Is the set $\{\delta_x\}_{x \in [a, \ b]}$ a basis for the set of distributions on $C^{\infty}_c([a, \ b])$?

$$\def\braket#1#2{\langle#1|#2\rangle}\def\bra#1{\langle#1}\def\ket#1{#1\rangle}$$

Is the set $$\{\delta_x\}_{x \in [a, b]}$$ a basis for the set of distributions on $$C^{\infty}_c([a, \ b])$$? Below are p.59 and p.60 from Shankar's book Principles of Quantum Mechanics.

According to Shankar, the set $$\{\delta_x\}_{x \in [a, \ b]}$$ is a basis for the space of functions on $$[a, \ b]$$ that vanishes on the set $$\{ a, b\}$$. Here, $$\delta_x(y) = \delta(x-y) = \braket{x}{y}$$ and $$\delta_x = |\ket{x}$$. However, Dirac deltas are not functions. Is the set $$\{\delta_x\}_{x \in (a, \ b)}$$ a basis for the set of distributions on $$C^{\infty}_c((a, \ b))$$?

$$\varphi:= \int_a^b | \ket{y} \bra{y} | dy$$

$$\varphi [f] = \int_a^b | \ket{y} \bra{y} | \ket{f} dy = \int_a^b | \ket{y} f(y) dy$$

$$\braket{x}{\varphi(f)} = \int_a^b \braket{x}{y} f(y) dy = \int_a^b \delta(x-y) f(y) dy = f(x) = \braket{x}{f}$$

Hence $$\varphi(f) = f$$, and $$\varphi = id$$.

$$\int_a^b | \ket{y} \bra{y} | dy = id \in \operatorname{Hom}(C_c^{\infty}((a, \ b), C_c^{\infty}((a, \ b))$$ (This is equation (1.10.11) in Shankar)

$$\varphi_g :=\int_a^b \ dy \ g(y) \bra{y}|$$

Then $$\varphi_g (|\ket{f}) = \int_a^b \ dy \ g(y) \braket{y}{f} = \int_a^b \ dy \ g(y) f(y) = \int_a^b g f \ dy = \braket{g}{f}$$

Hence $$g = \int_a^b \ dy \ g(y) \bra{y}|$$ in $$\operatorname{Hom}(C^{\infty}_c((a, \ b), \mathbb{R})$$?

My questions are:

1. Is the set $$\{\delta_x\}_{x \in (a, \ b)}$$ a basis for the set of distributions on $$C^{\infty}_c((a, \ b))$$?

2. If 1 is true, then in what sense (e.g. finite sum , countable sum ...) does $$\{\delta_x\}_{x \in (a, \ b)}$$ span the space of distributions on $$C^{\infty}_c((a, \ b))$$?

3. $$g = \int_a^b \ dy \ g(y) \bra{y}|$$ in $$\operatorname{Hom}(C^{\infty}_c((a, \ b), \mathbb{R})$$?

• Only a non-mathematician would use the term "basis" in the way Shankar does. In case of $(a,b)$, there are uncountably many $\left|{x}\right\rangle$ items, and $\varphi := \int_a^b$ whatever (an integral, not a sum) claims to write $\varphi$ as a combination of uncountably many of them. – GEdgar Jan 13 at 18:03

First of all, let me denote $$\mathcal D= C^\infty_c(a,b)$$ equipped with the topology of uniform convergence (of all its derivatives) on a compact set. Let $$\mathcal D'$$ be its dual, i.e. the set of distributions on $$\mathcal D$$. While $$S=\{\delta_x:x\in(a,b)\}$$ is not a "basis" of $$\mathcal D'$$ in the usual linear combination sense, we can approximate an element $$T\in \mathcal D'$$ using elements of $$S$$ in the following way:
It is known that $$\mathcal D$$ is sequenctially $$w^*$$-dense in $$\mathcal D'$$, so any $$T\in \mathcal D'$$ can be approximated arbitrarily well with a sequence $$\varphi_n\in \mathcal D$$. On the other hand, each $$\varphi_n$$ induces a (signed) Borel measure on $$(a,b)$$, and according to this, $$S \cup -S$$ is the set of extreme points of all such Borel measures with total variation norm 1. By Krein-Milman theorem, the scaled version of $$\varphi_n$$ (viewed as a measure) can be approximated by convex combination of $$S \cup -S$$. Hence any element of $$\mathcal D'$$ can be approximated (in $$w^*$$ sense) by $$\text{span } S$$.