# $C^{k}$ spaces and spaces of test functions

Jordan Bell
April 10, 2014

## 1 Notation

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

 $|\alpha|=\alpha_{1}+\cdots+\alpha_{n},$

and

 $\partial^{\alpha}=\partial_{1}^{\alpha_{1}}\cdots\partial_{n}^{\alpha_{n}}.$

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

## 2 Open sets

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

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 $\Omega=\bigcup_{j=1}^{\infty}K_{j}$; we call this an exhaustion of $\Omega$ by compact sets. For $f\in C^{k}(\Omega)$, we define

 $p_{k,N}(f)=\sup_{|\alpha|\leq\min(k,N)}\sup_{x\in K_{N}}|(\partial^{\alpha}f)(% x)|;$

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

 $V_{k,N}=\left\{f\in C^{k}(\Omega):p_{k,N}(f)<\frac{1}{N}\right\}$

form a local basis at $0$ for a topology on $C^{k}(\Omega)$, and because $p_{k,N}$ is a separating family of seminorms, with this topology $C^{k}(\Omega)$ is a locally convex space.11 1 Walter Rudin, Functional Analysis, second ed., p. 27, Theorem 1.37. Because $p_{k,N}$ is a countable separating family of seminorms, this topology is metrizable. We prove in the following theorem that $C(\Omega)$ is a Fréchet space.22 2 Walter Rudin, Functional Analysis, second ed., p. 33, Example 1.44.

###### Theorem 1.

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

###### Proof.

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

 $f_{i}-f_{j}\in V_{0,N}=\left\{f\in C(\Omega):\sup_{x\in K_{N}}|f(x)|<\frac{1}{% N}\right\}.$

For each $x\in\Omega$, eventually $x\in K_{N}$. If $x\in K_{N}$ and $i,j\geq i_{N}$, then

 $|f_{i}(x)-f_{j}(x)|<\frac{1}{N}.$

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

Let $K$ be a compact subset of $\Omega$, let $\epsilon>0$, and let $N$ be large enough both so that $K\subseteq K_{N}$ and so that $N\geq\frac{1}{\epsilon}$. For $i,j\geq i_{N}$,

 $\sup_{x\in K_{N}}|f_{i}(x)-f_{j}(x)|<\frac{1}{N}\leq\epsilon.$

Let $i\geq i_{N}$ and $x\in K_{N}$. There is some $j_{x}$ such that $j\geq j_{x}$ implies that $|f_{j}(x)-f(x)|<\epsilon$, and hence for $j\geq\max(i_{N},j_{x})$,

 $\displaystyle|f_{i}(x)-f(x)|$ $\displaystyle\leq$ $\displaystyle|f_{i}(x)-f_{j}(x)|+|f_{j}(x)-f(x)|$ $\displaystyle<$ $\displaystyle\epsilon+\epsilon.$

This shows that for $i\geq i_{N}$,

 $\sup_{x\in K}|f_{i}(x)-f(x)|\leq\sup_{x\in K_{N}}|f_{i}(x)-f(x)|\leq 2\epsilon.$

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

Let $x\in\Omega$, let $\epsilon>0$, and let $N$ be large enough both so that $x$ lies in the interior of $K_{N}$ and so that $N\geq\frac{1}{\epsilon}$. Because $\sup_{x\in K_{N}}|f_{i}(x)-f(x)|\to 0$ as $i\to\infty$, there is some $i_{0}$ so that $i\geq i_{0}$ implies

 $\sup_{x\in K_{N}}|f_{i}(x)-f(x)|<\epsilon.$

Let $i=\max(i_{0},i_{N})$. Because $f_{i}$ is continuous, there is some $\delta>0$ so that $|x-y|<\delta$ implies that $|f_{i}(x)-f_{i}(y)|<\epsilon$; take $\delta$ small enough so that the open ball with center $x$ and radius $\delta$ is contained in $K_{N}$. For $|y-x|<\delta$,

 $\displaystyle|f(x)-f(y)|$ $\displaystyle\leq$ $\displaystyle|f(x)-f_{i}(x)|+|f_{i}(x)-f_{i}(y)|+|f_{i}(y)-f(y)|$ $\displaystyle\leq$ $\displaystyle\sup_{z\in K_{N}}|f(z)-f_{i}(z)|+\frac{1}{N}+\sup_{z\in K_{N}}|f(% z)-f_{i}(z)|$ $\displaystyle<$ $\displaystyle\epsilon+\epsilon+\epsilon.$

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

We have already established that for any compact subset $K$ of $\Omega$, we have $\sup_{x\in K}|f_{i}(x)-f(x)|\to 0$ as $i\to\infty$. Thus, for any $N$, there is some $j_{N}$ so that if $i\geq j_{N}$ then $\sup_{x\in K_{N}}|f_{i}(x)-f(x)|<\frac{1}{N}$. In other words, if $i\geq j_{N}$, then $p_{0,N}(f_{i}-f)<\frac{1}{N}$, i.e. $f_{i}-f\in V_{0,N}$, showing that $f_{i}\to f$ in $C(\Omega)$. ∎

