$C^k$ spaces and spaces of test functions

1 Notation

Let $\mathbb{N}$ denote the set of nonnegative integers. For $\alpha \in {\mathbb{N}}^n$, we write

$$|\alpha |={\alpha }_1+\mathrm{\cdots }+{\alpha }_n,$$

and

$${\partial }^{\alpha }={\partial }_1^{{\alpha }_1}\mathrm{\cdots }{\partial }_n^{{\alpha }_n}.$$

We denote by $B_r(x)$ the open ball with center $x$ and radius $r$.

2 Open sets

Let $\mathrm{\Omega }$ be an open subset of ${ℝ}^n$ and let $k$ be either a nonnegative integer or $\mathrm{\infty }$. We define $C^k(\mathrm{\Omega })$ to be the set of those functions $f:\mathrm{\Omega }\to ℂ$ such that for each $\alpha \in {\mathbb{N}}^n$ with $|\alpha |\le k$, the derivative ${\partial }^{\alpha }f$ exists and is continuous. We write $C(\mathrm{\Omega })=C^0(\mathrm{\Omega })$.

One proves that there is a sequence of compact sets $K_j$ such that each $K_j$ is contained in the interior of $K_{j+1}$ and $\mathrm{\Omega }={\bigcup }_{j=1}^{\mathrm{\infty }}{K}_j$; we call this an exhaustion of $\mathrm{\Omega }$ by compact sets. For $f\in C^k(\mathrm{\Omega })$, we define

$$p_{k,N}(f)=\underset{|\alpha |\le \min (k,N)}{sup}\underset{x\in K_N}{sup}|({\partial }^{\alpha }f)(x)|;$$

this definition makes sense for $k=\mathrm{\infty }$. If $f$ is a nonzero element of $C^k(\mathrm{\Omega })$, then there is some $x\in \mathrm{\Omega }$ for which $f(x)\ne 0$ and then there is some $N$ for which $x\in K_N$, and hence $p_{k,N}(f)\ge {sup}_{y\in K_N}|f(y)|\ge |f(x)|>0$. Thus, $p_{k,N}$ is a separating family of seminorms on $C^k(\mathrm{\Omega })$. Those sets of the form

form a local basis at $0$ for a topology on $C^k(\mathrm{\Omega })$, and because $p_{k,N}$ is a separating family of seminorms, with this topology $C^k(\mathrm{\Omega })$ is a locally convex space. Because $p_{k,N}$ is a countable separating family of seminorms, this topology is metrizable. We prove in the following theorem that $C(\mathrm{\Omega })$ is a Fréchet space.

Theorem 1.

If $\mathrm{\Omega }$ is an open subset of ${\mathrm{R}}^n$, then $C\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{)}$ is a Fréchet space.

Proof.

Let $f_i\in C(\mathrm{\Omega })$ be a Cauchy sequence. That is, for every $N$ there is some $i_N$ such that if $i,j\ge i_N$ then

For each $x\in \mathrm{\Omega }$, eventually $x\in K_N$. If $x\in K_N$ and $i,j\ge i_N$, then

Therefore, $f_i(x)$ is a Cauchy sequence in $ℂ$ and hence converges to some $f(x)\in ℂ$. We have thus defined a function $f:\mathrm{\Omega }\to ℂ$. We shall prove that $f\in C(\mathrm{\Omega })$ and that $f_i\to f$ in $C(\mathrm{\Omega })$.

Let $K$ be a compact subset of $\mathrm{\Omega }$, let $ϵ>0$, and let $N$ be large enough both so that $K\subseteq K_N$ and so that $N\ge \frac{1}{ϵ}$. For $i,j\ge i_N$,

Let $i\ge i_N$ and $x\in K_N$. There is some $j_x$ such that $j\ge j_x$ implies that , and hence for $j\ge \max (i_N,j_x)$,

$|f_i(x)-f(x)|$ $\le$ $|f_i(x)-f_j(x)|+|f_j(x)-f(x)|$

$ϵ+ϵ.$

This shows that for $i\ge i_N$,

$$\underset{x\in K}{sup}|f_i(x)-f(x)|\le \underset{x\in K_N}{sup}|f_i(x)-f(x)|\le 2ϵ.$$

We have proved that for any compact subset $K$ of $\mathrm{\Omega }$, we have ${sup}_{x\in K}|f_i(x)-f(x)|\to 0$ as $i\to \mathrm{\infty }$.

