# Topologizing The Space Of Distributions

## Why do we need to require convergence of the derivatives when defining a topology on the set of distributions?

Steve Trettel

| Analysis

Let $C_c^\infty$ be the set of smooth, compactly supported, real-valued functions on a manifold $M$, and $(C_c^\infty)^\ast$ be its (algebraic) dual space, the set of all linear functionals $F\colon C_c^\infty\to\mathbb{R}$.

It is inside of this dual space that we carve out a subset called *distributions*, to serve as a suitable
generalization of functions for many purposes in analysis. Why don’t we just take this whole space?

The dual only knows $C_c^\infty$ *as a vector space*,
and so it has lost all knowledge of our underlying manifold and the behavior of its smooth functions!

For an example, lets take $M=\mathbb{S}^1$ realized as $\mathbb{R}/2\pi\mathbb{Z}$:
The set of functions ${\sin kx, \cos kx\mid k\in\mathbb{Z}_{\geq 0}}$ is linearly independent in $C_c^\infty$,
and so we may extend this to a basis $\mathcal{B}$ (note here I really mean an honest vector space basis - $\mathcal{B}$ is uncountable!
This is sometimes called a *Hamel basis*).
Since $\mathcal{B}$ is a basis, we can define a linear functional by declaring its values on $\mathcal{B}$. Let $F\colon C_c^\infty\to\mathbb{R}$
be the functional which sends $\sin kx$ and $\cos kx$ to $1$ and sends all other elements of $\mathcal{B}$ to $0$.
This functional has forgotten important information about the circle and its functions: for instance, if $\phi\in\mathcal{B}$
is any non-sinusoidal element of the basis, we may approximate $\phi$ arbitrary well by some partial sum of its fourier series
$$\phi\sim \sum a_k\cos kx+b_k\sin kx$$
But our linear functional $F$ sends $\phi$ to $0$, while sending this arbitrarily-good approximation of $\phi$ to the sum of its fourier
coefficients. Thus, while $\phi$ and its partial fourier series look very different “*through the eyes of $F$*”,
even though they are almost indistinguishable from the viewpoint of analysis.

The easiest way to force our definition to recall the nature of the functions in $C_c^\infty$
is to build a natural topology on $C_c^\infty$, and then restrict ourselves to linear functionals
which are *continuous*.
For example, we can topologize $C_c^\infty$ using the norm $|f|=\int_M|f|d\mathrm{vol}$,
and then see explicitly that the linear functional $F$ constructed above is not continuous:
the partial fourier sums $\phi_n$ converge to $\phi$, but $F(\phi_n)$ does not converge to $F(\phi)$.

In fact, a linear functional $F$ is continuous with respect to this topology on $C_c^\infty$ precisely if, for all convergent sequences $\phi_n\to\phi$ in $C_c^\infty$, we have $F(\phi_n)\to F(\phi)$. Doing a little more investigation, this rules out all functionals which are ‘badly behaved’ in ways similar to the above, while still useful things like delta ‘functions’.

Because_ it will prove useful later on, we review this definition here.
Given a point $p\in M$, the *delta distribution* $\delta_p$ is the linear functional such that for all $f\in C_c^\infty$,
$\delta_p(f)=f(p)$.
These are continuous with respect to the sup norm topology, as if $\phi_n\to \phi$, using smoothness (really, just continuity)
we see that these must also converge pointwise, so $\phi_n(p)\to\phi(p)$, and hence $\delta_p(\phi_n)\to\delta_p(\phi)$ for ny $p\in M$.
If $M$ has a distinguished point denoted $0$, we write $\delta$ for $\delta_0$.

HOWEVER - we inadvertently went too far in our desire to cut down the size of $(C_c^\infty)^\ast$!
There is a kind of inverse relationship between *how many functionals are continuous* and *how fine of a topology is imposed* on the domain.
We chose a rather weak topology, meaning it is rather easy for functions to converge in $C_c^\infty$, and through this inverse rerlationship,
continuity becomes a pretty strong requirement. So strong in fact, that our set of continuous linear functionals is not closed under useful operations like differentiation!
(Recall that the derivative $F^\prime$ of a linear functional $F$ is defined by analogy with integration by parts, so
$F^\prime(\phi)=-F(\phi^\prime)$).

As an explicit example, we see that the derivatives of delta distributions are not continuous. For concreteness let’s once more restrict ourselves to the circle $M=[0,2\pi]/\sim$ (the following argument goes through similarly for $M=\mathbb{R}$, after multiplying each of the sinusoids by a smooth function of compact support), and consider the sequence of functions $\phi_k=\tfrac{1}{k}\sin(kx)$. These functions converge to the constant function $0$ with respect to the sup norm, but applying $\delta^\prime$ gives $$\delta^\prime\left(\tfrac{1}{k}\sin(kx)\right)=-\delta\left(\tfrac{1}{k}\sin(kx)\right)^\prime=-\delta\left(\cos kx\right)=-1$$ So, even though $\phi_k\to 0$, we have $\delta^\prime(\phi_k)\to -1\neq \delta^\prime(0)=0$.

