \left (\frac{3}{4}\right )^n \phi \left (4^nx\right ) \leq \left (\frac{3}{4}\right )^n \\ & \Rightarrow \sum_{n=0}^{\infty} 0\leq \sum_{n=0}^{\infty} \left (\frac{3}{4}\right )^n \phi \left (4^nx\right ) \leq \sum_{n=0}^{\infty} \left (\frac{3}{4}\right )^n \\ & \Rightarrow 0\leq f(x) \leq \frac{1}{1-\frac{3}{4}} \\ & \Rightarrow 0\leq f(x) \leq \frac{1}{\frac{1}{4}} \\ & \Rightarrow 0\leq f(x) \leq 4\end{align*}

Therefore $f$ is bounded! #### Klaas van Aarsen

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I got stuck right now. We have at the definition that the range is $[0,1]$, yes. But at the definition we have that $x\in [-1,1]$, is $x$ from the same interval at $\phi \left (4^nx\right )$ ?
Don't forget that we "extend $ϕ$ to the whole $\mathbb R$."
Doesn't that mean that $\phi(x)$ is defined for any $x$ in $\mathbb R$? .

So taking that the range of $\phi$ is in general $[0,1]$, then we have the following:
\begin{align*}0\leq \phi \left (4^nx\right ) \leq 1 & \Rightarrow 0\leq \left (\frac{3}{4}\right )^n \phi \left (4^nx\right ) \leq \left (\frac{3}{4}\right )^n \\ & \Rightarrow \sum_{n=0}^{\infty} 0\leq \sum_{n=0}^{\infty} \left (\frac{3}{4}\right )^n \phi \left (4^nx\right ) \leq \sum_{n=0}^{\infty} \left (\frac{3}{4}\right )^n \\ & \Rightarrow 0\leq f(x) \leq \frac{1}{1-\frac{3}{4}} \\ & \Rightarrow 0\leq f(x) \leq \frac{1}{\frac{1}{4}} \\ & \Rightarrow 0\leq f(x) \leq 4\end{align*}

Therefore $f$ is bounded!
Yep. Can we deduce somehow that the series converges for any $x$? #### mathmari

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Don't forget that we "extend $ϕ$ to the whole $\mathbb R$."
Doesn't that mean that $\phi(x)$ is defined for any $x$ in $\mathbb R$? .
I got stuck right now. Does this mean that for each $x\in \mathbb{R}$ the range is $[0,1]$ ? Can we deduce somehow that the series converges for any $x$? From the comparison test with the series $\sum_{n=0}^{\infty} \left (\frac{3}{4}\right )^n$ do we not have that that the series converges? Or do we not look in that way if the series converges for any $x$ ? #### Klaas van Aarsen

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I got stuck right now. Does this mean that for each $x\in \mathbb{R}$ the range is $[0,1]$ ?
We extend $\phi$ such that $\phi(x+2)=\phi(x)$ don't we?
So what would for instance $\phi(x)$ be for $x=1.5$? From the comparison test with the series $\sum_{n=0}^{\infty} \left (\frac{3}{4}\right )^n$ do we not have that that the series converges? Or do we not look in that way if the series converges for any $x$ ?
The comparison tells us that the series has a lower bound of 0 and an upper bound of 4.
That is not sufficient for convergence though.
For instance a series like $2-1+1-1+1-1+\ldots$ is at all times within the same bounds, but it is not convergent is it? #### mathmari

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We extend $\phi$ such that $\phi(x+2)=\phi(x)$ don't we?
So what would for instance $\phi(x)$ be for $x=1.5$? We have that $\phi (1.5)=\phi(-0.5)=|0.5|=0.5\in [0,1]$. So we can apply that formula so many times till the result is in $[0,1]$, right? The comparison tells us that the series has a lower bound of 0 and an upper bound of 4.
That is not sufficient for convergence though.
For instance a series like $2-1+1-1+1-1+\ldots$ is at all times within the same bounds, but it is not convergent is it? Ok... What do we have to do then? Which convergence criteria do we have to use? #### Klaas van Aarsen

