Topologies on Functional Spaces

Metric Spaces

Distance

A distance $d$ on a set $X$ is a map

$$ d:X\times X\to[0,\infty) $$ such that:

$d(x,y)=0$ iff $x=y$;
$d(x,y)=d(y,x)$ for all $x,y\in X$;
$d(x,y)+d(y,z)\geq d(x,z)$ for all $x,y,z\in X$.

Natural topology on $(X,d)$

We set $$ B_r(x) =\left\{y\in X|d(x,y) < r\right\} $$ and we call it ball of radius $r$ centered at $x$. We say that a set $U_x$ is a neighborhood of $x$ if it contains a ball of some positive radius centered at $x$ and we say that a set $U$ is open if it contains a neighborhood of each of its point.

The set of all open sets of a metric space satisfies the axioms for a topology and so defines a topology on $X$. This is the weakest topology that makes $d$ continuous (namely it is the intersection of all topologies that make $d$ continuous).

Exercise

Examples

To be done, I accept volunteers :-)

Complete metric spaces

We say that a sequence $\{x_n\}$ is Cauchy when, for every $\epsilon>0$, there is a $N$ s.t. $$ d(x_n,x_m) < \epsilon $$ for all $n,m\geq N$.
We say that $(X,d)$ is complete when all of its Cauchy sequences converge in $X$.

The set of rationals $\Bbb Q$ is a metric space with respect to the distance $d(r,s)=|r-s|$ but it is not complete. For instance, the sequence $\{r_k\}$, where $r_k$ is the truncation of $\sqrt{2}$ in decimal to the $k$-th digit, is clearly Cauchy (prove it!) but it does not converge to any rational.

Complete metric spaces

The set of real numbers $\Bbb R$ is a complete metric space with respect to the distance $d(r,s)=|r-s|$.

Every Cauchy sequence is bounded (prove it!) and so (by Bolzano-Weierstrass) it has at least an accumulation point. Since Cauchy sequences cannot have more than one accumulation point (prove it!), the sequence converges.

Completion of a metric space

Given any metric space $(X,d)$, there is a complete metric space $(\hat X,\hat d)$ and an injection $i:X\to\hat X$ such that:

$\hat d(i(x),i(y)) = d(x,y)$ for all $x,y\in X$;
$i(X)$ is dense in $\hat X$;
$\hat X$ is unique modulo homeomorphisms.

TBW

The completion of $\Bbb Q$ with the "absolute value" distance is $\Bbb R$ with the same distance.

An exotic example: $p$-adic spaces

Every rational number $r$ writes as $r=\Pi_{n=1}^\infty p_n^{a_n}$, where $p_1,p_2,\dots$ is the list of all prime numbers and the $a_i$ are integers (almost all of the $a_n$ are of course equal to zero). This decomposition is unique. We denote by $ord_p r$ the exponent of $p$ in the decomposition of $r$. The function $$ d_p(r,s) = \begin{cases}p^{-ord_p(r-s)}&,r\neq s\\ 0&,r=s\\\end{cases} $$ is a metric on $\Bbb Q$ (prove it!).
Moreover, $\Bbb Q$ is not complete with respect to any of this norms. The relative completions are the sets $\Bbb Q_p$ of p-adic numbers, each of which is therefore a sort of "alternate version" of the set of real numbers. If you are curious about these spaces, a good reference on the topology of the $\Bbb Q_p$ is this set of notes from ETH (Swisse).

A few fundamental results

Continuous functions are enough to separate closed sets:

(Urysohn Lemma) Let $A,B$ two closed non-empty disjoint subsets of $(X,d)$. Then there exists a continous function $f:X\to[0,1]$ s.t.

$f|_A=0$;
$f|_B=1$;
$f(x)\in(0,1)$ for all $x\in X\setminus(A\cup B)$.

see these notes

Continuous functions defined on any closed subset can be extended to the whole space:

(Tietze extension thm) Let $A$ be a closed bounded subset of $(X,d)$ and $f:A\to[a,b]$ a continuous functions. Then there is a continuous function $g:X\to[a,b]$ s.t. $g|_A=f$.

see these notes

Normed Spaces

Norms

A norm $\|\cdot\|$ on a vector space $V$ over a field $F$ (in our case $F$ will be either $\Bbb R$ or, more rarely, $\Bbb C$) is a map

$$ \|\cdot\|:V\times V\to[0,\infty) $$ such that:

$\|v\|=0$ iff $v=0$;
$\|\lambda v\|=|\lambda|\cdot\|v\|$ for all $v\in V$, $\lambda|\in F$;
$\|v+w\|\leq\|v\|+\|w\|$.

