Continuous functions

A the beginning I want to remind you of things I think you already know and then go on to show the direction the course will be taking. Let me first try to set the context.

One basic notion I assume you are reasonably familiar with is that of a metric space ([5] p.9). This consists of a set, $X,$ and a distance function

$$d: X \times X = X^2 \longrightarrow [ 0, \infty ) ,$$

satisfying the following three axioms:

$$\begin{aligned}i) &d(x,y) = 0 \Leftrightarrow x = y, (\text... ...,y) \leq d (x,z) + d(z,y) \forall x,y,z \in X. \end{aligned}$$

The basic theory of metric spaces deals with properties of subsets (open, closed, compact, connected), sequences (convergent, Cauchy) and maps (continuous) and the relationship between these notions. Let me just remind you of one such result.

The basic example of a metric space is Euclidean space. Real $n$-dimensional Euclidean space, $\mathbb{R}^n$, is the set of ordered $n$-tuples of real numbers

$$x=(x_1 , \ldots , x_n) \in \mathbb{R}^n , x_j \in \mathbb{R} , j=1 , \ldots , n .$$

It is also the basic example of a vector (or linear) space with the operations

$$x+y =(x_1 + y_1 , x_2 + y_2 , \ldots , x_n + y_n )$$

$$cx = (cx_1 , \ldots , cx_n ) .$$

The metric is usually taken to be given by the Euclidean metric

$$\left\vert x \right\vert = (x^2_1 + \cdots + x^2_n )^{1/2} = (\sum^n_{j=1} x^2_j )^{1/2} ,$$

in the sense that

$$d(x,y) = \left\vert x-y \right\vert .$$

Let us abstract this immediately to the notion of a normed vector space, or normed space. This is a vector space $V$ (over $\mathbb{R}$ or $\mathbb{C}$) equipped with a norm, which is to say a function

$$\Vert \Vert : V \longrightarrow [0, \infty)$$

satisfying

$$\begin{aligned}i) &\Vert v \Vert = 0 \Longleftrightarrow v=0... ...ert v+w \Vert \leq \Vert v \Vert + \Vert w \Vert. \end{aligned}$$

This means that $(V,d)$, $d(v,w) = \Vert v-w \Vert$ is a vector space; I am also using $\mathbb{K}$ to denote either $\mathbb{R}$ or $\mathbb{C}$ as is appropriate.

The case of a finite dimensional normed space is not very interesting because, apart from the dimension, they are all ``the same''. We shall say (in general) that two norms $\Vert\bullet\Vert _1$ and $\Vert\bullet\Vert _2$ on $V$ are equivalent of there exists $C >0$ such that

$$\frac{1}{C} \Vert v \Vert _1 \leq \Vert v \Vert _2 \leq C \Vert v \Vert _1 \forall v \in V .$$

So, we are mainly interested in the infinite dimensional case. I will start the course, in a slightly unorthodox manner, by concentrating on one such normed space (really one class). Let $X$ be a metric space. The case of a continuous function, $f:X \to \mathbb{R}$ (or $\mathbb{C}$) is a special case of Proposition 1.1 above. We then define

$$C(X) = \left\{ f:X \to \mathbb{R}, f \text{ continuous} \right\} .$$

In fact the same notation is generally used for the space of complex-valued functions. If we want to distinguish between these two possibilities we can use the more pedantic notation $C(X;\mathbb{R})$ and $C(X; \mathbb{C})$. Now, the `obvious' norm on this linear space is the supremum (or `uniform') norm

$$\Vert f \Vert _{\infty} = \sup_{x \in X} \left\vert f(x) \right\vert .$$

Here $X$ is an arbitrary metric space. For the moment $X$ is supposed to be a ``physical'' space, something like $\mathbb{R}^n$. Corresponding to the finite-dimensionality of $\mathbb{R}^n$ we often assume (or demand) that $X$ is locally compact. This just means that every point has a compact neighborhood, i.e., is in the interior of a compact set. Whether locally compact or not we can consider

$$\mathcal{C}_0 (X) = \left\{ f \in\mathcal{C}(X) ; \forall \ep... ... s.t. \sup_{x \notin K} \left\vert f(x) \right\vert \leq \epsilon \right\} .$$ (3)

Here the notation $K \Subset X$ means `$K$ is a compact subset of $X$'.

