What Young measures can tell about weak-* convergence of non-linearly distorted functions

This entry is not precisely about some thing I stumbled upon but about some thing a that I wanted to learn for some time now, namely Young measures. Lately I had a several hour train ride and I had the book Kinetic Formulation of Conservation Laws with me.

While the book is about hyperbolic PDEs and their formulation as kinetic equation, it also has some pointers to Young measures. Roughly, Young measures are a way to describe weak limits of functions and especially to describe how these weak limits behave under non-linear functions, and hence, we start with this notation.

1. Weak convergence of functions

We are going to deal with sequences of function ${(f_n)}$ in spaces ${L^p(\Omega)}$ for some open bounded domain ${\Omega}$ and some ${1\leq p\leq \infty}$ .

For ${1\leq p < \infty}$ the dual space of ${L^p(\Omega)}$ is ${L^q(\Omega)}$ with ${1/p + 1/q = 1}$ and the dual pairing is

$\displaystyle \langle f,g\rangle_{L^p(\Omega)\times L^q(\Omega)} = \int_\Omega f\, g.$

Hence, a sequence ${(f_n)}$ converges weakly in ${L^p(\Omega)}$ to ${f}$ , if for all ${g\in L^p(\Omega)}$ it holds that

$\displaystyle \int f_n\, g \rightarrow \int f\, g.$

We denote weak convergence (if the space is clear) with ${f_n\rightharpoonup f}$ .

For the case ${p=\infty}$ one usually uses the so-called weak-* convergence: A sequence ${(f_n)}$ in ${L^\infty(\Omega)}$ converges weakly-* to ${f}$ , if for all ${g\in L^1(\Omega)}$ it holds that

$\displaystyle \int f_n\, g \rightarrow \int f\, g.$

The reason for this is, that the dual space of ${L^\infty(\Omega)}$ is not easily accessible as it can not be described as a function space. (If I recall correctly, this is described in “Linear Operators”, by Dunford and Schwarz.) Weak-* convergence will be denoted by ${f_n\rightharpoonup^* f}$ .

In some sense, it is enough to consider weak-* convergence in ${L^\infty(\Omega)}$ to understand what’s that about with Young measures and I will only stick to this kind of convergence here.

Example 1 We consider ${\Omega = [0,1]}$ and two values ${a,b\in{\mathbb R}}$ . We define a sequence of functions which jumps between these two values with an increasing frequency:

$\displaystyle f_n(x) = \begin{cases} a & \text{for }\ \tfrac{2k}{n} \leq x < \tfrac{2k+1}{n},\ k\in{\mathbb Z}\\ b & \text{else.} \end{cases}$

The functions ${f_n}$ look like this:

To determine the weak limit, we test with very simple functions, lets say with ${g = \chi_{[x_0,x_1]}}$ . Then we get

$\displaystyle \int f_n\, g = \int_{x_0}^{x_1} f_n \rightarrow (x_1-x_0)\tfrac{a+b}{2}.$

Hence, we see that the weak-* limit of the ${f_n}$ (which is, by the way, always unique) has no other chance than being

$\displaystyle f \equiv \frac{a+b}{2}.$

In words: the weak-* limit converges to the arithmetic mean of the two values between which the functions oscillate.

2. Non-linear distortions

Now, the norm-limit behaves well under non-linear distortions of the functions. Let’s consider a sequence ${f_n}$ which converges in norm to some ${f}$ . That is, ${\|f_n -f\|_\infty \rightarrow 0}$ . Since this means that ${\sup| f_n(x) - f(x)| \rightarrow 0}$ we see that for any boundedcontinuous function ${\phi:{\mathbb R}\rightarrow {\mathbb R}}$ we also have ${\sup |\phi(f_n(x)) - \phi(f(x))|\rightarrow 0}$ and hence ${\phi\circ f_n \rightarrow \phi\circ f}$ .

