Convergence in $L^2(1,\infty)$ – counterexample

Question

Consider the (complex) Hilbert space $L^2(1,\infty)$ and a real function $g(x)=x^2$ defined on the interval $(1,\infty)$.

Does there exist a sequence of functions $f_n \in L^2(1,\infty)$ and a function $f \in L^2(1,\infty)$ satisfying all the following conditions?

• $f_n \cdot g \in L^2(1,\infty)$ for every $n \in \mathbb{N}$
• $f \cdot g \in L^2(1,\infty)$
• $f_n \rightarrow f$ in the norm of $L^2(1,\infty)$
• $f_n \cdot g \rightarrow f \cdot g$ in the norm of $L^2(1,\infty)$
• $\int_1^{\infty}f_n(t) \mathrm{d}t=0$ for every $n \in \mathbb{N}$
• $\int_1^{\infty}f(t) \mathrm{d}t \neq 0$

This question was inspired by my recent thread (Adjoint of an operator on $L^2$), where I was shown that this should be possible. However, I'd like to see a straightforward (and perferably constructive) argument.

I've been thinking about this for more than hour now with no success.

Thanks for any help.

Answer 1

This is not possible. Let $h(x) = x^{-2}$. Then $h \in L^2(1,\infty)$, and if $f_n\cdot g, f\cdot g \in L^2(1,\infty)$ with $\lVert f_n\cdot g - f\cdot g\rVert_{L^2} \to 0$, then the Cauchy-Schwarz inequality shows $f_n, f \in L^1(1,\infty)$ and

\begin{align} \lVert f_n - f\rVert_{L^1} &= \lVert (f_n\cdot g - f\cdot g)\cdot h\rVert_{L^1} \\ & \leqslant \lVert f_n\cdot g - f\cdot g\rVert_{L^2}\cdot \lVert h\rVert_{L^2}, \end{align}

so $\lVert f_n - f\rVert_{L^1} \to 0$.

Since integration is continuous on $L^1(1,\infty)$, we then have

$$\int_1^{\infty} f(x)\,dx = \lim_{n\to \infty} \int_1^{\infty} f_n(x)\,dx.$$