A few fundamental results

Every normed vector space $(V,\|\cdot\|)$ is a metric space for the metric $$ d(v,w)=\|v-w\|. $$

Exercise

In a normed space, scalar multiplication and vectors addition are continuous maps. In fact, the associated topology (coming from the metric above) is the weaker one s.t. those operations are continuous.

Exercise

A few fundamental results

Every vector space is a normed space with respect to some norm. In particular, therefore:

it admits a distance $d$ such that
1. $d(\lambda v,\lambda w)=|\lambda|d(v,w)$ (homogeneous);
2. $d(v+u,w+u)=d(v,w)$ (translation invariant).
it admits a (homogeneous and translation invariant) Hausdorff topology where scalar multiplication and vectors sum are continuous.

See math.stackexchange.

Banach Spaces

A complete normed vector space is called a Banach Space.

Every finite-dimensional normed space $V$ is a Banach space. Moreover, all norms on it are equivalent in the sense that they all induce the same topology.

TBW

Spaces of differentiable functions

Notations

In case of differentiable functions, there are two kinds of sets that play a major role: compact sets and open sets.

From now on, we will denote the first by $K$ and the second by $\Omega$.

Compact sets

We start considering the case of a compact subset $K\subset\Bbb R^n$ and the set $C^\infty(K)$ of all continuous functions with continuous derivatives of all orders (e.g. all polynomials belong to it).

On $C^\infty(K)$ one can consider many inequivalent norms, below are the more relevant for differentiable functions:

$\displaystyle\|f\|_{C^0} = \sup_{x\in K}|f(x)|$
$\displaystyle\|f\|_{C^1} = \|f\|_{C^0} + \sum_{i=1}^n\|\partial_{x^i}f\|_{C^0}$
$\displaystyle\|f\|_{C^k} = \|f\|_{C^{k-1}} + \sum_{1\leq i_1,\dots,i_k\leq n}\|\partial^k_{x^{i_1}\dots x^{i_k}}f\|_{C^0}$, $k\in\Bbb N$

$(C^\infty(K),\|\cdot\|_{C^k})$ is not complete

With simple examples you can convince yourself that $C^\infty(K)$ is not complete in any of those norms. E.g. for $K=[-1,1]$ one can easily write explicitly (try!) a sequence of $C^\infty$ functions converging uniformly (namely in the $C^0$ norm) in $K$ to $|x|$, a function that belongs to $C^0(K)$ but not to $C^\infty(K)$.

In fact, there is no norm on $C^\infty(K)$ for which all inclusions $C^k(K)\to C^\infty(K)$ are continuous, since with that induced topology it happens that every closed bounded subset of $C^\infty(K)$ is compact, and this can happen in a normed space iff its dimension is finite (see math.stackexchange for details).

Uniform convergence

Let $\{f_i\}\subset C^k(K)$ be a sequence converging uniformly together with all of its derivatives up to order $k$. Then $\lim_{i\to\infty}f_n\in C^k(K)$.

It is enough to prove the claim for $k=1$ and $K=[0,1]$. Since by hypothesis both $\{f_i\}$ and $\{f'_i\}$ converge uniformly on $K$, then both sequences converge to continuous functions $f_i\to f$ and $f_i'\to g$. We need to show that $f'=g$.
By the fundamental theorem of calculus, $f_i(x)-f_i(x_0)=\displaystyle\int\limits_{x_0}^x f'_i(t)dt.$
Since the convergence of the $f'_i$ is uniform, $\displaystyle \lim_{i\to\infty}\int\limits_{x_0}^x f'_i(t)dt =\int\limits_{x_0}^xg(t)dt$.
Indeed, for all $\epsilon > 0$ there is some $N$ s.t. $|f'_i(t)-g(t)| < \epsilon$ for all $x\in K$ and $i\geq N$, so that $$ \left|\int\limits_{x_0}^x f'_i(t)dt-\int\limits_{x_0}^xg(t)dt\right| \leq \int\limits_{x_0}^x|f'_i(t)-g(t)|dt \leq|x-x_0|\epsilon. $$ Hence $f(x)-f(x_0) = \lim_{i\to\infty}f_i(x)-f_i(x_0) = \lim_{i\to\infty}\int\limits_{x_0}^x f'_i(x)dx = \int\limits_{x_0}^x g(x)dx.$
By the fundamental theorem of calculus, this means precisely that $f'=g$.

Completions of $(C^\infty(K),\|\cdot\|_{C^k})$

The theorem above shows that a sequence of $C^\infty(K)$ functions converging in the $C^k$ norm converges to a function in $C^k(K)$.

Together with the Stone-Weierstrass theorem, namely that polynomials can uniformly approximate any continuous function over a compact set, this shows that

$${\overline{C^\infty(K)}}^{\|\cdot\|_{C^k}}= C^k(K)$$

Topology of $C^\infty(K)$

We already mentioned that there is no norm on $C^\infty(K)$ compatible with the fact that we want $C^\infty(K)$ continuously embedded in all $C^k(K)$, namely so that two functions $C^\infty$ close will be also $C^k$ close for all $k\in\Bbb N$.