###### Theorem 2.

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

###### Proof.

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

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

For each $1\leq p\leq n$ we have $\partial_{p}f_{i}\in C^{k-1}(\Omega)$, and $\partial_{p}f_{i}$ is a Cauchy sequence in $C^{k-1}(\Omega)$. Because $C^{k-1}(\Omega)$ is a Fréchet space, for each $p$ there is some $g_{p}\in C^{k-1}(\Omega)$ such that $\partial_{p}f_{i}\to g_{p}$ in $C^{k-1}(\Omega)$. Fix $p$, and let $\alpha\in\mathbb{N}^{n}$ have $p$th entry $1$ and all other entries $0$. Then, fix $x\in\Omega$, 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 $C(\Omega)$ and $\partial_{p}f_{i}\to g_{p}$ in $C(\Omega)$, 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 $\Omega$, so $\partial_{p}g=g_{p}\in C^{k-1}(\Omega)$. Thus, for each $1\leq p\leq n$ we have $\partial_{p}g\in C^{k-1}(\Omega)$, from which it follows that $g\in C^{k}(\Omega)$. ∎

###### Theorem 3.

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

###### Proof.

Let $f_{i}\in C^{\infty}(\Omega)$ be a Cauchy sequence in $C^{\infty}(\Omega)$. Thus, for each $k$, $f_{i}$ is a Cauchy sequence in $C^{k}(\Omega)$, and so by Theorem 2 there is some $g_{k}\in C^{k}(\Omega)$ for which $f_{i}\to g_{k}$ in $C^{k}(\Omega)$. Define $g=g_{0}$, and check that $g_{0}=g_{1}=g_{2}=\cdots$, and hence that $g\in C^{\infty}(\Omega)$. ∎

## 3 Closed sets

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

 $\|f\|_{k}=\sup_{|\alpha|\leq k}\sup_{x\in\overline{\Omega}}|(\partial^{\alpha}% f)(x)|.$

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

###### Theorem 4.

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

###### Proof.

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

 $\sup_{x\in\overline{\Omega}}|f_{i}(x)-f_{j}(x)|<\epsilon.$

Then, for each $x\in\overline{\Omega}$ we have that $f_{i}(x)$ is a Cauchy sequence in $\mathbb{C}$ and hence converges to some $f(x)\in\mathbb{C}$, thus defining a function $f:\overline{\Omega}\to\mathbb{C}$. For $x\in\overline{\Omega}$ and $\epsilon>0$, because $f_{i}(x)\to f(x)$, there is some $j_{x}$ such that $j\geq j_{x}$ implies that $|f_{j}(x)-f(x)|<\epsilon$. For $i\geq i_{\epsilon}$ and $j\geq\max(i_{\epsilon},j_{x})$,

 $|f_{i}(x)-f(x)|\leq|f_{i}(x)-f_{j}(x)|+|f_{j}(x)-f(x)|<\epsilon+\epsilon.$

This shows that $\sup_{x\in\overline{\Omega}}|f_{i}(x)-f(x)|\to 0$ as $i\to\infty$.

Fix $x\in\Omega$ and let $\epsilon>0$. What we just proved shows that there is some $i_{0}$ for which $i\geq i_{0}$ implies that $\sup_{z\in\overline{\Omega}}|f_{i}(z)-f(z)|<\epsilon$. As $f_{i_{0}}:\overline{\Omega}\to\mathbb{C}$ is continuous, there is some $\delta>0$ such that for $y\in B_{\delta}(x)\cap\overline{\Omega}$, we have $|f_{i_{0}}(x)-f_{i_{0}}(y)|<\epsilon$. Then, for $y\in B_{\delta}(x)\cap\overline{\Omega}$,

 $\displaystyle|f(x)-f(y)|$ $\displaystyle\leq$ $\displaystyle|f(x)-f_{i_{0}}(x)|+|f_{i_{0}}(x)-f_{i_{0}}(y)|+|f_{i_{0}}(y)-f(y)|$ $\displaystyle<$ $\displaystyle\epsilon+\epsilon+\epsilon.$

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

###### Theorem 5.

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

###### Proof.

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

Let $f_{i}\in C^{k}(\overline{\Omega})$ be a Cauchy sequence. In particular, $f_{i}$ is a Cauchy sequence in $C(\overline{\Omega})$, and because $C(\overline{\Omega})$ is a Banach space, there is some $g\in C(\overline{\Omega})$ for which $\|f_{i}-g\|_{0}\to 0$. For each $1\leq p\leq n$ we have $\partial_{p}f_{i}\in C^{k-1}(\overline{\Omega})$. Because $C^{k-1}(\overline{\Omega})$ is a Banach space, for each $p$ there is some $g_{p}\in C^{k-1}(\overline{\Omega})$ for which $\|\partial_{p}f_{i}-g_{p}\|_{k-1}\to 0$.