If we wish to have our collection of distributions closed under differentiation, we somehow need to expand our current subset.
And, in light of the inverse relationship between continuity and topology, one way to do this is to *strengthen* the underlying topology,
and make it harder for sequences of functions to converge!
Looking at that calculation we can pinpoint exactly what went wrong: to compute $F^\prime(\phi_k)$ we actually compute
$F(\phi_k^\prime)$, and even though $\phi_k$ was convergent by hypothesis, we had no control over the behavior of $\phi_k^\prime$.

This suggests a new, stronger topology for $C_c^\infty$, we will say that $\phi_k\to\phi$ if and only if we have convergence not only of $\phi_k$ in the sup norm, but $\phi_k^\prime$ as well. With respect to this topology, we have fixed our immediate problem: our explicit example no longer poses a problem as the sequence $\tfrac{1}{k}\sin(kx)$ simply does not converge in $C_c^\infty$ anymore! This strengthening of the topology actually fixes all potential issues for $\delta^\prime$, which now is a continuous linear functional! Indeed, if $\phi_k\to\phi$ in this new topology, $$\delta^\prime(\phi_k)=-\delta(\phi_k^\prime)\to-\delta(\phi^\prime)=\delta^\prime(\phi).$$

Of course, we can’t stop here to celebrate for too long, as we have not fixed the underlying issue: our new space of continuous linear functionals is still not closed under differentiation. Indeed, managing to make $\delta^\prime$ continuous is kind of a 2-edged sword, on the one hand now $\delta$ has a derivative (yay!) but on the other hand, we now need $\delta^\prime$ to have a derivative as well (uh-oh).

From our experience above, its straightforward to show that $\delta^{\prime\prime}$ is not continuous: since by definition $$\delta^{\prime\prime}(\phi)=-\delta^\prime(\phi^\prime)=\delta(\phi^{\prime\prime}),$$ all we need to do is choose a sequence of functions $\phi_k$ which converge in our toplogy but $\phi_k^{\prime\prime}$ does not (for an explicit example, integrate our earlier sequence to get $\tfrac{1}{k^2}\cos(kx)$). So to widen our class of continuous functionals further, we need to again strengthen our topology, and prevent any pathological sequence like this from being continuous. Continuing our patchwwork fix, we may suggest requiring that $\phi_k,\phi_k^\prime,$ and $\phi_k^{\prime\prime}$ all converge in the sup norm for the sequence $\phi_k$ to count as convergent. And, of course, this will make $\delta^{\prime\prime}$ continuous, but simply push back the problem once more to its derivative.

Instead, we can attempt to be brave and ‘jump to the end of the line’ and think about what would happen if we applied this patchwork
process infinitely many times.
The resulting topology would require that if $\phi_k\to \phi$, all sequences of derivatives must converge, $\phi_k^{(n)}\to\phi^{(n)}$.
(Here, for 1-dimensional domains $(n)$ is just the number of derivatives, but in general is a multi-index: so for functions on a surface we
are requiring the convergence of things like $\partial_x^3\partial_y^7\phi_k\to \partial_x^3\partial_y^7\phi$).
This has certainly fixed all the problems we were aware of (for instance, $\delta^(n)$ is continuous for all $n$ now), but
this is real analysis, and its usually the problems that you *aren’t* yet aware of that mess everything up!

Happily, this time we have truly done it - there are no more unexpected problems out to get us, and the set of continuous linear functionals with respect to this new strengthened topology really is closed under differentiation! Let’s prove it:

Let $F$ be an arbitrary continuous linear functional, and $F^\prime$ be its derivative. We wish to show that $F^\prime$ is
continuous, by showing that for any sequence $\phi_k\to\phi$ we have $F^\prime(\phi_k)\to F^\prime(\phi)$.
Given such a sequence $\phi_k$, we compute for each $k$ the quantity $F^\prime(\phi_k)=-F(\phi_k^\prime)$ using the definition
of the derivative for linear functionals.
Crucially - our new topology implies that since $\phi_k$ converged to $\phi$, the sequence $\phi_k^\prime$ of functions
converges to $\phi^\prime$.
(Note we do not mean merely that $\phi_k^\prime\to\phi^\prime$ in the sup norm, but rather that $\phi_k^\prime$ *and all of its derivatives*
converge to $\phi^\prime$ *and all of its derivatives*, respectively).
Thus, using the assumed continuity of $F$ and the convergence of $\phi_k^\prime$, we see that $F(\phi_k^\prime)\to F(\phi^\prime)$.
Putting it all together,

$$F^\prime(\phi_k)=-F(\phi_k^\prime)\to-F(\phi^\prime)=F^\prime(\phi),$$

So $F^\prime$ is continuous as claimed.

This argument makes me think of the space of distributions as being defined by finding a ‘Goldilocks’ topology on $C_c^\infty$. At first we tried the entire dual space - but it was too big! Then we tried to cut it down to remember analytical properties of our function space, but the obvious way to cut it down made it too small! Then, after a sequence of patching the holes we created, we end up with a topology that’s ‘just right’: it remembers pointwise convergence but is also closed under differentiation.