Let $\;f:\mathbb R \to \mathbb R^m\;$ be a continuous function such that $\;\vert f(x)-a \vert \le e^{-kx}\;$ as $\;x \to +\infty\;$ and $\;\vert f(x)-b \vert \le e^{kx}\;$ as $\;x \to -\infty\;$(NOTE: $\;k \gt 0\;$ and $\;a\neq b\;$)
Take $\;y_n \in \mathbb R\;$ an unbounded sequence, in the sense that $\;y_n\to \infty\;$ as $\;n\to \infty\;$. If $\;f(x-y_n) \to b\;$ as $\;y_n \to \infty\;$, prove: $\;\int_{0}^{+\infty} {\vert f(x-y_n)-b \vert}^2 \;dx \to 0\;$ as $\;n \to \infty\;$
My Attempt:
$\;\int_{0}^{\infty} {\vert f(x-y_n) -b \vert}^2\;dx\le \int_{0}^{\infty} e^{2k(x-y_n)} \;dx\to 0\;$ as $\;y_n \to \infty\;$
However I'm a bit unsure if the above is valid since I used the estimate when $\;x-y_n \to -\infty\;$. I would really appreciate if somebody could confirm the above thought or in case it's wrong, fix it.
Thanks in advance!
For a fixed $n$, choose $K$ such that for $x \ge K$, $|f(x)-a| \le {1 \over 2} |a-b|$.
Then, if $x \ge K+y_n$, we have $|f(x-y_n)-a| \le {1 \over 2} |a-b|$.
Hence $|f(x-y_n) -b| \ge |a-b|-|f(x-y_n)-a| \ge {1 \over 2} |a-b| >0$.
In particular, the function $x \mapsto f(x-y_n) -b$ is not square integrable, and we have $\lim_n \int_0^\infty |f(x-y_n) -b| dx = \infty$.