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I find this pretty hard and it would be awesome if someone could help me.

The problem is the following (Problem 6/Chapter 3 from S&S's Real Analysis).

Suppose $F$ is a bounded measurable function on $\mathbb{R}$. If $F$ satisfies either one of the two following conditions:

(a) $\int_{\mathbb{R}}{|F(x+h)-F(x)|dx} \leq A|h|$, for some constant $A$ and all $h\in \mathbb{R}$;

(b) $|\int_{\mathbb{R}}{F(x)\phi '(x) dx}| \leq A$, where $\phi$ ranges over all $C^{1}$ functions of bounded support with $\sup_{x \in \mathbb{R}}{|\phi (x)|} \leq 1$; then $F$ can be modified on a set of measure zero as to become a function of bounded variation on $\mathbb{R}$.

Moreover, on $\mathbb{R}^{d}$ we have the following assertion. Suppose that $F$ is a bounded measurable function on $\mathbb{R}^{d}$. Then, the following two conditions on $F$ are equivalent:

(a') $\int_{\mathbb{R}^d}{|F(x+h)-F(x)|dx} \leq A|h|$, for some constant $A$ and all $h\in \mathbb{R}^d$;

(b') $|\int_{\mathbb{R}^d}{F(x)\frac{\partial {\phi}}{\partial{x_{j}}} dx}| \leq A$, for all $j=1,\ldots, d$, for all $\phi \in C^{1}$ of bounded support with $\sup_{x \in \mathbb{R}^d}{|\phi(x)|} \leq 1$.

I proved already that if $F$ is a BV function then (a) and (b) hold; so the first part should be a converse for that..

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To see that $(a')\Rightarrow (b')$, look here math.stackexchange.com/questions/116196/…. –  Davide Giraudo Apr 3 '12 at 10:10
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2 Answers

Here's a thought on how to prove $(b) \implies F \text{ of BV}$:

Define a linear functional $I: C_c^1(\mathbb R)\to \mathbb R$ by $I(\phi) = -\int_{\mathbb R} \phi'(x) F(x) \, dx$. By (b), we have $|I(\phi)|\le A\Vert \phi\Vert_\infty$ for all $\phi\in C_c^1(\mathbb R)$.

Since $C_c^1(\mathbb R)$ is dense in $C_0(\mathbb R)$ and $I$ is Lipschitz, we can extend $I$ to all of $C_0(\mathbb R)$ and obtain a bounded linear functional

$$I: C_0(\mathbb R) \to \mathbb R, \qquad \phi \mapsto I(\phi)$$

Now by Riesz' representation theorem, there is a finite regular Borel measure $\mu$ associated to $I$, such that $I(\phi) = \int_\mathbb{R} \phi \, d\mu$ for all $\phi \in C_0(\mathbb R)$. In particular, for all $\phi \in C_c^1(\mathbb R)$, we obtain $$\int_\mathbb{R} \phi \, d\mu = - \int_\mathbb{R} \phi'(x) F(x) \, dx$$ (the idea is that the RHS is "$=\int \phi(x) \, dF(x)$" if $F$ is of BV) So we next want to examine, whether $\mu([a,b]) = F(b) - F(a)$ for $a,b\in \mathbb R$.

For this, consider an approxmation $\phi_{\epsilon,\delta}(x)\in C_c^1(\mathbb R)$ to $\chi_{[a,b]}$, whose derivative is continuous, piecewise linear and satisfies

$$\phi'_{\epsilon,\delta}(x) = \begin{cases} (2\epsilon)^{-1} & x\in (a-\epsilon, a+\epsilon) \\ -(2\epsilon)^{-1} & x\in (b-\epsilon, b+\epsilon) \\ 0 & x\notin (a-\epsilon-\delta,a+\epsilon+\delta)\cup(b-\epsilon-\delta,b+\epsilon + \delta)\end{cases}$$

