2
$\begingroup$

By a distribution, I mean it is a linear functional of the space of smooth compactly supported functions over $\mathbb R^n$, i.e. $C_c^{\infty}(\mathbb R^n).$

I am reading a textbook by Strichartz, named A Guild to Distribution Theory and Fourier Transforms.

He wrote that linear functionals all tend to be continuous.

As we know, there are plenty of linear functionals which are not continuous. So what he referred to may be distributions. i.e. Distributions all tend to be continuous.

He gave a rough explanation which I think is not a proof.

His explanation:

Let $\varphi$ and $\varphi_1$ be in $C_c^{\infty}(\mathbb R^n),$ and $f$ be a distribution.

And let $\varphi_2:=\varphi_1-\varphi$

Then $\varphi_1=\varphi+\varphi_2$

Move $\varphi_1$ closer to $\varphi$ by considering $\varphi+t\,\varphi_2$ and let $t$ get small.

Then $\langle f,\varphi+t\,\varphi_2\rangle=\langle f,\varphi\rangle+t\,\langle f,\varphi_2\rangle$ by linearity, and as $t$ gets small this gets close to $\langle f,\varphi\rangle.$

End of explanation.

I think, to prove the continuity of $f$, we need to show $\langle f,\varphi_1\rangle\to\langle f,\varphi\rangle$ when $\varphi_1 \to \varphi.$

While in the explanation, what he proved is that $\langle f,\varphi+t\,\varphi_2\rangle\to\langle f,\varphi\rangle$ when $t\to 0.$

He also wrote that this does not constitute a proof of continuity, since the definition requires more "uniformly", but it should indicate that a certain amount of continuity is built into linearity.

And, all distribution you will ever encounter will be continuous.

So my problem is, are distributions all continuous?

Thanks in advance.

$\endgroup$
  • $\begingroup$ By the last italic phrase, it seems that continuity is not part of the definition of distribution.. $\endgroup$ – Berci Nov 3 '18 at 8:21
  • $\begingroup$ @Berci Sure, we never require that a distribution is continuous in the definition. My problem is that, can we prove that all distributions are continuous? $\endgroup$ – Sam Wong Nov 3 '18 at 8:25
  • $\begingroup$ No. It's a weaker property in the explanation, we could call it 'linear continuity' which is satisfied by linearity. $\endgroup$ – Berci Nov 3 '18 at 8:29
  • 1
    $\begingroup$ "A certain amount of continuity" - namely continuity along lines (or in any restriction to finite-dimensional subspaces) $\endgroup$ – Hagen von Eitzen Nov 3 '18 at 8:29
  • $\begingroup$ Part of the definition of a distribution is: $$\phi(f_n)\to\phi(f) \qquad \text{if }\|f_n-f\|\to0\text{ and }\|\partial_\alpha f_n- \partial_\alpha f\|\to0\text{ for all multi-indices $\alpha$}$$ It follows $\phi$ is continuous wrt the topology generated by the semi-norms $\|\partial_\alpha f\|=\sup_{x\in\Bbb R^d}|\partial_\alpha f|$. This is the usual topology for test-functions. $\endgroup$ – s.harp Nov 3 '18 at 9:31
2
$\begingroup$

This claim would be equivalent to saying that all linear functionals on the topological vector space $C_c^{\infty}(\mathbb{R}^n)$ are continuous.

This is not true, indeed, on any infinite dimensional topological vector space $X$ over a field $K$ there is a linear functional which is discontinuous.

Indeed, let $X$ be an infinite dimensional topological vector space, and let $\left\{e_n\right\}$ be a vanishing $(e_n\to 0)$ infinite sequence of linearly independent vectors on $X$, which may be easily constructed using continuity of the scalar product map. Define a linear functional $f$ on $X$ as any linear extension of the map $$f(e_n):=n $$ Then the sequence $\left\{e_n\right\}\subset X$ is such that $e_n\to 0$, but $f(e_n)\not\to 0$.

For instance, in $C_c^{\infty}(\mathbb{R})$ (but it is the same in higher dimensions), recall that we say that a sequence of test functions $\left\{\varphi_n\right\}$ converges to $0$ iff their supports $\operatorname{supp}\varphi_n$ are all contained on some compact set $K\subset \mathbb{R}$ and $\varphi_n^{(k)}\to 0$ uniformly on $K$ for each order of differentiation $k\in \mathbb{N}$.

EDIT: The above is the usual topology that is set in the space of test functions $C_c^{\infty}(\mathbb{R}^n)$. In order to discuss whether a functional $f:C_c^{\infty}(\mathbb{R}^n)\to \mathbb{R}$ is continuous we first need to fix a topology on $C_c^{\infty}(\mathbb{R}^n)$, so I assume you are referring to this one.

We may start from a function $0\neq \varphi \in C_c^{\infty}(\mathbb{R})$ and then set $$e_n(x):=\frac{1}{n}\varphi\left(x+\frac{1}{n}\right) $$ so that the $\left\{e_n\right\}$ are linearly independent (since their supports are disjoint), $\operatorname{supp}e_n\subset \operatorname{supp}\varphi+[-1,0)$ and $e_n^{(k)}=\frac{1}{n}\varphi^{(k)}\left(x+\frac{1}{n}\right)\to 0$ uniformly for all $k$ as $n\to +\infty$. Thus, $\left\{e_n\right\}$ is a vanishing sequence of linearly independent vectors, and the functional $f$ defined above in the general case is a discontinuous linear functional on $C_c^{\infty}(\mathbb{R})$.

On the other hand, this is basically a sequence of bump functions which quickly reduces in overall size as $n\to +\infty$ and slowly shifts towards the left. Can you imagine a reasonable linear functional which blows up on such a sequence of functions? As $n\to +\infty$, nothing meaningful 'blows up' here. I think this goes to show that you can usually expect your distributions to be continuous.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.