# Example of discontinuity of position operator $X\colon f(x) \mapsto xf(x)$ for $f(x) \in L^2(\mathbb{R})$

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?

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 $$X(g_n) \not\to X(f_0)$$.
• Very nice. A concrete example following your schema could be $g_n = \sqrt{\frac{2}{n}} x^{-\left(\frac{3}{2} + \frac{1}{n}\right)}$. Jun 8, 2020 at 18:19
• ... for $x \geq 1$, $0$ else; as above. I forgot that part. Jun 8, 2020 at 18:32