Test functions

So far we have largely been dealing with integration. One thing we have seen is that, by considering dual spaces, we can think of functions as functionals. Let me briefly review this idea.

Consider the unit ball in $\mathbb{R}^n$,

$$\overline{\mathbb{B}}^n = \left\{ x \in \mathbb{R}^n ; \left\vert x \right\vert \leq 1 \right\} .$$

I take the closed unit ball because I want to deal with a compact metric space. We have dealt with several Banach spaces of functions on $\overline{\mathbb{B}^n}$, for example

$$C(\overline{\mathbb{B}^n}) = \left\{ u: \overline{\mathbb{B}^n} \to \mathbb{C} ; u \hbox{ continuous} \right\}$$

$$L^2 (\overline{\mathbb{B}^n}) = \left\{ u: \overline{\mathbb{B}^n... ...orel measurable with } \int \left\vert u \right\vert^2 dx < \infty \right\}.$$

Here, as always below, $dx$ is Lebesgue measure and functions are identified if they are equal almost everywhere.

Since $\overline{\mathbb{B}^n}$ is compact we have a natural inclusion

$$C(\overline{\mathbb{B}^n}) \hookrightarrow L^2 (\overline{\mathbb{B}^n}) .$$ (58)

This is also a topological inclusion, i.e., is a bounded linear map, since

$$\Vert u \Vert _{L^2} \leq C \Vert u \vert\vert _{\infty}$$ (59)

where $C^2$ is the volume of the unit ball.

In general if we have such a set up then

Proof. By definition of the dual norm

$$\Vert \tilde{L} \Vert _{V'} = \sup \left\{ \left\vert \tilde{L} (v) \right\vert ; \Vert v \Vert _V \leq 1 , v \in V \right\}$$

$$\leq \sup \left\{ \left\vert \tilde{L} (v) \right\vert ; \Vert v \Vert _U \leq C , v \in V \right\}$$

$$\leq \sup \left\{ \left\vert L(u) \right\vert ; \Vert u \Vert _U \leq C , u \in U \right\}$$

$$=C \Vert L \Vert _{U'} .$$

If $V \subset U$ is dense then the vanishing of $L:U \to \mathbb{C}$ on $V$ implies its vanishing on $U$.

$\qedsymbol$

Going back to the particular case (5.1) we do indeed get a continuous map between the dual spaces

$$L^2 (\overline{\mathbb{B}^n}) \cong (L^2 (\overline{\mathbb{B}^n}))' \to (C(\overline{\mathbb{B}^n}))' = M(\overline{\mathbb{B}^n}) .$$

Here we use the Riesz representation theorem and duality for Hilbert spaces. The map use here is supposed to be linear not antilinear, i.e.,

$$L^2 (\overline{\mathbb{B}^n}) \ni g \longmapsto \int \cdot g dx \in (C(\overline{\mathbb{B}^n}))' .$$ (60)

So the idea is to make the space of `test functions' as small as reasonably possible, while still retaining density in reasonable spaces.

Recall that a function $u:\mathbb{R}^n \to \mathbb{C}$ is differentiable at $\overline{x} \in \mathbb{R}^n$ if there exists $a \in \mathbb{C}^n$ such that

$$\left\vert u (x) - u(\overline{x}) - a \cdot (x- \overline{x}) \right\vert = o( \left\vert x - \overline{x} \right\vert ) .$$ (61)

The `little oh' notation here means that given $\epsilon >0$ there exists $\delta > 0$ s.t.

$$\left\vert x - \overline{x} \right\vert < \delta \Rightarrow \lef... ...- \overline{x}) \right\vert < \epsilon \left\vert x - \overline{x} \right\vert.$$

The coefficients of $a= (a_1 , \ldots , a_n)$ are the partial derivations of $u$ at $\overline{x}$,

$$a_i= \frac{\partial u}{\partial x_j} (\overline{x})$$

since

$$a_i = \lim_{t \to 0} \frac{u(\overline{x} + t e_i) -u(\overline{x})}{t} ,$$ (62)

$e_i = (0, \ldots , 1, 0 , \ldots , 0)$ being the $i$th basis vector. The function $u$ is said to be continuously differentiable on $\mathbb{R}^n$ if it is differentiable at each point $\overline{x} \in \mathbb{R}^n$ and each of the $n$ partial derivatives are continuous,

$$\frac{\partial u}{\partial x_j}: \mathbb{R}^n \to \mathbb{C} .$$ (63)

Proof. That $\Vert \ \Vert _{\mathcal{C}^1}$ is a norm follows from the properties of $\Vert \Vert _{\infty}$. Namely $\Vert u \Vert _{\mathcal{C}^1} =0$ certainly implies $u=0$, $\Vert a u \Vert _{\mathcal{C}^1} = \left\vert a \right\vert \, \Vert u \Vert _{\mathcal{C}^1}$ and the triangle inequality follows from the same inequality for $\Vert \Vert _{\infty}$.