The same is totally untrue for weak-* (and also weak) limits:

Example 2 Consider the same sequence ${(f_n)}$ as in example~1which has the weak-* limit ${f\equiv\frac{a+b}{2}}$ . As a nonlinear distortion we take ${\phi(s) = s^2}$ which gives

$\displaystyle \phi\circ f_n(x) = \begin{cases} a^2 & \text{for }\ \tfrac{2k}{n} \leq x < \tfrac{2k+1}{n},\ k\in{\mathbb Z}\\ b^2 & \text{else.} \end{cases}$

Now we see

$\displaystyle \phi\circ f_n \rightharpoonup^* \frac{a^2 + b^2}{2} \neq \Bigl(\frac{a+b}{2}\Bigr)^2 = \phi\circ f.$

The example can be made a little bit more drastically by assuming ${b = -a}$ which gives ${f_n\rightharpoonup^* f\equiv 0}$ . Then, for every ${\phi}$ with ${\phi(0) = 0}$ we have ${\phi\circ f\equiv 0}$ . However, with such a ${\phi}$ we may construct any constant value ${c}$ for the weak-* limit of ${\phi\circ f_n}$ (take, e.g. ${\phi(b) = 0}$ , ${\phi(a) = 2c}$ ).

In fact, the relation ${\phi\circ f_n \rightharpoonup^* \phi\circ f}$ is only true for affine linear distortions ${\phi}$ (unfortunately I forgot a reference for this fact\dots).

It arises the question, if it is possible to describe the weak-* limits of distortions of functions and if fact, this will be possible with the notions of Young measure.

3. Young measures

In my understanding, Young measures are a method to view a function somehow a little bit more geometrically in giving more emphasis on the graph of the function rather than is mapping property.

We start with defining Young measures and illustrate how they can be used to describe weak(*) limits. In what follows we use ${\mathfrak{L}}$ for the Lebesgue measure on the (open and bounded) set ${\Omega}$ . A more through description in the spirit of this section is Variational analysis in Sobolev and BV spaces by Attouch, Buttazzo and Michaille.

Definition 1 (Young measure) A positive measure ${\mu}$ on ${\Omega\times {\mathbb R}}$ is called a Young measureif for every Borel subset ${B}$ of ${\Omega}$ it holds that

$\displaystyle \mu(B\times{\mathbb R}) = \mathfrak{L}(B).$

Hence, a Young measure is a measure such that the measure of every box ${B\times{\mathbb R}}$ is determined by the projection of the box onto the set ${\Omega}$ , i.e. the intersection on ${B\times{\mathbb R}}$ with ${\Omega}$ which is, of course, ${B}$ :

There are special Young measures, namely these, who are associated to functions. Roughly spoken, a Young measure associated to a function ${u:\Omega\rightarrow {\mathbb R}}$ is a measure which is equidistributed on the graph of ${u}$ .

Definition 2 (Young measure associated to ${u}$ ) For a Borel measurable function ${u:\Omega\rightarrow{\mathbb R}}$ we define the associated Young measure ${\mu^u}$ by defining for every continuous and bounded function ${\phi:\Omega\times{\mathbb R}\rightarrow{\mathbb R}}$

$\displaystyle \int_{\Omega\times{\mathbb R}}\phi(x,y){\mathrm d}{\mu^u(x,y)} = \int_\Omega \phi(x,u(x)){\mathrm d} \mathfrak{L}(x).$

It is clear that ${\mu^u}$ is a Young measure: Take ${B\subset \Omega}$ and approximate the characteristic function ${\chi_{B\times{\mathbb R}}}$ by smooth functions ${\phi_n}$ . Then

$\displaystyle \int_{\Omega\times{\mathbb R}}\phi_n(x,y){\mathrm d}{\mu^u(x,y)} = \int_\Omega \phi_n(x,u(x)){\mathrm d} \mathfrak{L}(x).$

The left hand side converges to ${\mu^u(B\times{\mathbb R})}$ while the right hand side converges to ${\int_B 1{\mathrm d}{\mathfrak{L}} = \mathfrak{L}(B)}$ as claimed.