The good news is that, though, we can get this topology with the following translation-invariant distance: $$ d(f,g) = \sum_{k=0}^\infty\frac{1}{2^k}\frac{\|f-g\|_{C^k}}{1+\|f-g\|_{C^k}} $$ This makes $C^\infty(K)$ a Frechet space, in particular a complete metric space.

Exercise: prove that, with this topology, all natural inclusions $C^\infty(K)\hookrightarrow C^k(K)$ are continuous.

The open case

We consider now the case of an open subset $\Omega\subset\Bbb R^n$. In this case, the $\sup$ norm is not well-defined anymore on the whole space.

To overcome this problem we must consider an "increasing" sequence of compact sets $K_i$ such that $\cup_i K_i=\Omega$ and use the same trick used to build a metric on $C^\infty(K)$, namely on $C^k(\Omega)$, $k=0,1,2,\dots$, we define the metric

$$ d(f,g) = \sum_{i=0}^\infty\frac{1}{2^i}\frac{\|f-g\|_{C^k(K_i)}}{1+\|f-g\|_{C^k(K_i)}} $$

This makes $C^k(\Omega)$ a Frechet space.

$C^\infty(\Omega)$

On $C^\infty(\Omega)$, $k=0,1,2,\dots$, we define the metric

$$ d(f,g) = \sum_{k=0}^\infty\frac{1}{2^k}\frac{\|f-g\|_{C^k(K_k)}}{1+\|f-g\|_{C^k(K_k)}} $$

This makes $C^\infty(\Omega)$ a Frechet space.

Spaces of measurable functions

Most important norms

In this case we consider only open sets $\Omega$. The duality compact/open is played here, in some weaker sense, by bounded/unbounded open set.

We consider the following norms on $C^\infty(\Omega)$:

$\displaystyle\|f\|_{L^p} = \displaystyle\left[\int\limits_\Omega|f(x)|^pdx\right]^{1/p}$, $p\in[1,\infty)$,
where $dx$ is the Lebesgue measure.
$\displaystyle\|f\|_{W^{p,1}} = \|f\|_{L^p} + \sum_{1\leq i\leq n}\|\partial_{x^{i}}f\|_{L^p}$
$\displaystyle\|f\|_{W^{p,k}} = \|f\|_{W^{p,k-1}} + \sum_{1\leq i_1\leq\dots\leq i_k\leq n}\|\partial^k_{x^{i_1}\dots x^{i_k}}f\|_{L^p}$, $k=1,2,\dots$

Restrictions of $C^\infty(\Omega)$

Unlike in the differentiable case, these norms are not well-defined on the whole $C^\infty(\Omega)$ since continuous functions on an open set are not necessarily bounded.

In this case, therefore, we must restrict the norm to the subsets
$C^{\infty;p}(\Omega)=\{f\in C^{\infty}(\Omega)\,|\,\|f\|_{L^p} < \infty\}$

and

$C^{\infty;p,k}(\Omega)=\{f\in C^{\infty}(\Omega)\,|\,\|f\|_{W^{p,k}} < \infty\}$.

Like in the differentiable case, these spaces are not complete in the corresponding norm. E.g. for $\Omega=(-1,1)$ one can easily write explicitly (try!) a sequence of $C^\infty$ functions converging in the $L^p$ norm to the function equal to 0 for $x<0$ and $1$ for $x>0$, that is not even continuous.

Completions of integrable $C^\infty(\Omega)$ functions

The completion of those spaces (we will see this when we will talk about mollifiers) gives the whole measurable space:

$\displaystyle\overline{C^{\infty;p}(\Omega)}^{\|\cdot\|_{L^p}}= L^p(\Omega)$, $p\in[1,\infty)$
(that's the space of all measurable functions with finite $L^p$ norm)
$\displaystyle\overline{C^{\infty;p,k}(\Omega)}^{\|\cdot\|_{W^{p,k}}}= W^{p,k}(\Omega)$, for all $p\in[1,\infty)$ and $k=1,2,\dots$
(that's the space of all $L^p$ functions with $L^p$ weak derivative up to order $k$)

All spaces $L^p(\Omega)$ and $W^{p,k}(\Omega)$ are Banach spaces.

