Let $\;f(x):\mathbb R \to \mathbb R^n\;$ be a smooth function such that $\;\vert f(x)-l_{\pm} \vert \le e^{-k\vert x \vert}\;$ as $\;\vert x \vert \to \infty\;$ ($\;k\;$ is a positive constant and $\;l_{-} \neq l_{+}\;$).
The above exponential estimate simply says that: $\;\lim_{x\to -\infty} f(x) =l_{-}\;,\;\lim_{x \to +\infty} f(x)=l_{+}\;$
I want to show $\; \int_{0}^{+\infty} {\vert f(x)-l_{-} \vert}^2\;dx\;=\infty\;$
I observed that $\;{\vert f(x) -l_{-} \vert}^2 \to {\vert l_{+} -l_{-} \vert}^2\;$ as $\;x \to +\infty\;$ and it is well known that $\;\int_{0}^{+\infty} {\vert l_{+} -l_{-} \vert}^2\;dx=\infty\;$. I wonder if I could use this fact somehow in order to prove that also $\; \int_{0}^{+\infty} {\vert f(x)-l_{-} \vert}^2\;dx\;=\infty\;$
Any help would be valuable!
Thanks in advance.
If $l_+ \ne l_-$, then $c:=|l_+- l_-|^2>0$. From $\;{\vert f(x) -l_{-} \vert}^2 \to {\vert l_{+} -l_{-} \vert}^2\;$ as $x \to \infty$ we get a number $x_0>0$ such that
$\;{\vert f(x) -l_{-} \vert}^2 \ge c/2;$ for $x \ge x_0$.
Hence $\; \int_{0}^{+\infty} {\vert f(x)-l_{-} \vert}^2\;dx \ge \; \int_{x_0}^{+\infty} {\vert f(x)-l_{-} \vert}^2\;dx \ge \int_{x_0}^{+\infty} \frac{c}{2}dx= \infty$.