derivativesreal-analysis
A book of mine says the following is true, and I am having some trouble proving it. (I've considered using the Lebesgue differentiation theorem and absolute continuity, as well as elementary analysis methods.)
Let $f: [0, \infty) \rightarrow \mathbb{R}$ be continuous and have right derivatives at each point in the domain, with the right derivative function being continuous. Then $f$ is differentiable.
What follows is taken (mostly) from more extensive discussions in the following sci.math posts:
http://groups.google.com/group/sci.math/msg/814be41b1ea8c024 [23 January 2000]
http://groups.google.com/group/sci.math/msg/3ea26975d010711f [6 November 2006]
http://groups.google.com/group/sci.math/msg/05dbc0ee4c69898e [20 December 2006]
Note: The term interval is restricted to nondegenerate intervals (i.e. intervals containing more than one point).
The continuity set of a derivative on an open interval $J$ is dense in $J.$ In fact, the continuity set has cardinality $c$ in every subinterval of $J.$ On the other hand, the discontinuity set $D$ of a derivative can have the following properties:
$D$ can be dense in $\mathbb R$.
$D$ can have cardinality $c$ in every interval.
$D$ can have positive measure. (Hence, the function can fail to be Riemann integrable.)
$D$ can have positive measure in every interval.
$D$ can have full measure in every interval (i.e. measure zero complement).
$D$ can have a Hausdorff dimension zero complement.
$D$ can have an $h$-Hausdorff measure zero complement for any specified Hausdorff measure function $h.$
More precisely, a subset $D$ of $\mathbb R$ can be the discontinuity set for some derivative if and only if $D$ is an $F_{\sigma}$ first category (i.e. an $F_{\sigma}$ meager) subset of $\mathbb R.$
This characterization of the discontinuity set of a derivative can be found in the following references: Benedetto [1] (Chapter 1.3.2, Proposition, 1.10, p. 30); Bruckner [2] (Chapter 3, Section 2, Theorem 2.1, p. 34); Bruckner/Leonard [3] (Theorem at bottom of p. 27); Goffman [5] (Chapter 9, Exercise 2.3, p. 120 states the result); Klippert/Williams [7].
Regarding this characterization of the discontinuity set of a derivative, Bruckner and Leonard [3] (bottom of p. 27) wrote the following in 1966: Although we imagine that this theorem is known, we have been unable to find a reference. I have found the result stated in Goffman's 1953 text [5], but nowhere else prior to 1966 (including Goffman's Ph.D. Dissertation).
Interestingly, in a certain sense most derivatives have the property that $D$ is large in all of the ways listed above (#1 through #7).
In 1977 Cliff Weil [8] published a proof that, in the space of derivatives with the sup norm, all but a first category set of such functions are discontinuous almost everywhere (in the sense of Lebesgue measure). When Weil's result is paired with the fact that derivatives (being Baire $1$ functions) are continuous almost everywhere in the sense of Baire category, we get the following:
(A) Every derivative is continuous at the Baire-typical point.
(B) The Baire-typical derivative is not continuous at the Lebesgue-typical point.
Note that Weil's result is stronger than simply saying that the Baire-typical derivative fails to be Riemann integrable (i.e. $D$ has positive Lebesgue measure), or even stronger than saying that the Baire-typical derivative fails to be Riemann integrable on every interval. Note also that, for each of these Baire-typical derivatives, $\{D, \; {\mathbb R} - D\}$ gives a partition of $\mathbb R$ into a first category set and a Lebesgue measure zero set.
In 1984 Bruckner/Petruska [4] (Theorem 2.4) strengthened Weil's result by proving the following: Given any finite Borel measure $\mu,$ the Baire-typical derivative is such that the set $D$ is the complement of a set that has $\mu$-measure zero.
In 1993 Kirchheim [5] strengthened Weil's result by proving the following: Given any Hausdorff measure function $h,$ the Baire-typical derivative is such that the set $D$ is the complement of a set that has Hausdorff $h$-measure zero.
[1] John J. Benedetto, Real Variable and Integration With Historical Notes, Mathematische Leitfäden. Stuttgart: B. G. Teubne, 1976, 278 pages. [MR 58 #28328; Zbl 336.26001]
[2] Andrew M. Bruckner, Differentiation of Real Functions, 2nd edition, CRM Monograph Series #5, American Mathematical Society, 1994, xii + 195 pages. [The first edition was published in 1978 as Springer-Verlag's Lecture Notes in Mathematics #659. The second edition is essentially unchanged from the first edition with the exception of a new chapter on recent developments (23 pages) and 94 additional bibliographic items.] [MR 94m:26001; Zbl 796.26001]
[3] Andrew M. Bruckner and John L. Leonard, Derivatives, American Mathematical Monthly 73 #4 (April 1966) [Part II: Papers in Analysis, Herbert Ellsworth Slaught Memorial Papers #11], 24-56. [MR 33 #5797; Zbl 138.27805]
[4] Andrew M. Bruckner and György Petruska, Some typical results on bounded Baire $1$ functions, Acta Mathematica Hungarica 43 (1984), 325-333. [MR 85h:26004; Zbl 542.26004]
[5] Casper Goffman, Real Functions, Prindle, Weber & Schmidt, 1953/1967, x + 261 pages. [MR 14,855e; Zbl 53.22502]
[6] Bernd Kirchheim, Some further typical results on bounded Baire one functions, Acta Mathematica Hungarica 62 (1993), 119-129. [94k:26008; Zbl 786.26002]
[7] John Clayton Klippert and Geoffrey Williams, On the existence of a derivative continuous on a $G_{\delta}$, International Journal of Mathematical Education in Science and Technology 35 (2004), 91-99.