If $V$ is a normed linear space we are particularly interested in the continuous linear functionals on $V$. Here `functional' just means function but $V$ is allowed to be `large' (not like $\mathbb{R}^n$) so `functional' is used for historical reasons.

Proof. $(1) \Longrightarrow (2)$ by definition. Then $(2)$ implies that $u^{-1} (-1,1)$ is an open neighborhood of $0 \in V$, so for some $\epsilon >0$, $u ( \left\{ f \in V ; \Vert f \Vert < \epsilon \right\}) \subset (-1,1)$. By linearity of $u$, $u ( \left\{ f \in V ; \Vert f \Vert < 1 \right\}) \subset (- \frac1{\epsilon} ,\frac1{\epsilon})$ is bounded, so $(2) \Longrightarrow (3).$ Then $(3)$ implies that

$$\left\vert u (f)\right\vert \leq C \forall f \in V , \Vert f \Vert \leq 1$$

for some $C$. Again using linearity of $u$, if $f\neq 0$,

$$\left\vert u(f) \right\vert \leq \Vert f \Vert u \left( \frac{f}{\Vert f \Vert} \right) \leq C \Vert f \Vert ,$$

giving $(4)$. Finally, assuming $(4)$,

$$\left\vert u(f) - u(g) \right\vert = \left\vert u(f-g) \right\vert \leq C \Vert f-g \Vert$$

shows that $u$ is continuous at any point $g \in V$. $\qedsymbol$

In view of this identification, continuous linear functionals are often said to be bounded. One of the important ideas that we shall exploit later is that of `duality'. In particular this suggests that it is a good idea to examine the totality of bounded linear functionals on $V$. The dual space is

$$V' =V^* = \left\{ u:V \longrightarrow \mathbb{K}\, , \, \hbox{ linear and bounded} \right\}.$$

This is also a normed linear space where the linear operations are

$$\begin{matrix}(u+v)(f)= u(f) + v(f)\ (cu)(f)= c (u(f))\end{matrix} \forall f \in V.$$ (4)

The natural norm on $V'$ is

$$\Vert u \Vert = \sup_{\Vert f\Vert\leq 1}\vert u(f)\vert.$$

This is just the `best constant' in the boundedness estimate,

$$\Vert u \Vert = \inf \left\{ C ; \left\vert u(f) \right\vert \leq C \Vert f \Vert \forall f \subset V \right\} .$$

One of the basic questions I wish to pursue in the first part of the course is: What is the dual of $\mathcal{C}_0 (X)$ for a locally compact metric space $X$? The answer is given by Riesz' representation theorem, in terms of (Borel) measures.

Let me give you a vague picture of `regularity of functions' which is what this course is about, even though I have not introduced most of these spaces yet. Smooth functions (and small spaces) are towards the top. Duality flips up and down and as we shall see $L^2,$ the space of Lebesgue square-integrable functions, is generally `in the middle'. What I will discuss first is the right side of the diagramme, where we have the space of continuous functions on $\mathbb{R}^n$ which vanish at infinity and its dual space, $M_{\text{fin}}(\mathbb{R}^n),$ the space of finite Borel measures. There are many other spaces that you may encounter, here I only include test functions, Schwartz functions, Sobolev spaces and their duals; $k$ is a general positive integer.

$$\xymatrix{ \mathcal{S}(\mathbb{R}^n)\ar@{^{(}->}[d]\ar@{^{(}->}[d... ...}(\mathbb{R}^n)\ar@{^{(}->}[dll]\ar@{_{(}->}[l]\ \mathcal{S}'(\mathbb{R}^n). }$$ (5)

I have set the goal of understanding the dual space $M(\bbR^n)$ of $\mathcal{C}_0 (X)$, where $X$ is a locally compact metric space. This will force me to go through the elements of measure theory and Lebesgue integration. It does require a little forcing!