Let $x\in \mathrm{\Omega }$, let $ϵ>0$, and let $N$ be large enough both so that $x$ lies in the interior of $K_N$ and so that $N\ge \frac{1}{ϵ}$. Because ${sup}_{x\in K_N}|f_i(x)-f(x)|\to 0$ as $i\to \mathrm{\infty }$, there is some $i_0$ so that $i\ge i_0$ implies

Let $i=\max (i_0,i_N)$. Because $f_i$ is continuous, there is some $\delta >0$ so that implies that ; take $\delta$ small enough so that the open ball with center $x$ and radius $\delta$ is contained in $K_N$. For ,

$|f(x)-f(y)|$ $\le$ $|f(x)-f_i(x)|+|f_i(x)-f_i(y)|+|f_i(y)-f(y)|$

$\le$ $\underset{z\in K_N}{sup}|f(z)-f_i(z)|+{\displaystyle \frac{1}{N}}+\underset{z\in K_N}{sup}|f(z)-f_i(z)|$

$ϵ+ϵ+ϵ.$

This shows that $f$ is continuous at $x$ and $x$ was an arbitrary point in $\mathrm{\Omega }$, hence $f\in C(\mathrm{\Omega })$.

We have already established that for any compact subset $K$ of $\mathrm{\Omega }$, we have ${sup}_{x\in K}|f_i(x)-f(x)|\to 0$ as $i\to \mathrm{\infty }$. Thus, for any $N$, there is some $j_N$ so that if $i\ge j_N$ then . In other words, if $i\ge j_N$, then , i.e. $f_i-f\in V_{0,N}$, showing that $f_i\to f$ in $C(\mathrm{\Omega })$. ∎

Theorem 2.

If $\mathrm{\Omega }$ is an open subset of ${\mathrm{R}}^n$ and $k$ is a positive integer, then $C^k\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{)}$ is a Fréchet space.

Proof.

We have proved in Theorem 1 that $C(\mathrm{\Omega })=C^0(\mathrm{\Omega })$ is a Fréchet space. We assume that $C^{k-1}(\mathrm{\Omega })$ is a Fréchet space, and using this induction hypothesis we shall prove that $C^k(\mathrm{\Omega })$ is a Fréchet space.

Let $f_i\in C^k(\mathrm{\Omega })$ be a Cauchy sequence in $C^k(\mathrm{\Omega })$. $f_i$ is in particular a Cauchy sequence in the Fréchet space $C(\mathrm{\Omega })$, hence there is some $g\in C(\mathrm{\Omega })$ such that $f_i\to g$ in $C(\mathrm{\Omega })$. We shall prove that $g\in C^k(\mathrm{\Omega })$ and that $f_i\to g$ in $C^k(\mathrm{\Omega })$.

For each $1\le p\le n$ we have ${\partial }_p{f}_i\in C^{k-1}(\mathrm{\Omega })$, and ${\partial }_p{f}_i$ is a Cauchy sequence in $C^{k-1}(\mathrm{\Omega })$. Because $C^{k-1}(\mathrm{\Omega })$ is a Fréchet space, for each $p$ there is some $g_p\in C^{k-1}(\mathrm{\Omega })$ such that ${\partial }_p{f}_i\to g_p$ in $C^{k-1}(\mathrm{\Omega })$. Fix $p$, and let $\alpha \in {\mathbb{N}}^n$ have $p$th entry $1$ and all other entries $0$. Then, fix $x\in \mathrm{\Omega }$, and take $N$ large enough so that $x$ lies in the interior of $K_N$. For each $i$, define $F_i(t)=f(x+t\alpha )$, for which

$$F_i^{\prime }(t)=(\nabla f)(x+t\alpha )\cdot \alpha =({\partial }_p{f}_i)(x+t\alpha ).$$

For nonzero $\tau$ small enough so that the line segment from $x$ to $x+\tau \alpha$ is contained in $K_N$,

$$F_i(\tau )-F_i(0)={\int }_0^{\tau }{F}_i^{\prime }(t)𝑑t,$$

i.e.

$$f_i(x+\tau \alpha )-f_i(x)={\int }_0^{\tau }({\partial }_p{f}_i)(x+t\alpha )𝑑t.$$

Because $f_i\to g$ in $C(\mathrm{\Omega })$ and ${\partial }_p{f}_i\to g_p$ in $C(\mathrm{\Omega })$, we have ${sup}_{y\in K_N}|f_i(y)-g(y)|\to 0$ and ${sup}_{y\in K_N}|({\partial }^p{f}_i)(y)-g_p(y)|\to 0$, from which it follows that

$$g(x+\tau \alpha )-g(x)={\int }_0^{\tau }{g}_p(x+t\alpha )𝑑t,$$

or

$$\frac{g(x+\tau \alpha )-g(x)}{\tau }=\frac{1}{\tau }{\int }_0^{\tau }{g}_p(x+t\alpha )𝑑t.$$

