Understanding a proof in Spivak

This is from Spivak's chapter 23. He wants to show that the function $$f(x) = \sum_{n=1}^{\infty} \frac{1}{10^n}\{10^nx\}$$ is continuous everywhere, but differentiable nowhere

I have only an excerpt of the proof and that is all I need because my question lies in his choice of $h_m = 10^{-m}$. I know he wants a decreasing sequence such that it converges to $0$ as he stated, but why $10^{-m}$ in particular? Why not something like $1/m^2$? And why is the difference quotient $$\lim_{m\to\infty}\dfrac{f(a+h_m)-f(a)}{h_m}$$, but not

$$\lim_{n\to\infty}\dfrac{f(a+h_n)-f(a)}{h_n}$$ Why the dependence on a new number $m$?

Also why does he think it is sufficient for $0 < a \leq 1$? Finally, I may be wrong but he set $$\{ 10^n (a + h_m)\} + \{ 10^n a\} = 0$$

for the reason that the partial sums of the convergent sequence (though it is finite) must go to $0$?

EDIT: I will write out the rest of the proof later today.

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What page is this on? – Ink Jan 9 '13 at 4:41
I have an older edition. This is in chapter 23 under "Uniform Convergence and Power series". It should just be after the M-test. In my book, it is page 422 (I have the 2nd edition) – Hawk Jan 9 '13 at 4:43
It is directly after the M-test in my edition too; I believe I have the 3rd edition in digital and a hardcover 4th edition. – Gyu Eun Lee Jan 9 '13 at 5:03

Edit: To condense the comments I have put many of my explanations in the actual answer.

For why Spivak uses this particular sequence (hm): "if $n\geq m$, then $10^nh_m$ is an integer, so..." You get no such nice property with $1/m^2$. This sequence works particularly well with this $f$, but of course there are functions where it won't be so pretty. Spivak can't use $n$ because $n$ is the index in the series, and $n$ and $m$ have nothing to do with each other. Besides, those difference quotients denote the same number.

$0\leq a < 1$ is sufficient because if the decimal expansion of a can be written $a=a_0.a_1a_2a_3...$, then we can separate this as $a=\phi+\theta$, where $\phi=a_0$ is an integer and $\theta=0.a_1a_2a_3...$... is in $[0,1)$. Now $\{10^na\}=\{10^n(\phi+\theta)\}=\{10^n\phi+10^n\theta\}=\{10^n\theta\}$ because $10^n\phi$ is an integer, so only the "decimal part" of the number $a$ actually affects the value of $f(a)$.

1) Spivak knows that the sequence $(h_m)$ he uses here works because he has worked out the proof prior to writing it down. He does not explain how he came up with $(h_m)$, which is admittedly unusual for his expository style within this book. However, this is a common occurrence in mathematical exposition, where final products (e.g. the proofs in books, articles, etc.) tend to be polished and the details of how the proof was discovered are superfluous. Moreover, sometimes it is simpler to just let the reader see that a technique works than to explain where it came from, especially is the technique is basically a clever trick like the one used here.

2) Spivak chooses the index $m$ for $(h_m)$ and the difference quotient $\displaystyle \frac{f(a+h_m)-f(a)}{h_m}$ instead of the series index $n$ because $m$ and $n$ are unrelated, so it wouldn't do for the indices to be the same. Note that

$\displaystyle \frac{f(a+h_m)-f(a)}{h_m} = \sum_{n=1}^\infty \frac{1}{10^n}\frac{\{10^n(a+h_m)\}-\{10^na\}}{\pm 10^{-m}}$ and this is not the same thing as

$\displaystyle \frac{f(a+h_n)-f(a)}{h_n} = \sum_{n=1}^\infty \frac{1}{10^n}\frac{\{10^n(a+h_n)\}-\{10^na\}}{\pm 10^{-n}}$

because in the first equation, $m$ is fixed and does not change as you vary $n$ during the summation.

