Consider $C_0^\infty(\mathbb R)$ as a real vector space. By choice, we can take a Hamel basis $\{e_\alpha\}_\alpha$ for $C_0^\infty(\mathbb R)$ such that every $f\in \mathbb C_0^\infty(\mathbb R)$ can be written as

$$f=\displaystyle\sum_\alpha c_\alpha e_\alpha, \quad c_\alpha \in \mathbb C.$$

Let $\{U_j\}_{j\geq 1}$ be a neighborhood basis to $0\in C_0^\infty(\mathbb R)$ in topology induced by open balls with respect to the local metric

$$d(\varphi_1, \varphi_2) := \displaystyle\sum_k 2^{-k} \frac{\|\varphi_1 - \varphi_2\|_k}{1+\|\varphi_1-\varphi_2\|_k}$$

where $$\|\varphi\|_k := \displaystyle \sum_{\alpha\leq k} \sup_{x\in K} |\partial^\alpha \varphi|, \quad \|\varphi\| := d(\varphi, 0).$$

The locality of the metric should be interpreted as defining the metric for test functions with support contained in a compact set $K \subseteq \mathbb R$.

Take now $ e_1\in \{e_α\}_α$ and $t\in \mathbb R\setminus\{0\}$ such that $t e_1 \in U_1$ and set

$$ g_1:=te_1.$$

Continue to choose $g_j\in U_j$ in the same manner. Then, by construction, all $g_j$ are linear independent and $\{ g_j\}_j$ a part of a Hamel basis. Now define the map $\lambda : C_0^\infty(\mathbb R)→\mathbb C$ with $$\lambda(g_j):= j$$ and extend $\lambda$ by linearity to be a linear map. Then $\lambda$ cannot be a distribution, since we have

$$g_j \to 0$$

in the topology on $C_0^\infty(\mathbb R)$ as defined above, but

$$\lambda(g_j) \not\to 0$$ as $\lambda$ will be an unbounded operator on any neighbourhood $U$ to $0\in C_0^\infty(\mathbb R)$.

So we have that

AoC implies the existence of a linear functional on $C_0^\infty(\mathbb R)$ which is not a distribution.

But does it require choice?

  • 3
    $\begingroup$ This definitely doesn't imply choice. Your proof uses axiom of choice for a very particular set (or two, since the extension of $\lambda$ uses choice as well). There is no way this implies full choice. A better question is whether existence of such a linear functional can be proven without the use of AC. $\endgroup$ – Wojowu Mar 8 at 21:17
  • 1
    $\begingroup$ Nothing, ever, which depends on a single, specific set is equivalent to choice. $\endgroup$ – Asaf Karagila Mar 8 at 21:55
  • 1
    $\begingroup$ But let me ask this first, is your question really asking about linear functionals which are discontinuous? Because if so, this was asked before. Yes, choice is necessary. $\endgroup$ – Asaf Karagila Mar 8 at 22:16
  • 1
    $\begingroup$ Well, I mean, there is a topology on that space, right? Which makes it some "reasonable TVS"... no? $\endgroup$ – Asaf Karagila Mar 8 at 22:33
  • 1
    $\begingroup$ Results by Wright in "All operators on a Hilbert space are bounded" provide a hint towards a yes answer: he proves that it is consistent with ZF+DC (indeed, follows from ZF+DC+"all sets of reals have the Baire property") that every linear functional on a Fréchet space is continuous. The space in question, however, is not Fréchet (it's not even metrizable, see here). However, this gives us, for instance, that every linear functional on the Schwartz space is a tempered distribution. $\endgroup$ – Wojowu Mar 8 at 23:30

Your Answer

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

Browse other questions tagged or ask your own question.