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Given a probability/measure space $(S, \Sigma, \mu) = ([0,1], \mathscr B([0,1]), \lambda)$, let random variable/measurable function $f$ and its derivative $f' \in \mathscr L^{1} (S, \Sigma, \mu)$.

Suppose f' is continuous.

Using Dominated Convergence Theorem to evaluate

$$\lim_{n \to \infty }\int_{S} f'(\frac{nx}{nx+1}) \cos(f(x)) dx$$

I guess we need to find $g$ s.t.

  1. $$|f'(\frac{nx}{nx+1}) \cos(f(x))| \le g$$

  2. $$g \in \mathscr L^{1} (S, \Sigma, \mu)$$

So:

$$|f'(\frac{nx}{nx+1}) \cos(f(x))|$$

$$\le |f'(\frac{nx}{nx+1})| |\cos(f(x))|$$

$$\le |f'(\frac{nx}{nx+1})|$$

$$\le |f'(a)|$$

where $a \in S$ s.t. $a$ maximises $f'(a)$

$g: = |f'(a)| \in L^{1} (S, \Sigma, \mu) \ \because f' \in L^{1} (S, \Sigma, \mu)$

Once a $g$ is found, we have

$$\lim_{n \to \infty} \int_{S} f'(\frac{nx}{nx+1}) \cos(f(x)) dx$$

$$ = \int_{S} \lim_{n \to \infty} f'(\frac{nx}{nx+1}) \cos(f(x)) dx$$

$$ = \int_{S} f'(\lim_{n \to \infty} \frac{nx}{nx+1}) \cos(f(x)) dx$$

$$ = \int_{S} f'(1) \cos(f(x)) dx$$

$$ = f'(1) \int_{S} \cos(f(x)) dx$$

If that's right, any further simplification possible? If that's wrong, how else can I approach this problem?