Now let $\epsilon_n \to 0$, and choose $\delta_n$ (small compared to $\epsilon_n$) in such a way that $\phi'_{\epsilon_n, \delta_n}$ approximates $\frac{1}{2\epsilon_n}(\chi_{(a-\epsilon_n, a+\epsilon_n)} - \chi_{(b-\epsilon_n, b+\epsilon_n)})$ sufficiently well for the following:

Assuming that $a,b$ are Lebesgue points of $F$, we obtain \begin{align} \mu([a,b]) &= \lim_{n\to \infty} \int \phi_{\epsilon_n, \delta_n} \, d\mu \\ &= \lim_{n\to\infty} - \int \phi'_{\epsilon_n, \delta_n}(x) F(x)\, dx \\ &= \lim_{n\to\infty} - \frac{1}{2\epsilon_n} \int_{B_{\epsilon_n}(a)} F(x) \, dx + \frac{1}{2\epsilon_n} \int_{B_{\epsilon_n}(b)} F(x) \, dx \\ &= F(b) - F(a) \end{align}

Since the first limit always exists and since almost every point of $\mathbb R$ is a Lebesgue point of $F$, we may define a function $\tilde F$ which is $=F$ a.e. by setting

$$\tilde F(x) = \mu([x_0,x]) - F(x_0)$$

where $x_0$ is a Lebesgue point of $F$. Then $\mu([a,b]) = \tilde F(b) - \tilde F(a)$ for all $a,b$.

But then for any partition $x_0 < x_1<\dots < x_n$, we have $$\sum_{i=1}^n \left|\tilde F(x_i) -\tilde F(x_{i-1}) \right| \le 2\Vert \mu\Vert = 2\Vert I\Vert <\infty$$ (I guess one could avoid the factor of $2$, if one defined $\tilde F$ in such a way that $\tilde F(b) - \tilde F(a) =\mu((a,b])$...)

In particular, $\tilde F$ is of bounded variation.

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There is typo: "But then for any partition x_0&ltx_1<\dots < x_n, we have", must be x_0\lt x_1<\dots < x_n –  leo Apr 15 '12 at 7:06
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I'll try to prove $(b') \implies (a')$.

Denote $x\in\mathbb{R}$ by $x=(x^1,x^2,\ldots,x^d)$. Let $$\Phi=\left\{\phi\in C^1:\phi \text{ has bounded support and } \sup_{x\in\mathbb{R}^d}|\phi(x)|\leq 1\right\}.$$ I'll use $\int f$ to denote the integral of $f$ over the whole space.

Fix $h\in\mathbb{R}^d$.