The bounded case

Functions in $C^\infty(\Omega)$ can be very wild close to $\partial\Omega$. In particular, they might not extend to $C^\infty(\Bbb R^n)$, which is often inconvenient.
The good news is that, when $\Omega$ is bounded and $\partial\Omega$ is $C^1$ (or even just Lipschitz), every function in $L^p(\Omega)$ and $W^{p,k}(\Omega)$ can be approximated by a function that is extendable, namely:

$\displaystyle{\overline{C^\infty(\overline{\Omega})}}^{\|\cdot\|_{L^p}}= L^p(\Omega)$, for all $p\in[1,\infty)$
$\displaystyle{\overline{C^\infty(\overline{\Omega})}}^{\|\cdot\|_{W^{p,k}}}= W^{p,k}(\Omega)$,for all $p\in[1,\infty)$ and $k=1,2,\dots$

For a proof that such functions can be extended to the whole $\Bbb R^n$, see this discussion on mathoverflow and this classic paper by Withney.
Moreover, every such smooth function can be extended to a global smooth function with compact support, namely functions in $L^p(\Omega)$ and $W^{p,k}(\Omega)$ can be approximated by a sequence in $C_c^\infty(\Bbb R^n)$.

Closure of $C_c^\infty(\Omega)$

In case of $C^k$ norms, it can be provesd that $\overline{C_c^\infty(\Omega)}^{\|\cdot\|_{C^k}}=C_0^k(\Omega)$, the space of all $C^k$ functions going to zero at $\partial\Omega$ (or at infinity if $\Omega=\Bbb R^n$).

In case of integrable functions it makes no sense talking about limits and the corresponding definition is given in terms of approximations by $C^\infty_c$ functions:

$\displaystyle L^p_0(\Omega)={\overline{C^\infty_c(\Omega)}}^{\|\cdot\|_{L^p}}$, for all $p\in[1,\infty)$
$\displaystyle W^{p,k}_0(\Omega)={\overline{C^\infty_c(\Omega)}}^{\|\cdot\|_{W^{p,k}}}$, for all $p\in[1,\infty)$ and $k=1,2,\dots$

All spaces $L^p_0(\Omega)$ and $W^{p,k}_0(\Omega)$ are Banach spaces. Finally, note that, due to the previous slide, ${\overline{C^\infty_c(\Bbb R^n)}}^{\|\cdot\|_{W^{p,k}}}=W^{p,k}(\Bbb R^n)$.

Spaces of locally integrable functions

When $\Omega$ has infinite volume, we can also define local versions of all these spaces. For instance, $L^p_{loc}(\Omega)$ is the space of all measurable functions whose $L^p$ norm on every compact $K$ inside $\Omega$ is finite.

Similarly to the case of the $C^k(\Omega)$ with $\Omega$ open, the $L^p_{loc}(\Omega)$ are not normed spaces but only metric ones, with the metric $$ d(f,g) = \sum_{i=0}^\infty\frac{1}{2^i}\frac{\|f-g\|_{L^p(\Omega_i)}}{1+\|f-g\|_{L^p(\Omega_i)}}\;, $$ where $\Omega_i$ is an increasing sequence of compact sets such that $\cup_i \Omega_i=\Omega$.

All spaces $L^p_{loc}(\Omega)$ and $W^{p,k}_{loc}(\Omega)$ are Frechet spaces. Note also that, when the volume of $\Omega$ is finite, these local versions coincide with their global counterpart, so they are Banach spaces.

The operators $\frac{\partial\phantom{x}}{\partial x^\alpha}$

Continuity of $\frac{\partial\phantom{x}}{\partial x^\alpha}$

The operators

$\frac{\partial\phantom{x}}{\partial x^\alpha}:C^k\to C^{k-1}$, $k=1,2,\dots$;
$\frac{\partial\phantom{x}}{\partial x^\alpha}:C^\infty\to C^\infty$;
$\frac{\partial\phantom{x}}{\partial x^\alpha}:W^{p,k}\to W^{p,k-1}$, $p\in[1,\infty)$, $k=1,2,\dots$;

are all continuous together with their composition up to order $k$.

This is a trivial corollary of the structure of the norm function in those spaces.

Discontinuity of $\frac{\partial\phantom{x}}{\partial x^\alpha}$

The derivative operators on $C^k$ and $W^{p,k}$ are discontinuous in all other cases, exactly because the corresponding norms do not contain higher derivatives in its expression.

For instance, consider $\frac{d\phantom{x}}{dx}$ defined on $(C^0([0,1]),\|\cdot\|_{C^0})$. Of course the derivative operator is defined only over its dense subset $(C^1([0,1]),\|\cdot\|_{C^0})$ but it is not continuous because we can have sequences of functions $\{f_i\}\subset C^1([0,1])$ converging to a function $f\in C^0([0,1])$ that is not differentiable at some point, so that the sequence $\{f'_i\}$ does not converge in $(C^0([0,1]),\|\cdot\|_{C^0})$. On the other side, if $d/dx$ were a continuous map then we would have that $\lim d/dx(f_i) = d/dx(\lim f_i)=d/dx f$.

The very same phenomenon happens to the derivative operators on the spaces $L^p$.