The intuition that a Young measure associated to a function is an equidistributed measure on the graph can be made more precise by “slicing” it:

Definition 3 (Slicing a measure) Let ${\mu}$ be a positive measure on ${\Omega\times{\mathbb R}}$ and let ${\sigma}$ be its projection onto ${\Omega}$ (i.e. ${\sigma(B) = \mu(B\times{\mathbb R})}$ ). Then ${\mu}$ is sliced into measures ${(\sigma_x)_{x\in\Omega}}$ , i.e. it holds:

Each ${\mu_x}$ is a probability measure.

The mapping ${x\mapsto \int_{\mathbb R} \phi(x,y){\mathrm d}{\mu_x(y)}}$ is measurable for every continuous ${\phi}$ and it holds that
$\displaystyle \int_{\Omega\times{\mathbb R}} \phi(x,y){\mathrm d}{\mu(x,y)} = \int_\Omega\int_{\mathbb R} \phi(x,y){\mathrm d}{\mu_x(y)}{\mathrm d}{\sigma(x)}.$

The existence of the slices is, e.g. proven in Variational analysis in Sobolev and BV spaces, Theorem 4.2.4.

For the Young measure ${\mu^u}$ associated to ${u}$ , the measure ${\sigma}$ in Definition~3is ${\mathfrak{L}}$ and hence:

$\displaystyle \int_{\Omega\times{\mathbb R}}\phi(x,y){\mathrm d}{\mu^u(x,y)} = \int_\Omega\int_{\mathbb R} \phi(x,y){\mathrm d}{\mu^u_x(y)}{\mathrm d}{\mathfrak{L}(x)}.$

On the other hand:

$\displaystyle \int_{\Omega\times{\mathbb R}}\phi(x,y){\mathrm d}{\mu^u(x,y)} = \int_\Omega\phi(x,u(x)){\mathrm d}{\mathfrak{L}} = \int_\Omega\int_{\mathbb R} \phi(x,y) {\mathrm d}{\delta_{u(x)}(y)}{\mathrm d}{\mathfrak{L}(x)}$

and we see that ${\mu^u}$ slices into

$\displaystyle \mu^u_x = \delta_{u(x)}$

and this can be vaguely sketched:

4. Narrow convergence of Young measures and weak* convergence in ${L^\infty(\Omega)}$

Now we ask ourself: If a sequence ${(u^n)}$ converges weakly* in ${L^\infty(\Omega)}$ , what does the sequence of associated Young measures do? Obviously, we need a notion for the convergence of Young measures. The usual notion here, is that of narrow convergence:

Definition 4 (Narrow convergence of Young measures) A sequence ${(\mu_n)}$ of Young measures on ${\Omega\times{\mathbb R}}$ converges narrowly to ${\mu}$ , if for all bounded and continuous functions ${\phi}$ it holds that

$\displaystyle \int_{\Omega\times{\mathbb R}}\phi(x,y){\mathrm d}{\mu_n(x,y)} \rightarrow \int_{\Omega\times{\mathbb R}} \phi(x,y){\mathrm d}{\mu(x,y)}.$

Narrow convergence will also be denoted by ${\mu_n\rightharpoonup\mu}$ .

One may also use the non-continuous test functions of the form ${\phi(x,y) = \chi_B(x)\psi(y)}$ with a Borel set ${B\subset\Omega}$ and a continuous and bounded ${\psi}$ , leading to the same notion.

The set of Young measures is closed under narrow convergence, since we may test with the function ${\phi(x,y) = \chi_B(x)\chi_{\mathbb R}(y)}$ to obtain:

$\displaystyle \mathfrak{L}(B) = \lim_{n\rightarrow\infty} \int_{\Omega\times{\mathbb R}}\phi(x,y){\mathrm d}{\mu_n(x,y)} = \int_{\Omega\times{\mathbb R}}\phi(x,y){\mathrm d}{\mu(x,y)} = \mu(B\times E).$

The next observation is the following:

Proposition 5 Let ${(u^n)}$ be a bounded sequence in ${L^\infty(\Omega)}$ . Then the sequence ${(\mu^{u_n})}$ of associated Young measures has a subsequence which converges narrowly to a Young measure ${\mu}$ .

