$f \mapsto f(0)$ is not continuous on $L^2$

Question

Show that the the linear map $f \mapsto f(0)$ is not a continuous on $L^2(\mathbb{R},m)$.

In order to show that the map $L^2(\mathbb{R}) \ni f \mapsto f(0) \in \mathbb{R}$ is not continuous, it suffices to show that there exists $f \in L^2(\mathbb{R})$ so that the mapping is not continuous at $f$. Let $f \in L^2(\mathbb{R})$ wlog assume that either $-\infty < f(0) < \infty$ or $f(x) = \infty$. Define

$$g_n(x) = \begin{cases} f(x)-\sqrt{n} \quad \, : \, x \in [-\frac{1}{2n^2}, \frac{1}{2n^2}] \\ f(x) \quad \, : \, \text{ else } \end{cases}.$$

Then,

\begin{eqnarray*} \| f-g_n \|_{L^2} = \int_\mathbb{R} |f-g_n|^2 \, dm &=& \int_{\mathbb{R}} \left|\sqrt{n} \chi_{[-\frac{1}{2n^2}, \frac{1}{2n^2}]} \right|^2 \, dm \\ &=& \frac{1}{n}. \end{eqnarray*} \ Now, let $\epsilon = 1$, then for any $\delta > 0$, there exists an $N \in \mathbb{N}$ such that, if $n \geq N$, $$\|f-g_n\|_{L^2} < \delta$$ but $$|f(0) - g_n(0)| >1.$$ Therefore the mapping is not continuous at $f \in L^2(\mathbb{R})$, as desired.

Any issues here? Dear Downvote, I know this problem is not well-defined, but it's homework.

Answer 1

A precise way to present this problem would be as follows:

The linear functional $f\mapsto f(0)$, defined on the linear subspace $L^2(\mathbb{R})\cap C(\mathbb{R})$, is not continuous in the $L^2$ norm.

(Hence, it cannot be extended to a continuous linear functional on $L^2$, no matter how we try to make sense of $f(0)$ there.)

The proof of the statement follows by considering the sequence $f_n(x)=\max(0, 1-n|x|)$, which satisfies $f_n(0)=1$ for all $n$ despite $f_n\to0$ in $L^2$ norm.