# Some questions about Banach's proof of the existence of continuous nowhere differentiable functions

Im reading Continuity and Category of Chapter11, Carothers' Real Analysis, 1ed. Here is a reading material at the end of this chapter, talking about Banach's proof of the existence of continuous nowhere differentiable functions,

I have 4 questions here,

1. I cannot understand the sentence in paragraph3 that is " In particular, any f∈$C$[0,1] having a right-hand derivative at most n in magnitude at even one point in [0, 1-(1/n)] is in $E_n$". I mean what is "at most n in magnitude "? Im not an english native speaker.

2. How did he guarantee that |f(x+h)-f(x)| =< nh for all 0 < h < 1-x ?

3. Why does the proof merely focus on right-hand derivatives?

4. What is the quotients involved in?

• @Matt E: This question has already been adequately answered, but I thought it would be worth while to mention a common misunderstanding about Baire category proofs of existence, at least some of them. Quite often these proofs are sufficiently constructive, and the Banach proof you mention is one such proof, that the proof provides an explicit construction of an object, in much the same way that a procedure that allows you to calculate the first $n$ decimal digits of a real number, for each positive integer $n,$ provides an explicit construction of that real number. continued – Dave L. Renfro Jan 28 '14 at 16:26
• @Matt E: (continuation) See, for instance, pp. 31-32 of Kuratowski's freely available 1980 paper Some remarks on the origins of of the theory of functions of a real variable and of the descriptive set theory. – Dave L. Renfro Jan 28 '14 at 16:30

• "at most $n$ in magnitude" means "of absolute value $\leq n$".
• If $f$ has a right hand derivative at some point $x$ whose absolute value $< n$, then $|f(x+h) - f(x)| \leq n h$ if $h$ is small enough, say $0 < h < epsilon$. On the other hand, if $\epsilon > 0$, then since $f$ is continuous, we can find $n' > 0$ so that $|f(x + h) - f(x)| \leq n'h$ for $\epsilon \leq h \leq 1 - x$ (just because $| f(x+h) - f(x)|$ is bounded on the interval $\epsilon \leq h \leq 1 - x$, as $f$ is continuous, and $h$ is bounded below by $\epsilon$). So, letting $n'' = \max\{n,n'\}$, we find that if $f$ has a right hand derivative $<n$ at some point, then $f \in E_{n''}$. (The authors actually asserts this with $n'' = n$ and with $\leq n$ rather than $< n$ as the condition; I don't see this right now, but I could well be missing something, and it doesn't matter anyway; in the end all we care about is that $f \in E$.)
• Focusing on right-hand derivatives gives an even stronger result (differentiable means that the two-sided derivative exists, and this implies that the right-hand derivatie exists). Presumably it also makes the analsis of the set $E$ a bit easier, because the set $E_n$ is simpler to define than if we tried to write down an analogous set that fited well with two-sided derivatives.
• "difference quotient" mean the expression $\bigl(f(x+h) - f(x)\bigr)/h$. "Right hand difference quotient" means we're taking $h > 0$. If $f$ is in $E_n$ then the abs. value of this is bounded by $n$, just by definition of $E_n$. So $f \in E$ if and only if it has bounded r.h.d.q.'s at some point $x \in [0,1).$ (Just take $n$ to be $\geq$ whatever the bound on the abs. value is.)