First suppose that $F\in L^1$. You can pick a function $K\in\Phi$ so that $\int K=1$. Define $$K_n=\frac{1}{1/n}K\left( \frac{x}{1/n} \right).$$ Then you know that $$||F\star K_n-F||_1\to 0,$$ where the star denotes convolution. Recall that convergence in $L^1$ of the sequence $(F\star K_n)_{n\in\mathbb{N}}$ to the function $F$, implies $F\star K_n\stackrel{m}{\longrightarrow} F$, and then there is a subsequence $(F\star K_{n_j})_{j\in\mathbb{N}}$ such that $$F\star K_{n_j}\to F$$ almost everywhere. Now, notice that for all $z\in\mathbb{R}^d$ and all $n\in\mathbb{N}$ $$\begin{align*} K_n(z+h)-K_n(z) &= K_n(z+(h^1,h^2,\ldots,h^d))-K_n(z)\\ &= K_n(z+(h^1,h^2,\ldots,h^d)) - K_n(z+(0,h^2,\ldots,h^d))+\\ &\phantom{=} K_n(z+(0,h^2,\ldots,h^d))-K_n(z+(0,0,h^3,\ldots,h^d))+\\ &\phantom{=} K_n(z+(0,0,h^3,h^4,\ldots,h^d))-K_n(z+(0,0,0,h^4,\ldots,h^d))+\\ &\phantom{=} \ldots +\\ &\phantom{=} K_n(z+(0,0,\ldots,h^d))-K_n(z). \end{align*}$$ By the mean value theorem for derivatives in one variable you get, for example, that there exist a $\theta_1(z)$ strictly between $z^1$ and $z^1+h^1$ so that $$K_n(z+(h^1,h^2,\ldots,h^d)) - K_n(z+(0,h^2,\ldots,h^d))=h^1\frac{\partial K_n}{\partial x_1}(\theta_1(z)).$$ Repeating the same argument you can find analogous $\theta_j$ so that $$ K_n(z+h)-K_n(z) = \sum_{j=1}^d h^j\frac{\partial K_n}{\partial x_j} (\theta_j(z)), $$ for all $z\in\mathbb{R}^d$ and $n\in\mathbb{N}$. Then $$\begin{align*} \left| F\star K_{n_i}(x+h)-F\star K_{n_i}(x) \right|&= \left| \int F(t)K_{n_i}(x+h-t)\textrm{d}t - \int F(t)K_{n_i}(x-t)\textrm{d}t \right|\\ &= \left| \int F(t)[K_{n_i}(x+h-t)-K_{n_i}(x-t)]\textrm{d}t \right|\\ &=\left| \int F(t)\sum_{j=1}^d h^j\frac{\partial K_{n_i}}{\partial x_j} (\theta_j(x-t)) \textrm{d}t\right|\\ &=\left| \sum_{j=1}^d h^j\int F(t)\frac{\partial K_{n_i}}{\partial x_j} (\theta_j(x-t)) \textrm{d}t \right|\\ &\leq \sum_{j=1}^d \left|h^j\right|\left| \int F(t)\frac{\partial K_{n_i}}{\partial x_j} (\theta_j(x-t)) \textrm{d}t \right|\\ &\leq |h|\sum_{j=1}^d \left| \int F(t)\frac{\partial K_{n_i} }{\partial x_j} (\theta_j(x-t)) \textrm{d}t \right|\\ &\leq |h|\sum_{j=1}^d A\\ &= |h|Ad, \end{align*}$$ i.e. $$\left| F\star K_{n_i}(x+h)-F\star K_{n_i}(x) \right|\leq |h|Ad\quad \forall i\in\mathbb{N}.$$ Therefore, by letting $i\to\infty$, $$|F(x+h)-F(x)|\leq |h|M,$$ for some constant $M$.

Now consider the case when $F$ is just a bounded function. For each $n\in\mathbb N$, define $F_n:\mathbb{R}^d\to\mathbb{R}$ by $$F_n(x)=F(x)\chi_{B(0,n)}(x).$$ Then $F_n\uparrow F$. Since $F$ is bounded and $F_n$ is of bounded support, we have that $F_n\in L^1$. Apply what we have already proved to get $$|F_n(x+h)-F_n(x)|\leq |h|M.$$ By the continuity of the absolute value we get $$|F(x+h)-F(x)|=\left| \lim_{n\to\infty} F_n(x+h)-F_n(x)\right|\leq |h|M,$$ as promised.

For the first part. I don't have worked out the details but here are some thoughts. By what has been proved above you can see that $(b)\implies (a)$. I think that from $(a)$ it is possible to deduce that $F$ is absolutely continuous in each bounded interval, absolute continuity of $f$ implies that $f$ is of BV.

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@DavideGiraudo, if my last argument is correct there is one –  leo Apr 3 '12 at 17:20
    
Thanks! What about the first part (i.e. (a) or (b) implies $F$ is of BV)... –  Anna Apr 4 '12 at 3:20
    
You used $\int |F(t) \frac{\partial K}{\partial x_j}(\theta(t)_j)|dt \leq A$, but that's not the given inequality in the question... it's $|\int F(t) \frac{\partial K}{\partial x_j}(\theta(t)_j)dt| \leq A$, which is weaker. how do you correct that? –  user28340 Apr 4 '12 at 20:02
    
@nerv, you are right, let me check that and I'll fixit tomorrow probably. –  leo Apr 7 '12 at 14:41
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@leo Looks good. I can't upvote you another time ;-). –  Jonas Teuwen Apr 11 '12 at 11:54
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