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

 $f_{i}(x+\tau\alpha)-f_{i}(x)=\int_{0}^{\tau}(\partial_{p}f_{i})(x+t\alpha)dt.$

Because $\|f_{i}-g\|_{0}\to 0$ and $\|\partial_{p}f_{i}-g_{p}\|_{0}\to 0$ (the latter because $\|\partial_{p}f_{i}-g_{p}\|_{k-1}\to 0$), 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 $x\in\Omega$, and so $\partial_{p}g=g_{p}\in C^{k-1}(\overline{\Omega})$. Because this is true for each $1\leq p\leq n$, we obtain $g\in C^{k}(\overline{\Omega})$. ∎

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

 $C^{\infty}(\overline{\Omega})=\bigcap_{k=0}^{\infty}C^{k}(\overline{\Omega}).$

It can be proved that $C^{\infty}(\overline{\Omega})$ is the projective limit of the Banach spaces $C^{k}(\overline{\Omega})$, $k=0,1,\ldots$.33 3 See Paul Garrett, Banach and Fréchet spaces of functions, http://www.math.umn.edu/~garrett/m/fun/notes_2012-13/02_spaces_fcns.pdf A projective limit of a countable projective system of Banach spaces is a Fréchet space, and thus $C^{\infty}(\overline{\Omega})$ is a Fréchet space.

## 4 Test functions

Let $\Omega$ be an open subset of $\mathbb{R}^{n}$. If $f:\Omega\to\mathbb{C}$ is a function, the support of $f$ is the closure of the set $\{x\in\Omega:f(x)\neq 0\}$. 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 $C_{c}^{\infty}(\Omega)$ the set of all elements of $C^{\infty}(\Omega)$ with compact support. We write $\mathscr{D}(\Omega)=C_{c}^{\infty}(\Omega)$.

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

 $\|f\|_{N}=\sup_{|\alpha|\leq N}\sup_{x\in\Omega}|(\partial^{\alpha}f)(x)|.$

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

 $\mathscr{D}(K)=\{f\in C_{c}^{\infty}(\Omega):\mathrm{supp}\,f\subseteq K\}.$

The restriction of these norms to $\mathscr{D}(K)$ are norms, in particular seminorms. Hence, with the topology for which a local basis at $0$ is the collection of sets of the form $\{f\in\mathscr{D}(K):\|f\|_{N}<\frac{1}{N}\}$, we have that $\mathscr{D}(K)$ is a locally convex space, and because there are countably many seminorms $\|\cdot\|_{N}$, the space is metrizable. One checks that the topology on $\mathscr{D}(K)$ is equal to the subspace topology it inherits from $C^{\infty}(\Omega)$.44 4 Walter Rudin, Functional Analysis, second ed., p. 151. Theorem 3 tells us that $C^{\infty}(\Omega)$ is a Fréchet space, and in the following theorem we show that $\mathscr{D}(K)$ is a closed subspace of this Fréchet space, and hence is a Fréchet space itself.

###### Theorem 6.

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

###### Proof.

Let $f_{i}\in\mathscr{D}(K)$, $f\in C^{\infty}(\Omega)$, and suppose that $f_{i}\to f$ in $C^{\infty}(\Omega)$. If $x\in\Omega\setminus K$, 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

 $|f(x)|=|0-f(x)|=|f_{i}(x)-f(x)|\leq\sup_{y\in K_{N}}|f_{i}(y)-f(y)|\to 0,$

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

Let $K_{j}$ be an exhaustion of $\Omega$ by compact sets. Check that $\mathscr{D}(K_{j})$ is a closed subspace of $\mathscr{D}(K_{j+1})$ and that the inclusion $\mathscr{D}(K_{j})\hookrightarrow\mathscr{D}(K_{j+1})$ is a homeomorphism onto its image. We define the following topology on the set $\mathscr{D}(\Omega)$. Let $\mathscr{B}$ be the collection of all convex balanced subsets $V$ of $\mathscr{D}(U)$ such that for all $j$, the set $V\cap\mathscr{D}(K_{j})$ is open in $\mathscr{D}(K_{j})$. (To be balanced means that $\alpha V\subseteq V$ if $|\alpha|\leq 1$.) We define $\mathscr{T}$ be the collection of all subsets $U$ of $\mathscr{D}(\Omega)$ such that $x_{0}\in U$ implies that there is some $V\in\mathscr{B}$ for which $x_{0}+V\subseteq U$. We check that $\mathscr{T}$ is a topology on $\mathscr{D}(\Omega)$, which we call the strict inductive limit topology. One proves55 5 John B. Conway, A Course in Functional Analysis, second ed., pp. 116–123, chap. IV, §5; this is presented without using the language of inductive limits in Walter Rudin, Functional Analysis, second ed., p. 152, Theorem 6.4. that with this topology, $\mathscr{D}(\Omega)$ is a locally convex space. With the strict inductive limit topology, we call the locally convex space $\mathscr{D}(\Omega)$ the strict inductive limit of the Fréchet spaces $\mathscr{D}(K_{1})\hookrightarrow\mathscr{D}(K_{2})\hookrightarrow\cdots$, and write

 $\mathscr{D}(\Omega)=\varinjlim\mathscr{D}(K_{j}).$