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We have that $\phi (1.5)=\phi(-0.5)=|0.5|=0.5\in [0,1]$. So we can apply that formula so many times till the result is in $[0,1]$, right?
Yep. Ok... What do we have to do then? Which convergence criteria do we have to use?
Isn't there a theorem that says that an increasing sequence that has an upper bound is convergent? #### mathmari

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Isn't there a theorem that says that an increasing sequence that has an upper bound is convergent? Ohh yes We have the following, or not?
\begin{align*}f(x+1)&=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x+1)\right ) =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n\right )\\ & =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n-2\right )\\ & =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n-2-2\right )=\ldots \\ & =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n-2^{2n}\right )=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )\\ & =f(x)\end{align*} #### Klaas van Aarsen

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We have the following, or not?
\begin{align*}f(x+1)&=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x+1)\right ) =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n\right )\\ & =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n-2\right )\\ & =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n-2-2\right )=\ldots \\ & =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n-2^{2n}\right )=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )\\ & =f(x)\end{align*}
Yes. #### mathmari

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Yes. So since $f(x+1)=f(x)$ it also holds that $f(x+1)\geq f(x)$, which implies that $f$ is increasing.

So since $f$ is increasing and bounded, the series converges.

Does it follow from that that $f$ is continuous? #### Klaas van Aarsen

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So since $f(x+1)=f(x)$ it also holds that $f(x+1)\geq f(x)$, which implies that $f$ is increasing.

So since $f$ is increasing and bounded, the series converges.
Hold on. We have a series to evaluate $f(x)$ at a specific value of $x$, don't we?
I'm afraid that the value of $f(x)$ at other values of $x$ is not involved.
Moreover, we cannot tell that $f$ itself is increasing from that, and actually it is not. Let's examine $f$ at a specific value of $x$. Say $x_0$.
So we have:
$$f(x_0)=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right ) =\phi(x_0) + \frac 34\phi(4x_0)+\left (\frac{3}{4}\right )^2\phi(4^2x_0)+\ldots$$
We know that each partial sum has a lower bound of 0 and an upper bound of 4.
And each term is non-negative isn't it?
So the sequence of partial sums is increasing - all for a specific value of $x=x_0$ - isn't it?
Doesn't that mean that the series converges so that $f(x_0)$ is well-defined? Does it follow from that that $f$ is continuous?
Let's start with continuity of $f$ at some point $x_0$.

First off, we need that $f(x_0)$ is well-defined.
That is, if we evaluate the series at $x=x_0$ it must converge to some value.

The next step is that we evaluate $\lim\limits_{x\to x_0} f(x)$.
Is it the same as $f(x_0)$? #### mathmari

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We have a series to evaluate $f(x)$ at a specific value of $x$, don't we?
I'm afraid that the value of $f(x)$ at other values of $x$ is not involved.
Moreover, we cannot tell that $f$ itself is increasing from that, and actually it is not. Let's examine $f$ at a specific value of $x$. Say $x_0$.
So we have:
$$f(x_0)=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right ) =\phi(x_0) + \frac 34\phi(4x_0)+\left (\frac{3}{4}\right )^2\phi(4^2x_0)+\ldots$$
We know that each partial sum has a lower bound of 0 and an upper bound of 4.
And each term is non-negative isn't it?
So the sequence of partial sums is increasing - all for a specific value of $x=x_0$ - isn't it?
Doesn't that mean that the series converges so that $f(x_0)$ is well-defined? So we have that for a specific value of $x$ the series is increasing and bounded and so it converges. The resulting function must be then well-defined? Let's start with continuity of $f$ at some point $x_0$.

First off, we need that $f(x_0)$ is well-defined.
That is, if we evaluate the series at $x=x_0$ it must converge to some value.