[8] Clifford Weil, The space of bounded derivatives, Real Analysis Exchange 3 (1977-78), 38-41. [Zbl 377.26005]
Yes.
The keystone is:
Lemma. Let $f\colon [a,b]\to\mathbb R$ be continuous and assume that $f'_+(x)$ exists and is $>0$ for all $x\in [a,b)$. Then $f$ is strictly increasing.
Assume otherwise, i.e. $f(a)\ge f(b)$. We recursively define a map $g\colon \operatorname{Ord}\to [a,b)$ such that $g$ and $f\circ g$ are strictly inreasing. Since the class $\operatorname{Ord}$ of ordinals is a proper class and $g$ is injective, we arrive at a contradiction, thus showing the claim.
$\square$
Corollary 1. (something like a one-sided Rolle theorem) Let $f\colon [a,b]\to\mathbb R$ be continuous with $f(a)=f(b)$. Assume $f_+$ exists and is continuos in $[a,b)$. Then $f'_+(x)=0$ for some $x\in[a,b)$.
Proof. Assume otherwise. Then either $f_+(x)>0$ for all $x$ or $f_+(x)<0$ for all $x$. In the first case the lemma applies and gives us a contradiction to $f(a)=f(b)$; in the other case, we consider $-f$ instead of $f$. $\square$
Corollary 2. (something like a one-sided IVT) Let $f\colon [a,b]\to\mathbb R$ be continuous. Assume $f_+$ exists and is continuos in $[a,b)$. Then $f'_+(x)=\frac{f(b)-f(a)}{b-a}$ for some $x\in[a,b)$.
Proof. Apply the previous corollary to $f(x)-\frac{f(b)-f(a)}{b-a}x$. $\square$
By symmetry, we have
Corollary 3. Let $f\colon [a,b]\to\mathbb R$ be continuous. Assume $f_-$ exists and is continuos in $(a,b]$. Then $f'_-(x)=\frac{f(b)-f(a)}{b-a}$ for some $x\in(a,b]$. $\square$
Theorem. Let $f\in C(\mathbb R)$ be a function with $f'_-$ continuous on $\mathbb R$. Then $f\in C^1(\mathbb R)$.
Proof. Consider aribtrary $a\in \mathbb R$. Let $\epsilon>0$ be given. Then by continuity of $f'_-$, for some $\delta>0$ we have $|f'_-(x)-f'_-(a)|<\epsilon$ for all $x\in(a,a+\delta)$. Thus for $0<h<\delta$ we have $\left|\frac{f(a+h)-f(a)}{h}-f'_-(a)\right|<\epsilon$ by corollary 3. We conclude that $f'_+(a)=f'_-(a)$, i.e. $f$ is differentiable at $a$. $\square$
Best Answer
Let us denote the right derivative of $f$ by $g$.
Proof: Define $$L(a)=g(a)\quad\text{and}\quad L(x)=\frac{f(x)-f(a)}{x-a},\ x\in(a,b].$$ By definition, $L$ is continuous on $[a,b]$, and it suffices to show that for every $\delta>0$, $$E_\delta:=\Big\{x\in[a,b]\,\Big|\, m-\delta\le L(y)\le M+\delta, \forall y\in[a,x] \Big\}=[a,b].$$ By definition and the continuity of $L$, we know that $E_\delta=[a,c]$ for some $c\in[a,b]$, and from $m\le g(a)\le M$ we know $c>a$. Then from $c\in E_\delta $ and $m\le g(c)\le M$ it is easy to see that $c<b$ is impossible. Therefore, $c=b$ and the lemma follows. $\quad\square$
Now let us show that $f$ is differentiable for any $x>0$. Since $g$ is continuous, given $0<h<x$, we can define $$m_h=\min_{y\in[x-h,x]}g(y),\quad M_h=\max_{y\in[x-h,x]}g(y),$$ and we know that $$\lim_{h\to 0^+}m_h=\lim_{h\to 0^+}M_h=g(y).$$ Due to the lemma, for $a=x-h$, $b=x$, $m=m_h$ and $M=M_h$, we have $$m_h\le\frac{f(x-h)-f(x)}{-h}\le M_h.$$ Let $h\to 0^+$, it follows that the left derivative of $f$ at $x$ exists and is equal to $g(x)$, i.e. $f$ is differentiable at $x$. $\quad\square$