Similarly, the main part of the completeness of $\mathcal{C}^1_0 (\mathbb{R}^n)$ follows from the completeness of $\mathcal{C}^0_0 (\mathbb{R}^n)$. If $\left\{ u_n \right\}$ is a Cauchy sequence in $\mathcal{C}^1_0 (\mathbb{R}^n)$ then $u_n$ and the $\frac{\partial u_n}{\partial x_j}$ are Cauchy in $\mathcal{C}^0_0 (\mathbb{R}^n)$. It follows that there are limits of these sequences,

$$u_n \to v \, , \, \frac{\partial u_n}{\partial x_j} \to v_j \in \mathcal{C}^0_0 (\mathbb{R}^n) \, .$$

However we do have to check that $v$ is continuously differentiable and that $\frac{\partial v}{\partial x_j}=v_j$.

One way to do this is to use the Fundamental Theorem of Calculus in each variable. Thus

$$u_n (\overline{x} + t e_i)= \int^t_0 \frac{\partial u_n}{\partial x_j} (\overline{x} + s e_i) ds + u_n (\overline{x}) .$$

As $n \to \infty$ all terms converge and so, by the continuity of the integral,

$$u( \overline{x} + t e_i)= \int^t_0 v_j (\overline{x} + s e_i) ds + u (\overline{x}) .$$

This shows that the limit in (5.6) exists, so $v_i(\overline{x})$ is the partial derivation of $u$ with respect to $x_i.$ It remains only to show that $u$ is indeed differentiable at each point and I leave this to you in Problem 17.

$\qedsymbol$

So, almost by definition, we have an example of Lemma 5.1, $\mathcal{C}^1_0 (\mathbb{R}^n) \hookrightarrow \mathcal{C}^0_0 (\mathbb{R}^n)$. It is in fact dense but I will not bother showing this (yet). So we know that

$$(\mathcal{C}^0_0(\mathbb{R}^n))' \to (\mathcal{C}^1_0(\mathbb{R}^n))'$$

and we expect it to be injective. Thus there are more functionals on $\mathcal{C}^1_0 (\mathbb{R}^n)$ including things that are `more singular than measures'.

An example is related to the Dirac delta

$$\delta (\overline{x}) (u) = u (\overline{x}) \, , \ u \in \mathcal{C}^0_0(\mathbb{R}^n) \, ,$$

namely

$$\mathcal{C}^1_0 (\mathbb{R}^n) \ni u \longmapsto \frac{\partial u}{\partial x_j} (\overline{x}) \in \mathbb{C}\, .$$

This is clearly a continuous linear functional which it is only just to denote $\frac{\partial}{\partial x_j} \delta (\overline{x})$.

Of course, why stop at one derivative?

These are all Banach spaces, since if $\left\{ u_n \right\}$ is Cauchy in $\mathcal{C}^k_0 (\mathbb{R}^n)$, it is Cauchy and hence convergent in $\mathcal{C}^{k-1}_0 (\mathbb{R}^n)$, as is $\partial u_n / \partial x_j$, $j=1 , \ldots , n-1.$ Furthermore the limits of the $\partial u_n / \partial x_j$ are the derivatives of the limits by Proposition 5.3.

This gives us a sequence of spaces getting `smoother and smoother'

$$\mathcal{C}^0_0 (\mathbb{R}^n) \supset \mathcal{C}^1_0 (\mathbb{R}^n) \supset \cdots \supset \mathcal{C}^k_0 (\mathbb{R}^n) \supset \cdots \, ,$$

with norms getting larger and larger. The duals can also be expected to get larger and larger as $k$ increases.

As well as looking at functions getting smoother and smoother, we need to think about `infinity', since $\mathbb{R}^n$ is not compact. Observe that an element $g \in L^1 (\mathbb{R}^n)$ (with respect to Lebesgue measure by default) defines a functional on $\mathcal{C}^0_0 (\mathbb{R}^n)$ -- and hence all the $\mathcal{C}^k_0 (\mathbb{R}^n)$s. However a function such as the constant function $1$ is not integrable on $\mathbb{R}^n.$ Since we certainly want to talk about this, and polynomials, we consider a second condition of smallness at infinity. Let us set

$$\langle x \rangle = (1+ \left\vert x \right\vert^2)^{1/2}$$ (64)

a function which is the size of $\left\vert x \right\vert$ for $\left\vert x \right\vert$ large, but has the virtue of being smooth¹⁰

Notice that the definition just says that $u= \langle x \rangle^{- l} v$, with $v \in \mathcal{C}^k_0 (\mathbb{R}^n)$. It follows immediately that $\langle x \rangle^{- l} \mathcal{C}^k_0 (\mathbb{R}^n)$ is a Banach space with this norm.
¹¹

It is not immediately apparent that this space is non-empty (well 0 is in there but...); that

$$\exp (- \left\vert x \right\vert^2) \in \mathcal{S} (\mathbb{R}^n)$$

is Problem 19.