The next step is that we evaluate $\lim\limits_{x\to x_0} f(x)$.
Is it the same as $f(x_0)$? We have that $$\lim\limits_{x\to x_0} f(x)=\lim\limits_{x\to x_0}\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\lim\limits_{x\to x_0}\phi \left (4^nx\right )$$
Does it hold that $\lim\limits_{x\to x_0}\phi \left (4^nx\right )=\phi \left (4^nx_0\right )$ ? #### Klaas van Aarsen

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So we have that for a specific value of $x$ the series is increasing and bounded and so it converges. The resulting function must be then well-defined?
Yes. It means that $f(x)$ is defined for every $x\in\mathbb R$.

So we now have that $f$ is a function with domain $\mathbb R$.
And we have that $f$ is bounded. We have that $$\lim\limits_{x\to x_0} f(x)=\lim\limits_{x\to x_0}\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\lim\limits_{x\to x_0}\phi \left (4^nx\right )$$
Does it hold that $\lim\limits_{x\to x_0}\phi \left (4^nx\right )=\phi \left (4^nx_0\right )$ ?
Yep. #### mathmari

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Yes. It means that $f(x)$ is defined for every $x\in\mathbb R$.

So we now have that $f$ is a function with domain $\mathbb R$.
And we have that $f$ is bounded. Ahh ok! Yep. Ok! So this means that $f$ is continuous at $x_0$.
Can we say that since this holds for any $x_0$, the function is continuous? #### Klaas van Aarsen

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Ok! So this means that $f$ is continuous at $x_0$.
Can we say that since this holds for any $x_0$, the function is continuous?
Yep. #### mathmari

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For $x_0\in \mathbb{R}$ and $k\in \mathbb{N}^{\star}$ take $\delta_k=\pm \frac{1}{2}4^k$, where the sign is chosen such that there is no integer between $4^kx_0$ and $4^k(x_0+\delta_k)$ (it is $4^k\mid \delta_k$).
Define $\gamma_n=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]$.
Is there a specific reason that we take $\delta_k$ in such a way? For $n>k$ it is $\gamma_n=0$ and for $0\leq n\leq k$ it is $|\gamma_n|\leq 4^n$.
Let $n>k$ then $n=k+m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k+m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^k}{2}\left [\phi \left (4^{k+m}(x_0\pm \frac{1}{2}4^k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^k}{2}\left [\phi \left (4^{k+m}x_0\pm \frac{1}{2}4^{2k+m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

Then $\frac{1}{2}4^{2k+m}$ is a multiple $2$ and so using several times the property $\phi (x+2)=\phi (x)$ it holds that $\phi \left (4^{k+m}x_0\pm \frac{1}{2}4^{2k+m}\right )=\phi \left (4^nx_0\right )$ and so we get that $\gamma_n=0$.

Is this part correct? How do we get that $|\gamma_n|\leq 4^n$ if $0\leq n\leq k$ ? #### Klaas van Aarsen

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Is there a specific reason that we take $\delta_k$ in such a way?
I'm guessing we are trying to prove that $f$ is not differentiable anywhere?
Then we want a $\delta >0$ that is 'small' and gives us some special result depending on whether $n$ is greater than $k$ or smaller.

Anyway, it says: "the sign is chosen such that there is no integer between $4^kx_0$ and $4^k(x_0+δ_k)$".
But there are always integers in between, aren't there? Can it be that there is a typo?
Was perhaps $\delta_k=\pm \frac{1}{2}4^{-k}$ intended? How do we get that $|\gamma_n|\leq 4^n$ if $0\leq n\leq k$ ?
That would become true if we have $\delta_k=\pm \frac{1}{2}4^{-k}$ instead wouldn't it? #### mathmari

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I'm guessing we are trying to prove that $f$ is not differentiable anywhere?
Then we want a $\delta >0$ that is 'small' and gives us some special result depending on whether $n$ is greater than $k$ or smaller.

Anyway, it says: "the sign is chosen such that there is no integer between $4^kx_0$ and $4^k(x_0+δ_k)$".
But there are always integers in between, aren't there? Can it be that there is a typo?
Was perhaps $\delta_k=\pm \frac{1}{2}4^{-k}$ intended? Ahh because with $\delta_k=\pm \frac{1}{2}4^{k}$ we would have $4^kx_0$ and $4^k(x_0+δ_k)=4^kx_0+\frac{1}{2}4^{2k}$ and between them there are integers.