As $\tau$ tends to $0$, the right hand side tends to $g_{\alpha }(x)$, showing that $({\partial }_{p}g)(x)=g_p(x)$. But $x$ was an arbitrary point in $\mathrm{\Omega }$, so ${\partial }_{p}g=g_p\in C^{k-1}(\mathrm{\Omega })$. Thus, for each $1\le p\le n$ we have ${\partial }_{p}g\in C^{k-1}(\mathrm{\Omega })$, from which it follows that $g\in C^k(\mathrm{\Omega })$. ∎

Theorem 3.

If $\mathrm{\Omega }$ is an open subset of ${\mathrm{R}}^n$, then $C^{\mathrm{\infty }}\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{)}$ is a Fréchet space.

Proof.

Let $f_i\in C^{\mathrm{\infty }}(\mathrm{\Omega })$ be a Cauchy sequence in $C^{\mathrm{\infty }}(\mathrm{\Omega })$. Thus, for each $k$, $f_i$ is a Cauchy sequence in $C^k(\mathrm{\Omega })$, and so by Theorem 2 there is some $g_k\in C^k(\mathrm{\Omega })$ for which $f_i\to g_k$ in $C^k(\mathrm{\Omega })$. Define $g=g_0$, and check that $g_0=g_1=g_2=\mathrm{\cdots }$, and hence that $g\in C^{\mathrm{\infty }}(\mathrm{\Omega })$. ∎

3 Closed sets

Let $\mathrm{\Omega }$ be an open subset of ${ℝ}^n$ such that $\overline{\mathrm{\Omega }}$ is compact, i.e. $\mathrm{\Omega }$ is a bounded open subset of ${ℝ}^n$. If $k$ is a nonnegative integer, let $C^k(\overline{\mathrm{\Omega }})$ be those elements $f$ of $C^k(\mathrm{\Omega })$ such that for each $\alpha \in {\mathbb{N}}^n$ with $|\alpha |\le k$, the function ${\partial }^{\alpha }f$ is continuous $\mathrm{\Omega }\to ℂ$ and can be extended to a continuous function $\overline{\mathrm{\Omega }}\to ℂ$; if there is such a continuous function $\overline{\mathrm{\Omega }}\to ℂ$ it is unique, and it thus makes sense to talk about the value of ${\partial }^{\alpha }f$ at points in $\partial \mathrm{\Omega }$, and thus to write ${\partial }^{\alpha }f:\overline{\mathrm{\Omega }}\to ℂ$. We write $C(\overline{\mathrm{\Omega }})=C^0(\overline{\mathrm{\Omega }})$. For $f\in C^k(\overline{\mathrm{\Omega }})$, we define

$${\parallel f\parallel }_k=\underset{|\alpha |\le k}{sup}\underset{x\in \overline{\mathrm{\Omega }}}{sup}|({\partial }^{\alpha }f)(x)|.$$

It is straightforward to check that this is a norm on $C^k(\overline{\mathrm{\Omega }})$.

Theorem 4.

If $\mathrm{\Omega }$ is a bounded open subset of ${\mathrm{R}}^n$, then $C\mathit{}\mathrm{(}\overline{\mathrm{\Omega }}\mathrm{)}$ is a Banach space.

Proof.

Let $f_i\in C(\overline{\mathrm{\Omega }})$ be a Cauchy sequence. Thus, $f_i:\overline{\mathrm{\Omega }}\to ℂ$ are continuous, and for any $ϵ>0$ there is some $i_{ϵ}$ such that if $i,j\ge i_{ϵ}$ then

Then, for each $x\in \overline{\mathrm{\Omega }}$ we have that $f_i(x)$ is a Cauchy sequence in $ℂ$ and hence converges to some $f(x)\in ℂ$, thus defining a function $f:\overline{\mathrm{\Omega }}\to ℂ$. For $x\in \overline{\mathrm{\Omega }}$ and $ϵ>0$, because $f_i(x)\to f(x)$, there is some $j_x$ such that $j\ge j_x$ implies that . For $i\ge i_{ϵ}$ and $j\ge \max (i_{ϵ},j_x)$,

This shows that ${sup}_{x\in \overline{\mathrm{\Omega }}}|f_i(x)-f(x)|\to 0$ as $i\to \mathrm{\infty }$.

