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It is well known that continuous linear operators are bounded and vice versa. It is also well known that the position operator (which I shall call $X$) causes many headaches in quantum mechanics due to its unboundedness, which results in the construction of rigged Hilbert spaces and so on.

I know that $$f(x) = \begin{cases} \frac{1}{x}, & x \geq 1\\ 0, & x < 1\end{cases}$$ can be used as an example for a function $ f(x) \in L^2(\mathbb{R})$ that demonstrates the unboundedness of $X$. However, I would be interested in an example that explicitly shows the/a discontinuity of $X$.

By linearity, it should even be possible (for example) to find a sequence that converges to the everywhere zero function where the $X$ operator is discontinuous, however, after playing around with a few Gaussian function sequences and trying to look up some canonical example, I couldn't come up with such a case. What would an example be?

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If $X$ is unbounded, then there exists some sequence $(f_n)$ such that $\|X(f_n)\| \to \infty$ but $\|f_n\| \le 1$. Let $g_n = f_n / \|X(f_n)\|$. Then $$0 \le \|g_n\| = \frac{\|f_n\|}{\|X(f_n)\|} \le \frac{1}{\|X(f_n)\|} \to 0.$$ Thus, $g_n \to 0$, but $$\|X(g_n)\| = \frac{\|X(f_n)\|}{\|X(f_n)\|} = 1,$$ and hence $X(g_n)$ does not converge to $0$. This provides your counterexample.

In this situation, $X$ is not just discontinuous at $0$, but discontinuous everywhere. Simply shift this sequence by any $f_0$, and you'll get a sequence $g_n \to f_0$, but $X(g_n) \not\to X(f_0)$.

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  • $\begingroup$ Very nice. A concrete example following your schema could be $g_n = \sqrt{\frac{2}{n}} x^{-\left(\frac{3}{2} + \frac{1}{n}\right)}$. $\endgroup$ Jun 8, 2020 at 18:19
  • $\begingroup$ ... for $x \geq 1$, $0$ else; as above. I forgot that part. $\endgroup$ Jun 8, 2020 at 18:32

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