If we have $\delta_k=\pm \frac{1}{2}4^{-k}$ then we get $4^kx_0$ and $4^k(x_0+δ_k)=4^kx_0+\frac{1}{2}$ which is possible that in between there are no integers.

So you are right that it must be $\delta_k=\pm \frac{1}{2}4^{-k}$ intended. So, we have the following:

Let $n>k$ then $n=k+m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k+m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^{k+m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

Then $\frac{1}{2}4^{m}$ is a multiple of $2$ and so using several times the property $\phi (x+2)=\phi (x)$ it holds that $\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )=\phi \left (4^nx_0\right )$ and so we get that $\gamma_n=0$.

Is this part correct? For $0\leq n\leq k$ then $n=k-m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k-m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^{k-m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

How do we ontinue? #### Klaas van Aarsen

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Ahh because with $\delta_k=\pm \frac{1}{2}4^{k}$ we would have $4^kx_0$ and $4^k(x_0+δ_k)=4^kx_0+\frac{1}{2}4^{2k}$ and between them there are integers.

If we have $\delta_k=\pm \frac{1}{2}4^{-k}$ then we get $4^kx_0$ and $4^k(x_0+δ_k)=4^kx_0+\frac{1}{2}$ which is possible that in between there are no integers.

So you are right that it must be $\delta_k=\pm \frac{1}{2}4^{-k}$ intended. So, we have the following:

Let $n>k$ then $n=k+m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k+m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^{k+m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

Then $\frac{1}{2}4^{m}$ is a multiple of $2$ and so using several times the property $\phi (x+2)=\phi (x)$ it holds that $\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )=\phi \left (4^nx_0\right )$ and so we get that $\gamma_n=0$.

Is this part correct?
Shouldn't we have:
$$\frac{1}{\delta_k}=\frac{1}{\pm \frac 12 4^{-k}}=\pm 2\cdot 4^k$$
? For $0\leq n\leq k$ then $n=k-m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k-m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^{k-m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

How do we continue?
Isn't the slope of $\phi$ either $+1$ or $-1$ everywhere?
And it only flips when we pass through an integer doesn't it? We picked $\delta_k$ such that there was no integer in between, so it wouldn't flip in between would it? #### mathmari

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Shouldn't we have:
$$\frac{1}{\delta_k}=\frac{1}{\pm \frac 12 4^{-k}}=\pm 2\cdot 4^k$$
? Ahh yes!

So, we have the following:

Let $n>k$ then $n=k+m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k+m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm2 \cdot 4^{k}\left [\phi \left (4^{k+m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm2 \cdot 4^{k}\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

Then $\frac{1}{2}4^{m}$ is a multiple of $2$ and so using several times the property $\phi (x+2)=\phi (x)$ it holds that $\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )=\phi \left (4^nx_0\right )$ and so we get that $\gamma_n=0$.

Right? Isn't the slope of $\phi$ either $+1$ or $-1$ everywhere?
And it only flips when we pass through an integer doesn't it? We picked $\delta_k$ such that there was no integer in between, so it wouldn't flip in between would it? Do you mean that the derivative is either $1$ or $-1$ and so we have that ? For $0\leq n\leq k$ then $n=k-m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k-m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm 2\cdot 4^k\left [\phi \left (4^{k-m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm 2\cdot 4^k\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

From IVT we have that $$\phi' (\xi)=\frac{\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )}{4^nx_0\pm \frac{1}{2}4^{-m}-4^nx_0}=\frac{\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )}{\pm \frac{1}{2}4^{-m}}=\pm 2\cdot 4^m\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]$$ for some $4^nx_0<\xi <4^nx_0\pm \frac{1}{2}4^{-m}$.

