# Weierstrass function

#### Klaas van Aarsen

##### MHB Seeker
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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.
I believe you are right. For $n<k$ there won't be an integer in between either, so we get equality as well.

Still, doesnt $|\gamma_n|=4^n$ imply that $|\gamma_n|\le 4^n$?
And doesn't it suffice for the purpose of the proof that we have the inequality?

#### mathmari

##### Well-known member
MHB Site Helper
I believe you are right. For $n<k$ there won't be an integer in between either, so we get equality as well.

Still, doesnt $|\gamma_n|=4^n$ imply that $|\gamma_n|\le 4^n$?
And doesn't it suffice for the purpose of the proof that we have the inequality?
Ok! I just wanted to understand the differnece of $|\gamma_n|$and $|\gamma_k|$, why at the one we had inequality and at the other one equality.

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....
Do we apply here the IVT for $f$ ?

If yes, then we have for $x_0<\xi <x_0+\delta_k$ that \begin{align*}f'(\xi)&=\frac{f(x_0+\delta_k)-f(x_0)}{x_0+\delta_k-x_0}=\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\\ \Rightarrow |f'(\xi)| &=\left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\frac{\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x_0+\delta_k)\right )-\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )}{\delta_k}\right | \\ & = \left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\frac{\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]}{\delta_k}\right | =\left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | \\ & = \left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\cdot 0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\end{align*}

Is it correct so far?

We have that $|\gamma|\leq 4^n$, so isn't the inequality symbol above at $$\left |\sum_{n=0}^k\left (\frac{3}{4}\right )^n\gamma_n\right |\geq 3^k-\sum_{n=0}^{k-1}3^n$$ wrong?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Do we apply here the IVT for $f$ ?

If yes, then we have for $x_0<\xi <x_0+\delta_k$ that \begin{align*}f'(\xi)&=\frac{f(x_0+\delta_k)-f(x_0)}{x_0+\delta_k-x_0}=\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\\ \Rightarrow |f'(\xi)| &=\left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\frac{\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x_0+\delta_k)\right )-\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )}{\delta_k}\right | \\ & = \left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\frac{\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]}{\delta_k}\right | =\left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | \\ & = \left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\cdot 0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\end{align*}
We can't because we do not know if $f'$ exists.
Worse, we are trying to prove that $f'$ does not exist, not for any value of $x$, aren't we?

Instead I believe we should just substitute our results so far, which amounts to the same thing, except that we can't say that it is equal to $|f'(\xi)|$.

We have that $|\gamma|\leq 4^n$, so isn't the inequality symbol above at $$\left |\sum_{n=0}^k\left (\frac{3}{4}\right )^n\gamma_n\right |\geq 3^k-\sum_{n=0}^{k-1}3^n$$ wrong?
Don't we have for $n < k$:
$$\left |\left (\frac{3}{4}\right )^n\gamma_n\right | \le 3^n \implies \left (\frac{3}{4}\right )^n\gamma_n \ge -3^n$$

#### mathmari

##### Well-known member
MHB Site Helper
We can't because we do not know if $f'$ exists.
Worse, we are trying to prove that $f'$ does not exist, not for any value of $x$, aren't we?

Instead I believe we should just substitute our results so far, which amounts to the same thing, except that we can't say that it is equal to $|f'(\xi)|$.
So do we take the right side of IVT and we show that this doesn't converge and so it cannot be equal to the derivative?

Don't we have for $n < k$:
$$\left |\left (\frac{3}{4}\right )^n\gamma_n\right | \le 3^n \implies \left (\frac{3}{4}\right )^n\gamma_n \ge -3^n$$
We have the following:
\begin{align*}&\frac{f(x_0+\delta_k)-f(x_0)}{x_0+\delta_k-x_0}=\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\\ & \Rightarrow \left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\frac{\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x_0+\delta_k)\right )-\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )}{\delta_k}\right | \\ & = \left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\frac{\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]}{\delta_k}\right | =\left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | \\ & = \left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\cdot 0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\end{align*}

So for $0\leq n\leq k$ we have that $|\gamma_n|\leq 4^n$ so we get that \begin{align*}-4^n\leq \gamma_n\leq 4^n &\Rightarrow -\left (\frac{3}{4}\right )^n4^n\leq \left (\frac{3}{4}\right )^n\gamma_n\leq \left (\frac{3}{4}\right )^n4^n \\ & \Rightarrow -3^n\leq \left (\frac{3}{4}\right )^n\gamma_n\leq 3^n \\ & \Rightarrow \sum_{n=0}^{k}\left (-3^n\right )\leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \sum_{n=0}^{k}3^n \\ & \Rightarrow -\sum_{n=0}^{k}3^n \leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \sum_{n=0}^{k}3^n \\ & \Rightarrow -\frac{3^{k+1}-1}{2} \leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \frac{3^{k+1}-1}{2}\end{align*}

Doesn't this means that $$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\leq \frac{3^{k+1}-1}{2}$$ ? I got stuck how we get the other inequality symbol.

Or do we take it as follows?

$$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\geq -\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So do we take the right side of IVT and we show that this doesn't converge and so it cannot be equal to the derivative?
More specifically, we have a sub sequence and we are showing that it diverges.
Consequently that means that the limit that we need for $f'$ does not exist, and thus $f'$ does not exist.