The basic case of interest is $\mathbb{R}^n.$ Then an obvious example of a continuous linear functional on $\mathcal{C}_0 (\mathbb{R}^n)$ is given by Riemann integration, for instance over the unit cube $[0,1]^n$:

$$u(f) = \int_{[0,1]^n} f(x) dx .$$

In some sense we must show that all continuous linear functionals on $\mathcal{C}_0 (X)$ are given by integration. However, we have to interpret integration somewhat widely since there are also evaluation functionals. If $z \in X$ consider the Dirac delta

$$\delta_z (f) = f(z) .$$

This is also called a point mass of $z$. So we need a theory of measure and integration wide enough to include both of these cases.

One special feature of $\mathcal{C}_0 (X)$, compared to general normed spaces, is that there is a notion of positivity for its elements. Thus $f \geq 0$ just means $f(x) \geq 0 \forall x \in X$.

Proof. Simply define

$$f_{\pm} (x) = \left\{ \begin{array}{ccl} \pm f(x) & \hbox{ if } & \pm f (x) \geq 0 \ 0 & \hbox{ if } & \pm f(x) < 0 \end{array} \right.$$

for the same sign throughout. Then (1.6) holds. Observe that $f_+$ is continuous at each $y \in X$ since, with $U$ an approximate neighborhood of $y$, in each case

$$f(y) >0 \Longrightarrow f(x) >0 \hbox{ for } x \in U \Longrightarrow f_+ = f \hbox{ in } U$$

$$f(y) <0 \Longrightarrow f(x) <0 \hbox{ for } x \in U \Longrightarrow f_+ = 0 \hbox{ in } U$$

$$f(y) =0 \Longrightarrow \hbox{ given } \epsilon >0 \exists U \hbox{ s.t. } \left\vert f(x) \right\vert < \epsilon \hbox{ in } U$$

$$\Longrightarrow \left\vert f_+ (x) \right\vert < \epsilon \hbox{ in } U .$$

Thus $f_- = f-f_+ \in \mathcal{C}_0 (X)$, since both $f_+$ and $f_-$ vanish at infinity. $\qedsymbol$

We can similarly split elements of the dual space into positive and negative parts although it is a little bit more delicate. We say that $u \in (\mathcal{C}_0(X))'$ is positive if

$$u(f) \geq 0 \forall 0 \leq f \in \mathcal{C}_0 (X) .$$ (6)

For a general (real) $u \in (\mathcal{C}_0(X))'$ and for each $0 \leq f \in \mathcal{C}_0(X)$ set

$$u_+ (f)= \sup \left\{ u(g) ; g \in \mathcal{C}_0 (X) , 0 \leq g(x) \leq f(x) \forall x \in X \right\} .$$ (7)

This is certainly finite since $u(g) \leq C \Vert g \Vert _{\infty} \leq C \Vert f \Vert _{\infty}$. Moreover, if $0< c \in \mathbb{R}$ then $u_+ (cf) = cu_+ (f)$ by inspection. Suppose $0 \leq f_i \in \mathcal{C}_0(X)$ for $i=1,2$. Then given $\epsilon >0$ there exist $g_i \in \mathcal{C}_0(X)$ with $0 \leq g_i (x) \leq f_i(x)$ and

$$u_+(f_i) \leq u(g_i) + \epsilon .$$

It follows that $0 \leq g(x) \leq f_1(x) + f_2(x)$ if $g=g_1 +g_2$ so

$$u_+ (f_1 + f_2) \geq u(g) = u(g_1) +u(g_2) \geq u_+ (f_1) + u_+ (f_2) - 2 \epsilon .$$

Thus

$$u_+ (f_1 + f_2) \geq u_+ (f_1) + u_+ (f_2).$$

Conversely, if $0 \leq g(x) \leq f_1(x) + f_2(x)$ set $g_1 (x) = \min (g,f_1) \in \mathcal{C}_0(X)$ and $g_2=g-g_1$. Then $0 \leq g_i \leq f_i$ and $u_+ (f_1) + u_+ (f_2) \geq u(g_1) + u (g_2) = u(g)$. Taking the supremum over $g$, $u_+ (f_1 + f_2) \leq u_+ (f_1) + u_+ (f_2)$, so we find

$$u_+ (f_1 + f_2) = u_+ (f_1) + u_+ (f_2) .$$ (8)