So we get $$\pm 2 \left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]=\frac{\phi '(\xi)}{4^m}$$

Therefore we get $$\gamma_n=4^k\cdot \frac{\phi '(\xi)}{4^m}=4^{k-m}\cdot \phi '(\xi)=4^{n}\cdot \phi '(\xi) \Rightarrow |\gamma_n|=4^n\cdot |\phi '(\xi)|\leq 4^n\cdot 1=4^n$$

#### Klaas van Aarsen

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Ahh yes!

So, we have the following:

Let $n>k$ then $n=k+m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k+m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm2 \cdot 4^{k}\left [\phi \left (4^{k+m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm2 \cdot 4^{k}\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

Then $\frac{1}{2}4^{m}$ is a multiple of $2$ and so using several times the property $\phi (x+2)=\phi (x)$ it holds that $\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )=\phi \left (4^nx_0\right )$ and so we get that $\gamma_n=0$.

Right? Do you mean that the derivative is either $1$ or $-1$ and so we have that ? For $0\leq n\leq k$ then $n=k-m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k-m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm 2\cdot 4^k\left [\phi \left (4^{k-m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm 2\cdot 4^k\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

From IVT we have that $$\phi' (\xi)=\frac{\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )}{4^nx_0\pm \frac{1}{2}4^{-m}-4^nx_0}=\frac{\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )}{\pm \frac{1}{2}4^{-m}}=\pm 2\cdot 4^m\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]$$ for some $4^nx_0<\xi <4^nx_0\pm \frac{1}{2}4^{-m}$.

So we get $$\pm 2 \left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]=\frac{\phi '(\xi)}{4^m}$$

Therefore we get $$\gamma_n=4^k\cdot \frac{\phi '(\xi)}{4^m}=4^{k-m}\cdot \phi '(\xi)=4^{n}\cdot \phi '(\xi) \Rightarrow |\gamma_n|=4^n\cdot |\phi '(\xi)|\leq 4^n\cdot 1=4^n$$
Yep. All correct. #### mathmari

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Yep. All correct. For $n=k$ we have the following:
\begin{align*}\gamma_k&=\frac{1}{\delta_k}\left [\phi \left (4^k(x_0+\delta_k)\right )-\phi \left (4^kx_0\right )\right ]\\ & =\pm 2\cdot 4^k\left [\phi \left (4^{k}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^kx_0\right )\right ]\\ & =\pm 2\cdot 4^k\left [\phi \left (4^kx_0\pm \frac{1}{2}\right )-\phi \left (4^kx_0\right )\right ]\end{align*}

From IVT we get \begin{equation*}\phi' (\xi)=\frac{\phi \left (4^kx_0\pm \frac{1}{2}\right )-\phi \left (4^kx_0\right )}{4^kx_0\pm \frac{1}{2}-4^kx_0}=\frac{\phi \left (4^kx_0\pm \frac{1}{2}\right )-\phi \left (4^kx_0\right )}{\pm \frac{1}{2}}=\pm 2\cdot \left [\phi \left (4^kx_0\pm \frac{1}{2}\right )-\phi \left (4^kx_0\right )\right ]\end{equation*}

So we get \begin{equation*}\gamma_k=4^k\cdot \phi' (\xi) \Rightarrow |\gamma_k|=|4^k\cdot \phi' (\xi)|=|4^k|\cdot |\phi' (\xi)|\end{equation*} Isn't it $|\phi '(\xi)|\leq 1$ ? How do we get $|\gamma_k|=4^k$ ? #### Klaas van Aarsen

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So we get \begin{equation*}\gamma_k=4^k\cdot \phi' (\xi) \Rightarrow |\gamma_k|=|4^k\cdot \phi' (\xi)|=|4^k|\cdot |\phi' (\xi)|\end{equation*} Isn't it $|\phi '(\xi)|\leq 1$ ? How do we get $|\gamma_k|=4^k$ ?
Don't we have that $|\phi'(\xi)|=1$ everywhere? Except that it's not defined for integers? #### mathmari

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Don't we have that $|\phi'(\xi)|=1$ everywhere? Except that it's not defined for integers? But at post #21 how did we get the inequality $|\gamma_n|\leq 4^n$ ? Shouldn't we get there also the equality then? I got stuck right now. 