We have the following:
\begin{align*}&\frac{f(x_0+\delta_k)-f(x_0)}{x_0+\delta_k-x_0}=\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\\ & \Rightarrow \left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\frac{\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x_0+\delta_k)\right )-\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )}{\delta_k}\right | \\ & = \left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\frac{\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]}{\delta_k}\right | =\left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | \\ & = \left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\cdot 0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\end{align*}

So for $0\leq n\leq k$ we have that $|\gamma_n|\leq 4^n$ so we get that \begin{align*}-4^n\leq \gamma_n\leq 4^n &\Rightarrow -\left (\frac{3}{4}\right )^n4^n\leq \left (\frac{3}{4}\right )^n\gamma_n\leq \left (\frac{3}{4}\right )^n4^n \\ & \Rightarrow -3^n\leq \left (\frac{3}{4}\right )^n\gamma_n\leq 3^n \\ & \Rightarrow \sum_{n=0}^{k}\left (-3^n\right )\leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \sum_{n=0}^{k}3^n \\ & \Rightarrow -\sum_{n=0}^{k}3^n \leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \sum_{n=0}^{k}3^n \\ & \Rightarrow -\frac{3^{k+1}-1}{2} \leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \frac{3^{k+1}-1}{2}\end{align*}

Doesn't this means that $$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\leq \frac{3^{k+1}-1}{2}$$ ? I got stuck how we get the other inequality symbol.

Or do we take it as follows?

$$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\geq -\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n$$
Can't we split it up like:
$$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right | \geq \left |\left (\frac{3}{4}\right )^k\gamma_k\right | -\sum_{n=0}^{k-1}\left (\frac{3}{4}\right )^n|\gamma_n|$$

#### mathmari

##### Well-known member
MHB Site Helper
Can't we split it up like:
$$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right | \geq \left |\left (\frac{3}{4}\right )^k\gamma_k\right | -\sum_{n=0}^{k-1}\left (\frac{3}{4}\right )^n|\gamma_n|$$
Ahh ok! So we split it in that way to get a lower bound that goes to infinity if $k\rightarrow \infty$ and so the limit of IVT that we need for $f'$ does not exist, and thus $f'$ does not exist and so the proof is complete, right?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Ahh ok! So we split it in that way to get a lower bound that goes to infinity if $k\rightarrow \infty$ and so the limit of IVT that we need for $f'$ does not exist, and thus $f'$ does not exist and so the proof is complete, right?
Yep.

#### mathmari

##### Well-known member
MHB Site Helper
Great!! Thank you so much!!

#### mathmari

##### Well-known member
MHB Site Helper
For $x\in [-1,1]$ we define $\phi (x)=|x|$ and then we extend $\phi$ to the whole $\mathbb{R}$ such that $\phi (x+2)=\phi (x)$.
How does this look graphically?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
How does this look graphically?
\begin{tikzpicture}
\draw[help lines] (-4,-2) grid (4,2);
\draw[->] (-4.4,0) -- (4.4,0);
\draw[->] (0,-2.2) -- (0,2.2);
\draw foreach \i in {-4,...,4} { (\i,0.1) -- (\i,-0.1) node[below] {$\i$} };
\draw foreach \i in {-2,...,2} { (0.1,\i) -- (-0.1,\i) node[ left ] {$\i$} };
\draw[domain=-4.2:4.2, variable=\x, blue, ultra thick, samples=200] plot ({\x}, {abs(\x-floor((\x-1)/2)*2-2)}) node[above right] {$\phi$};
\end{tikzpicture}

#### mathmari

##### Well-known member
MHB Site Helper
We have that $$\lim\limits_{x\to x_0} f(x)=\lim\limits_{x\to x_0}\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\lim\limits_{x\to x_0}\phi \left (4^nx\right )$$
Is it correct that we can take the limit into the sum?
Can we do that because the series converges?

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Is it correct that we can take the limit into the sum?
Can we do that because the series converges?
It's because Tannery's theorem says:

Let $S_n = \sum\limits_{k=0}^\infty a_k(n)$ and suppose that $\lim\limits_{n\rightarrow\infty} a_k(n) = b_k$. If $|a_k(n)| \le M_k$ and $\sum\limits_{k=0}^\infty M_k < \infty$, then $\lim\limits_{n\rightarrow\infty} S_n = \sum_{k=0}^{\infty} b_k$.

Can we find such $M_k$?

#### mathmari

##### Well-known member
MHB Site Helper
It's because Tannery's theorem says:

Let $S_n = \sum\limits_{k=0}^\infty a_k(n)$ and suppose that $\lim\limits_{n\rightarrow\infty} a_k(n) = b_k$. If $|a_k(n)| \le M_k$ and $\sum\limits_{k=0}^\infty M_k < \infty$, then $\lim\limits_{n\rightarrow\infty} S_n = \sum_{k=0}^{\infty} b_k$.

Can we find such $M_k$?
So we want to find an upper bound of $\left |\left (\frac{3}{4}\right )^n\phi (4^nx)\right |$, right?
We have that $\left |\left (\frac{3}{4}\right )^n\phi (4^nx)\right |\leq \left |\left (\frac{3}{4}\right )^n\right |\cdot \left |\phi (4^nx)\right |\leq 1\cdot 1=1$ or not?
But this $M_k$ does not satisfy the condition $\sum\limits_{k=0}^\infty M_k < \infty$.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So we want to find an upper bound of $\left |\left (\frac{3}{4}\right )^n\phi (4^nx)\right |$, right?
We have that $\left |\left (\frac{3}{4}\right )^n\phi (4^nx)\right |\leq \left |\left (\frac{3}{4}\right )^n\right |\cdot \left |\phi (4^nx)\right |\leq 1\cdot 1=1$ or not?
But this $M_k$ does not satisfy the condition $\sum\limits_{k=0}^\infty M_k < \infty$.
How about $M_k=\left (\frac{3}{4}\right )^k$?