The proof uses the notion of tightness of sets of measures and the Prokhorov compactness theorem for Young measures (Theorem 4.3.2 in Variational analysis in Sobolev and BV spaces).

Example 3 (Convergence of the Young measures associated to Example 1) Consider the functions ${f_n}$ from Example~1and the associated Young measures ${\mu^{f_n}}$ . To figure out the narrow limit of these Young measures we test with a function ${\phi(x,y) = \chi_B(x)\psi(y)}$ with a Borel set ${B}$ and a bounded and continuous function ${\psi}$ . We calculate

$\displaystyle \begin{array}{rcl} \int_{[0,1]\times{\mathbb R}}\phi(x,y){\mathrm d}{\mu^{f_n}(x,y)} &= &\int_0^1\phi(x,f_n(x)){\mathrm d}{\mathfrak{L}(x)}\\ & = &\int_B\psi(f_n(x)){\mathrm d}{\mathfrak{L}(x)}\\ & \rightarrow &\mathfrak{L}(B)\frac{\psi(a)+\psi(b)}{2}\\ & = & \int_B\frac{\psi(a)+\psi(b)}{2}{\mathrm d}{\mathfrak{L}(x)}\\ & = & \int_{[0,1]}\int_{\mathbb R}\phi(x,y){\mathrm d}{\bigl(\tfrac{1}{2}(\delta_a+\delta_b)\bigr)(y)}{\mathrm d}{\mathfrak{L}(y)}. \end{array}$

We conclude:

$\displaystyle \mu^{f_n} \rightharpoonup \tfrac{1}{2}(\delta_a+\delta_b)\otimes\mathfrak{L}$

i.e. the narrow limit of the Young measures ${\mu^{f_n}}$ is notthe constant function ${(a+b)/2}$ but the measure ${ \mu = \tfrac{1}{2}(\delta_a+\delta_b)\otimes\mathfrak{L}}$ . This expression may be easier to digest in sliced form:

$\displaystyle \mu_x = \tfrac{1}{2}(\delta_a+\delta_b)$

i.e. the narrow limit is something like the “probability distribution” of the values of the functions ${f_n}$ . This can be roughly put in a picture:

Obviously, this notion of convergence goes well with nonlinear distortions:

$\displaystyle \mu^{\phi\circ f^n} \rightharpoonup \tfrac{1}{2}(\delta_{\phi(a)} + \delta{\phi(b)})\otimes\mathfrak{L}.$

Recall from Example~1: The weak-* limit of ${\phi\circ f_n}$ was the constant function ${\tfrac{\phi(a)+\phi(b)}{2}}$ , i.e.

$\displaystyle \phi\circ f_n \rightharpoonup^* \tfrac{\phi(a)+\phi(b)}{2}\chi_{[0,1]}.$

The observation from the previous example is in a similar way true for general weakly-* converging sequences ${f_n}$ :

Theorem 6 Let ${f_n\rightharpoonup^* f}$ in ${L^\infty(\Omega)}$ with ${\mu^{f_n}\rightharpoonup\mu}$ . Then it holds for almost all ${x}$ that

$\displaystyle f(x) = \int_{\mathbb R} y{\mathrm d}{\mu_x(y)}.$

In other words: ${f(x)}$ is the expectation of the probability measure ${\mu_x}$ .

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The chi function tells the same thing about weak-* convergence as Young measures « regularize Says:

May 22, 2011 at 10:36 pm

[...] However, this type of convergence does not occur in bounded domains, and hence, can not be treated with Young measures as they have been introduced the in my previous entry. [...]

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