Having shown this effective linearity on the positive functions we can obtain a linear functional by setting

$$u_+ (f)= u_+ (f_+) - u_+ (f_-) \forall f \in\mathcal{C}_0 (X) .$$ (9)

Note that (1.9) shows that $u_+(f)=u_+(f_1)-u_+(f_2)$ for any decomposiiton of $f=f_1-f_2$ with $f_i\in\mathcal{C}_0(X),$ both positive. [Since $f_1+f_-=f_2+f_+$ so $u_+(f_1)+u_+(f_-)=u_+(f_2)+u_+(f_+).$] Moreover,

$$\left\vert u_+ (f) \right\vert \leq \max (u_+ (f_+) , u (f_-)) \leq \Vert u \Vert \Vert f \Vert _{\infty}$$

$$\Longrightarrow \Vert u_+ \Vert \leq \Vert u \Vert .$$

The functional

$$u_- = u_+ -u$$

is also positive, since $u_+ (f) \geq u(f)$ for all $0 \leq f \in \mathcal{C}_0 (x)$. Thus we have proved

The idea behind the definition of $u_+$ is that $u$ itself is, more or less, ``integration against a function'' (even though we do not know how to interpret this yet). We are trying to throw away the negative part of that function. The next step is to show that a positive functional corresponds to a `measure' meaning a function measuring the size of sets. To define this we really want to evaluate $u$ on the characteristic function of a set

$$\chi_E (x) = \left\{ \begin{array}{ll} 1 & \hbox{ if } x \in E \ 0 & \hbox{ if } x \notin E . \end{array} \right.$$

The problem is that $\chi_E$ is not continuous. Instead we use an idea similar to (1.8).

If $0 \leq u \in (\mathcal{C}_0 (X))'$ and $U\subset X$ is open, set¹

$$\mu (U) = \sup \left\{ u (f) ; 0 \leq f(x) \leq 1 , f \in \mathcal{C}_0(X) , \operatorname{supp} (f) \Subset U \right\} .$$ (10)

Here the support of $f$, $\operatorname{supp} (f)$, is the closure of the set of points where $f(x) \neq 0$. Thus $\operatorname{supp} (f)$ is always closed, in (1.11) we only admit $f$ if its support is a compact subset of $U.$ The reason for this is that, only then do we `really know' that $f\in \mathcal{C}_0(X)$.

Suppose we try to measure general sets in this way. We can do this by defining

$$\mu^* (E) = \inf\left\{ \mu (U) ; U \supset E , U \hbox{ open} \right\} .$$ (11)

Already with $\mu$ it may happen that $\mu (U) = \infty$, so we think of

$$\mu^* : {\mathcal P}(X) \to [0, \infty ]$$ (12)

as defined on the power set of $X$ and taking values in the extended positive real numbers.

To prove this we need to find enough continuous functions. I have relegated the proof of the following result to Problem 2.

Proof. [Proof of Lemma 1.7] We have to prove (1.14). Suppose first that the $A_i$ are open, then so is $A= \bigcup_i A_i$. If $f \in C(X)$ and $\operatorname{supp} (f) \Subset A$ then $\operatorname{supp} (f)$ is covered by a finite union of the $A_i$s. Applying Lemma 1.8 we can find $f_i$'s, all but a finite number identically zero, so $\operatorname{supp} (f_i) \Subset A_i$ and $\sum_i f_i =1$ in a neighborhood of $\operatorname{supp} (f)$.

Since $f= \sum_i f_i f$ we conclude that

$$u (f) = \sum_i u (f_if) \Longrightarrow \mu^* (A) \leq \sum_i \mu^* (A_i)$$

since $0 \leq f_if \leq 1$ and $\operatorname{supp} (f_if) \Subset A_i$.

Thus (1.14) holds when the $A_i$ are open. In the general case if $A_i \subset B_i$ with the $B_i$ open then, from the definition,

$$\mu^* ( \bigcup_i A_i) \leq \mu^* (\bigcup_i B_i) \leq \sum_i \mu^* (B_i) .$$

Taking the infimum over the $B_i$ gives (1.14) in general. $\qedsymbol$