Question regarding closeability of the operator $\Lambda:L^2(\mathbb{R})\to \mathbb{R}$ given by $\Lambda f:= f(0)$?

Question

I have a question on the closeability of an operator. First is the definition I am using for the closeable property:

Theorem

Let $X$ and $Y$ be Banach spaces, let $\text{dom}(A)\subset X$ be a linear subspace, and let $A:\text{dom}(A)\to Y$ be a linear operator. Then the following are equivalent.

1. $A$ is closeable.
2. The projection $\pi_X:\overline{\text{graph}(A)} \to X$ onto the first factor is injective.
3. If $(x_n)_{n\in \mathbb{N}}$ is a sequence in $\text{dom}$ and $y\in Y$ is a vector such that $\lim_{n\to\infty} x_n = 0$, and $\lim_{n\to\infty} Ax_n = y$, then $y=0$.

Proposition

Let $H=L^2(\mathbb{R})$ and define $\Lambda:\text{dom}(\Lambda)\to \mathbb{R}$ by $$ \text{dom}(\Lambda):= C_c(\mathbb{R}), \qquad \Lambda f: = f(0), $$ for $f\in C_c(\mathbb{R})$ (the space of compactly supported continuous real valued functions $f:\mathbb{R}\to \mathbb{R}$). This linear functional is not closeable due to the above theorem, because there is a sequence of continuous functions $f_n$ with compact support such that $f_n(0)=1$ and $||f_n||_{L^2}\le\frac{1}{n}$ for all $n\in \mathbb{N}$.

Question

This statement seems to be using point 3. of the theorem to show that there is the operator $\Lambda$ is not closeable. But why does it state that $||f_n||_{L^2}\le\frac{1}{n}$ for all $n\in \mathbb{N}$ instead of just saying $\lim_{n\to \infty}f_n =0$ which is the actual statement in the theorem? Furthermore, what is the sequence $f_n$ such that $||f_n||_{L^2}\le\frac{1}{n}$, it seems very arbitrary in the proposition to just say 'there exists a sequence' without specifying it?

Answer 1

For such a sequence take a sequence $f_n$ of isosceles triagles where $f_n(0)=1$ meaning that the heights of these triangles always remain $1$.

So these triangles have base of lenth $\frac{2}{n^2}$ and $f_n$ are continuous functions and converge to zero function under the $L^2$ norm.

Every $f_n$ is $0$ outside of an interval $(-\frac{1}{n^2},\frac{1}{n^2})$ so their compact supports are $cl((-\frac{1}{n^2},\frac{1}{n^2}))=[-\frac{1}{n^2},\frac{1}{n^2}]$

Also note that $C_c(\mathbb{R})$ is dense in $L^2(\mathbb{R})$ and in $L^2$ we have a certain norm.

So when we say $\lim f_n=0$ in $L^2$ we mean $||f_n||_2=||f_n-0||_2=(\int |f_n-0|^2)^{\frac{1}{2}} \rightarrow 0$

Answer 2

Stating $\|f_n\|_{L^2}\le\frac1n$ is just being slightly more specific than stating "3 holds", while not being quite as specific as an explicit counterexample. In this case it's all kind of moot, because $L^2$ is an equivalence class of functions, so $\Lambda f$ is not really well-defined.