BCLC
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    In what sense do you want to take $f'\left(x\right)$? – asomog Dec 06 '15 at 15:45
  • @ÁkosSomogyi Don't know :( Is $f'(x)$ meaningless if f is a random variable? – BCLC Dec 06 '15 at 15:51
  • @ÁkosSomogyi Is that measurable? – BCLC Dec 06 '15 at 16:03
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    The hypotheses don't imply that the dominated convergence theorem is applicable. If $f'$ is not continuous at $1$, then $\lim\limits_{y\to 1^-} f'(y)$ doesn't exist, and $\lim\limits_{n\to \infty} f'\bigl(\frac{nx}{nx+1}\bigr)$ need not exist for any $x \in (0,1]$. – Daniel Fischer Dec 06 '15 at 22:08
  • @DanielFischer Can a continuous function have a derivative that is not continuous? – BCLC Dec 07 '15 at 04:54
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    Sure. We wouldn't speak of "continuously differentiable" functions otherwise. For an example, let $$g(x) = \begin{cases} 0 &, x = 0 \ \sin \frac{1}{x} &, x \neq 0\end{cases}$$ and $G(x) = \int_0^x g(t),dt$. Then $G$ is differentiable with $G' = g$, and $G'$ is not continuous at $0$. See Volterra's function for a function whose derivative is discontinuous on a set of positive measure. – Daniel Fischer Dec 07 '15 at 10:14
  • @DanielFischer Sorry I asked the wrong question. $f$ and $f'$ are measurable. That means they are continuous? Or not? – BCLC Dec 07 '15 at 11:30
  • @DanielFischer Also why does DCT not apply exactly? – BCLC Dec 07 '15 at 11:31
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    Measurability does not imply continuity. Differentiability however implies continuity, so $f$ is continuous. A derivative is always (Borel) measurable, but not always continuous. The informal description of Luzin's theorem in wikipedia is a bit misleading, a measurable function can be discontinuous everywhere [that can't happen for derivatives, a derivative is always continuous on a large set]. One mustn't confuse $f$ being continuous at an $x\in E$ with the restriction $f\lvert_E$ of $f$ to $E$ being continuous (at $x$, or on all of $E$). – Daniel Fischer Dec 07 '15 at 12:08
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    And the DCT doesn't apply a) because the hypotheses don't imply the existence of a pointwise a.e. limit of the integrand, and b) they don't imply the existence of an integrable dominating function. – Daniel Fischer Dec 07 '15 at 12:14
  • @DanielFischer Ummmmm do all our problems go away if f' is continuous? – BCLC Dec 07 '15 at 12:32
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    Yes, if $f'$ is continuous, everything is straightforward. – Daniel Fischer Dec 07 '15 at 12:34
  • @DanielFischer okay thanks. Edited. How about now? – BCLC Dec 07 '15 at 12:53
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    The first part doesn't look like you know what's going on. The last line "$g: = |f'(a)| \in L^{1} (S, \Sigma, \mu) \ \because f' \in L^{1} (S, \Sigma, \mu)$" doesn't make much sense. Why can an $a$ be chosen so that the constant $\lvert f'(a)\rvert$ is a dominating function? (Aside, use words to explain what you're doing. If what you're doing is correct, it's easier to follow, and if it's incorrect, it's easier to find the mistake.) – Daniel Fischer Dec 07 '15 at 13:12
  • @DanielFischer I don't know if or why we can choose such a. What do you suggest for dominating function? – BCLC Dec 07 '15 at 13:17
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    Since we have a space of finite measure, a constant is great if it works. Do the new assumptions guarantee that there is a constant that works? If yes, why/how, if no, why not? – Daniel Fischer Dec 07 '15 at 13:22
  • @DanielFischer Continuous function on a closed interval has a max and a min? – BCLC Dec 07 '15 at 13:48
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    Closed and bounded. $[1,+\infty)$ may also be called a closed interval. Yes, that's the reason. – Daniel Fischer Dec 07 '15 at 13:50
  • @DanielFischer Is that a closed interval? I know it's a closed set. Depends on the text I think >:P anyway thanks ^-^ Is the rest correct as well? – BCLC Dec 07 '15 at 13:54
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    It's an interval, and it's closed. But some people reserve the word interval only for bounded intervals, for those "closed and bounded interval" would be redundant, like "strictly positive natural number" would be redundant for those who believe $0 \notin \mathbb{N}$. The latter people are numerous enough that there is no standard convention regarding the question whether $0\in \mathbb{N}$. Yes, the rest is correct, with the continuity assumptions we have a pointwise limit and a dominating function, so all works smoothly. – Daniel Fischer Dec 07 '15 at 14:04
  • @DanielFischer Post as answer? – BCLC Dec 07 '15 at 14:10

1 Answers1

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With the new assumption that $f'$ is continuous, the argument works. Since $[0,1]$ is compact, every continuous function on $[0,1]$ is bounded (and attains its maximum and minimum, but we don't need that), and since $\lvert \cos \varphi\rvert \leqslant 1$ for all $\varphi \in \mathbb{R}$, a bound for $\lvert f'\rvert$ is also a (uniform) bound for

$$f'\biggl(\frac{nx}{nx+1}\biggr) \cos \bigl(f(x)\bigr).$$

So we have a dominating function, and by the continuity of $f'$ at $1$ we have

$$\lim_{n\to\infty} f'\biggl(\frac{nx}{nx+1}\biggr) = \begin{cases} f'(0) &, x = 0 \\ f'(1) &, x \in (0,1]. \end{cases}$$

Since $\{0\}$ is a null set, the dominated convergence theorem thus implies

$$\lim_{n\to\infty} \int_0^1 f'\biggl(\frac{nx}{nx+1}\biggr)\cos \bigl(f(x)\bigr)\,dx = \int_0^1 f'(1)\cos \bigl(f(x)\bigr)\,dx.$$

It is sufficient to reach the same conclusion - with basically the same argument - that the derivative $f'$ is bounded, and continuous at $1$.

However, if $f'$ is not continuous at $1$, then [in general] a pointwise a.e. limit of the integrand does not exist, and if $f'$ is not bounded, then [in general] there is no integrable dominating function, so both these hypotheses are necessary for an application of the dominated convergence theorem in the abstract situation.

Daniel Fischer
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