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

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.

Shankar Principles of Quantum Mechanics p.59 Shankar Principles of Quantum Mechanics p.60

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 $\operatorname{Hom}(C^{\infty}_c((a, \ b), \mathbb{R})$?

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    $\begingroup$ 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. $\endgroup$ – GEdgar Jan 13 at 18:03
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This is probably not a complete answer to your question (since I don't understand many parts of the physics textbook that you quoted). Perhaps you'll need to wait for someone more knowledgeable to give a better viewpoint.

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 $\text{span } S$.

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