3) $\{10^n(a+h_m)\}-\{10^na\}=0$, $n \geq m$: why is this true? First, we get an idea of the $\{\cdot\}$ operation. $\{\cdot\}$ is the distance to the nearest integer: $\{1.3\}=\{1.7\}=0.3$, $\{1.1\}=\{1.9\}=0.1$, $\{n\}=0$ for any integer $n$, and $\{a+n\} = \{a\}$ for any integer $n$. Well, we note that $10^nh_m$ is an integer when $n \geq m$. Now, $\{10^n(a+h_m)\}=\{10^na+10^nh_m)\} = \{10^na\}$ since $10^nh_m$ is an integer.

4) $(a_m\pm 1)$ comes from $(a+h_m)$. $h_m=\pm 10^{-m}$, so if $a=a_0.a_1a_2a_3...a_m...$ then $a+h_m = (a_0.a_1a_2a_3...a_m...)\pm 10^{-m} = a_0.a_1a_2a_3...a_m\pm 1...$ .

5) Referring to the proof of non-differentiability for your other equation: $y_i$ most likely is shorthand for "$y_0$ or $y_1$." I would have written "$y_i$, $i=0,1$."

6) This step is a consequence of the reverse triangle inequality (you can find its statement and proof on Wikipedia).

We have $\displaystyle |f(y_0)-f(y_1)| = \left|\sum_{k=1}^\infty f_k(y_0)-f_k(y_1)\right| = \left|(f_n(y_0)-f_n(y_1)) - \left(\sum_{k \neq n} f_k(y_1)-f_k(y_0)\right)\right|\\ \geq \displaystyle \left| |f_n(y_0)-f_n(y_1)| - \left| \sum_{k \neq n} f_k(y_1)-f_k(y_0)\right| ~ \right|$ where the last part follows from the reverse triangle inequality. Now the last term is $\displaystyle \geq |f_n(y_0)-f_n(y_1)| - \left| \sum_{k \neq n} f_k(y_0)-f_k(y_1)\right| \geq |f_n(y_0)-f_n(y_1)| - \sum_{k \neq n} |f_k(y_0)-f_k(y_1)|$ via the regular triangle inequality, so we conclude that $\displaystyle |f(y_0)-f(y_1)| \geq |f_n(y_0)-f_n(y_1)| - \sum_{k \neq n} |f_k(y_0)-f_k(y_1)|$.

The next step applies the estimates we've obtained so far. The sum goes from $k=1$ to $k=n-1$ because for $k > n$, we have $f_k(y_0)=f_k(y_1)$, so $f_k(y_0)-f_k(y_1)=0$ and thus we can ignore all the terms in the sum with $k > n$. The $k=n$ term is not even in the sum, since we're summing over $k \neq n$; we took it out in the beginning in the $f_n(y_0)-f_n(y_1)$ term.

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Do you know why in the very end he has $(a_n \pm 1)$ in the end? – Hawk Jan 9 '13 at 5:12
Referring to the first comment I made under this post. If $a$ = $2$. I have a book that shows that it is not differentiable. Here is an excerpt. s9.postimage.org/n2xlu03z3/… why does it say that for $k > n$, $10^k y_i \pi$ is an integer multiple of $2\pi$ What is $i$ here? – Hawk Jan 10 '13 at 2:49
Same problem, the proof continues here. s9.postimage.org/dvfh7j9jz/…. Why are they substracting the sum? And why is it going from $k = 1$ to $n - 1$? Why $n-1$ in particular? – Hawk Jan 10 '13 at 3:37
Yes. They never really mentioned for $n$ is. But I assumed it was just some random integer strictly equal or greater than 1 – Hawk Jan 10 '13 at 4:05
I am more interested in this equality $\left|(f_n(y_0)-f_n(y_1)) - \left(\sum_{k \neq n} f_k(y_1)-f_k(y_0)\right)\right|$. How did you 'factor" the $n$th term? – Hawk Jan 10 '13 at 5:44