Schwartz idea is that the dual of $\mathcal{S}(\mathbb{R}^n)$ should contain all the `interesting' objects, at least those of `polynomial growth'. The problem is that we do not have a norm on $\mathcal{S}(\mathbb{R}^n)$. Rather we have a lot of them. Observe that

$$\langle x \rangle^{- l} \mathcal{C}^k_0 (\mathbb{R}^n) \subset \l... ...hcal{C}^{k'}_0 (\mathbb{R}^n) \hbox{ if } l \geq l' \hbox{ and } k \geq k' \, .$$

Thus we see that as a linear space

$$\mathcal{S} (\mathbb{R}^n) = \bigcap_k \langle x \rangle^{-k} \mathcal{C}^k_0 (\mathbb{R}^n) \, .$$ (65)

Since these spaces are getting smaller, we have a countably infinite number of norms. For this reason $\mathcal{S}(\mathbb{R}^n)$ is called a countably normed space.

Proof. The series in (5.12) certainly converges, since

$$\frac{\Vert u-v \Vert _{(k)}}{1+ \Vert u-v \Vert _{(k)}} \leq 1 .$$

The first two conditions on a metric are clear,

$$d(u,v)=0 \, \Rightarrow \Vert u-v \Vert _{\mathcal{C}_0} =0 \Rightarrow u=v,$$

and symmetry is immediate. The triangle inequaly is perhaps more mysterious!

Certainly it is enough to show that

$$\tilde{d} (u,v) = \frac{\Vert u-v \Vert}{1+ \Vert u-v \Vert}$$ (66)

is a metric on any normed space, since then we may sum over $k$. Thus we consider

$$\begin{multline*} \frac{\Vert u-v \Vert}{1+ \Vert u-v \Vert} + \frac{\Ver... ...-v \Vert )} {(1+ \Vert u-v \Vert ) ( 1+ \Vert v-w \Vert )} . \end{multline*}$$

Comparing this to $\tilde{d} (v,w)$ we must show that

$$\begin{multline*} (1+ \Vert u-v \Vert )(1+ \Vert v-w \Vert ) \Vert u-w \Vert \\... ...\Vert v-w \Vert (1+ \Vert u-v \Vert)) (1+ \Vert u-w \Vert) . \end{multline*}$$

Starting from the LHS and using the triangle inequality,

$$\hbox{LHS } \leq \Vert u-w \Vert + ( \Vert u-v \Vert + \Vert v-w \Vert + \Vert u-v \Vert \Vert v-w \Vert) \Vert u-w \Vert$$

$$\leq ( \Vert u-v \Vert + \Vert v-w \Vert + \Vert u-v \Vert \Vert v-w \Vert) (1+ \Vert u-w \Vert)$$

$$\leq \hbox{RHS} .$$

Thus, $d$ is a metric.

Suppose $u_n$ is a Cauchy sequence. Thus, $d(u_n , u_m) \to 0$ as $n,m \to \infty$. In particular, given

$$\epsilon >0 \exists N \hbox{ s.t. } n,m >N \hbox{ implies }$$

$$d(u_n ,u_m ) < \epsilon 2^{-k} \forall n,m >N.$$

The terms in (5.12) are all positive, so this implies

$$\frac{\Vert u_n - u_m \Vert _{(k)}}{1+ \Vert u_n - u_m \Vert _{(k)}} < \epsilon \forall n,m >N .$$

If $\epsilon < 1/2$ this in turn implies that

$$\Vert u_n - u_m \Vert _{(k)} < 2 \epsilon ,$$

so the sequence is Cauchy in $\langle x \rangle^{-k} \mathcal{C}^k_0 (\mathbb{R}^n)$. From the completeness of these spaces it follows that $u_n \to u$ in $\langle x \rangle^{-k} \mathcal{C}^k_0 (\mathbb{R}^n)_j$ for each $k.$ Given $\epsilon >0$ choose $k$ so large that $2^{-k} < \epsilon /2$. Then $\exists$ $N$ s.t. $n>N$

$$\Rightarrow \Vert u-u_n \Vert _{(j)} < \epsilon /2 n>N , j \leq k .$$

Hence

$$d(u_n , u) = \sum_{j \leq k} 2^{-j} \frac{\Vert u-u_n \Vert _{(j)}}{1+ \Vert u-u_n \Vert _{(j)}}$$

$$+ \sum_{j>k} 2^{-j} \frac{\Vert u-u_n \Vert _{(j)}}{1+ \Vert u-u_n \Vert _{(j)}}$$

$$\leq \epsilon / 4 + 2^{-k} < \epsilon .$$

This $u_n \to u$ in $\mathcal{S}(\mathbb{R}^n)$. $\qedsymbol$

A discussion of $\mathcal{C}_c(\bbR^n)$ should go here.