Fix $x\in \mathrm{\Omega }$ and let $ϵ>0$. What we just proved shows that there is some $i_0$ for which $i\ge i_0$ implies that . As $f_{i_0}:\overline{\mathrm{\Omega }}\to ℂ$ is continuous, there is some $\delta >0$ such that for $y\in B_{\delta }(x)\cap \overline{\mathrm{\Omega }}$, we have . Then, for $y\in B_{\delta }(x)\cap \overline{\mathrm{\Omega }}$,

$|f(x)-f(y)|$ $\le$ $|f(x)-f_{i_0}(x)|+|f_{i_0}(x)-f_{i_0}(y)|+|f_{i_0}(y)-f(y)|$

$ϵ+ϵ+ϵ.$

This proves that $f$ is continuous at $x$, and because $x$ was an arbitrary point in $\overline{\mathrm{\Omega }}$, we have that $f\in C(\overline{\mathrm{\Omega }})$. ∎

Theorem 5.

If $\mathrm{\Omega }$ is a bounded open subset of ${\mathrm{R}}^n$ and $k$ is a positive integer, then $C^k\mathit{}\mathrm{(}\overline{\mathrm{\Omega }}\mathrm{)}$ is a Banach space.

Proof.

We proved in Theorem 4 that $C(\overline{\mathrm{\Omega }})=C^0(\overline{\mathrm{\Omega }})$ is a Banach space. We assume that $C^{k-1}(\overline{\mathrm{\Omega }})$ is a Banach space, and using this induction hypothesis we shall prove that $C^k(\overline{\mathrm{\Omega }})$ is a Banach space.

Let $f_i\in C^k(\overline{\mathrm{\Omega }})$ be a Cauchy sequence. In particular, $f_i$ is a Cauchy sequence in $C(\overline{\mathrm{\Omega }})$, and because $C(\overline{\mathrm{\Omega }})$ is a Banach space, there is some $g\in C(\overline{\mathrm{\Omega }})$ for which ${\parallel f_i-g\parallel }_0\to 0$. For each $1\le p\le n$ we have ${\partial }_p{f}_i\in C^{k-1}(\overline{\mathrm{\Omega }})$. Because $C^{k-1}(\overline{\mathrm{\Omega }})$ is a Banach space, for each $p$ there is some $g_p\in C^{k-1}(\overline{\mathrm{\Omega }})$ for which ${\parallel {\partial }_p{f}_i-g_p\parallel }_{k-1}\to 0$.

Let $\alpha \in {\mathbb{N}}^n$ have $p$th entry $1$ and all other entries $0$, and let $x\in \mathrm{\Omega }$. For nonzero $\tau$ small enough so that the line segment from $x$ to $x+\tau \alpha$ is contained in $\mathrm{\Omega }$,

$$f_i(x+\tau \alpha )-f_i(x)={\int }_0^{\tau }({\partial }_p{f}_i)(x+t\alpha )𝑑t.$$

Because ${\parallel f_i-g\parallel }_0\to 0$ and ${\parallel {\partial }_p{f}_i-g_p\parallel }_0\to 0$ (the latter because ${\parallel {\partial }_p{f}_i-g_p\parallel }_{k-1}\to 0$), we obtain

$$g(x+\tau \alpha )-g(x)={\int }_0^{\tau }{g}_p(x+t\alpha )𝑑t,$$

or

$$\frac{g(x+\tau \alpha )-g(x)}{\tau }=\frac{1}{\tau }{\int }_0^{\tau }{g}_p(x+t\alpha )𝑑t.$$

As $\tau$ tends to $0$ the right hand side tends to $g_p(x)$, which shows that $({\partial }_{p}g)(x)=g_p(x)$. We did this for all $x\in \mathrm{\Omega }$, and so ${\partial }_{p}g=g_p\in C^{k-1}(\overline{\mathrm{\Omega }})$. Because this is true for each $1\le p\le n$, we obtain $g\in C^k(\overline{\mathrm{\Omega }})$. ∎

If $\mathrm{\Omega }$ is a bounded open subset of ${ℝ}^n$, then

$$C^{\mathrm{\infty }}(\overline{\mathrm{\Omega }})=\bigcap _{k=0}^{\mathrm{\infty }}{C}^k(\overline{\mathrm{\Omega }}).$$

It can be proved that $C^{\mathrm{\infty }}(\overline{\mathrm{\Omega }})$ is the projective limit of the Banach spaces $C^k(\overline{\mathrm{\Omega }})$, $k=0,1,\mathrm{\dots }$. A projective limit of a countable projective system of Banach spaces is a Fréchet space, and thus $C^{\mathrm{\infty }}(\overline{\mathrm{\Omega }})$ is a Fréchet space.

