Weierstrass Function: Continuous and Bounded on $\mathbb{R}$

[mathmari]

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

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? (Wondering)

 

[I like Serena]

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. (Nod)

 

[mathmari]

Great! Thank you so much!

 

[mathmari]

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? :unsure:

 

[I like Serena]

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}
:geek:

 

[mathmari]

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? :unsure:
Can we do that because the series converges?

 

[I like Serena]

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]

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 $.

:unsure:

 

[I like Serena]

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 $M_k=\left (\frac{3}{4}\right )^k$?