4 Test functions

Let $\mathrm{\Omega }$ be an open subset of ${ℝ}^n$. If $f:\mathrm{\Omega }\to ℂ$ is a function, the support of $f$ is the closure of the set $\{x\in \mathrm{\Omega }:f(x)\ne 0\}$. We denote the support of $f$ by $\mathrm{supp}f$. If $\mathrm{supp}f$ is a compact set, we say that $f$ has compact support, and we denote by $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ the set of all elements of $C^{\mathrm{\infty }}(\mathrm{\Omega })$ with compact support. We write $𝒟(\mathrm{\Omega })=C_c^{\mathrm{\infty }}(\mathrm{\Omega })$.

For $f\in 𝒟(\mathrm{\Omega })$, we define

$${\parallel f\parallel }_N=\underset{|\alpha |\le N}{sup}\underset{x\in \mathrm{\Omega }}{sup}|({\partial }^{\alpha }f)(x)|.$$

If $K$ is a compact subset of $\mathrm{\Omega }$, we define

$$𝒟(K)=\{f\in C_c^{\mathrm{\infty }}(\mathrm{\Omega }):\mathrm{supp}f\subseteq K\}.$$

The restriction of these norms to $𝒟(K)$ are norms, in particular seminorms. Hence, with the topology for which a local basis at $0$ is the collection of sets of the form , we have that $𝒟(K)$ is a locally convex space, and because there are countably many seminorms $\parallel \cdot {\parallel }_N$, the space is metrizable. One checks that the topology on $𝒟(K)$ is equal to the subspace topology it inherits from $C^{\mathrm{\infty }}(\mathrm{\Omega })$. Theorem 3 tells us that $C^{\mathrm{\infty }}(\mathrm{\Omega })$ is a Fréchet space, and in the following theorem we show that $𝒟(K)$ is a closed subspace of this Fréchet space, and hence is a Fréchet space itself.

Theorem 6.

If $\mathrm{\Omega }$ is an open subset of ${\mathrm{R}}^n$ and $K$ is a compact subset of $\mathrm{\Omega }$, then $\mathrm{D}\mathit{}\mathrm{(}K\mathrm{)}$ is a closed subspace of the Fréchet space $C^{\mathrm{\infty }}\mathit{}\mathrm{(}\mathrm{\Omega }\mathrm{)}$.

Proof.

Let $f_i\in 𝒟(K)$, $f\in C^{\mathrm{\infty }}(\mathrm{\Omega })$, and suppose that $f_i\to f$ in $C^{\mathrm{\infty }}(\mathrm{\Omega })$. If $x\in \mathrm{\Omega }\setminus K$, then $f_i(x)=0$. There is some $K_N$ that contains $K$, and the fact that $f_i\to f$ gives us in particular that

$$|f(x)|=|0-f(x)|=|f_i(x)-f(x)|\le \underset{y\in K_N}{sup}|f_i(y)-f(y)|\to 0,$$

hence $f(x)=0$. This shows that $\mathrm{supp}f\subseteq K$, and hence that $f\in 𝒟(K)$. ∎

Let $K_j$ be an exhaustion of $\mathrm{\Omega }$ by compact sets. Check that $𝒟(K_j)$ is a closed subspace of $𝒟(K_{j+1})$ and that the inclusion $𝒟(K_j)\hookrightarrow 𝒟(K_{j+1})$ is a homeomorphism onto its image. We define the following topology on the set $𝒟(\mathrm{\Omega })$. Let $ℬ$ be the collection of all convex balanced subsets $V$ of $𝒟(U)$ such that for all $j$, the set $V\cap 𝒟(K_j)$ is open in $𝒟(K_j)$. (To be balanced means that $\alpha V\subseteq V$ if $|\alpha |\le 1$.) We define $𝒯$ be the collection of all subsets $U$ of $𝒟(\mathrm{\Omega })$ such that $x_0\in U$ implies that there is some $V\in ℬ$ for which $x_0+V\subseteq U$. We check that $𝒯$ is a topology on $𝒟(\mathrm{\Omega })$, which we call the strict inductive limit topology. One proves that with this topology, $𝒟(\mathrm{\Omega })$ is a locally convex space. With the strict inductive limit topology, we call the locally convex space $𝒟(\mathrm{\Omega })$ the strict inductive limit of the Fréchet spaces $𝒟(K_1)\hookrightarrow 𝒟(K_2)\hookrightarrow \mathrm{\cdots }$, and write

$$𝒟(\mathrm{\Omega })=\underrightarrow{\lim }𝒟(K_j).$$