Give an example of a sequence of function in $C^1(\mathbb{R})$ to show that the Poincaré inequality does not hold on a general non-compact domain.
I tried to find the sequence in $C^1(\mathbb{R})$ such that it is not belong to $L^2$ on unbounded domain in but its dirivative is in $L^2$ in that unbounded domain but this sequence has to vanish on the boundary.
Consider
$f(x)= \begin{cases} x^{-1/3} &\text{if}\,\, x\neq 0\\ 0 & \text{if}\,\, x = 0 \end{cases}$ and $f'(x)= \begin{cases} -\dfrac{1}{3}x^{-4/3} &\text{if}\,\, x\neq 0\\ 0 & \text{if}\,\, x = 0 \end{cases}$ on the domain $(1, \infty).$
$$\int_1^{\infty}x^{-2/3}dx=+\infty$$
but
$$\int_1^{\infty} -\dfrac{1}{3}[x^{-4/3}]^2dx=\dfrac{5}{9}.$$
It is easy to find an example that it is not integrable on unbounded domain but its dirivative is integrable in that unbounded domain. But the problem is in my example $f$ is not continuously differentiable at $0$ at not vanish on boundary.
So I have trouble to find the function belongs to $C^1(\mathbb{R}),$ counte-examples that I found only countinuously differentiable on limit domains not for $\mathbb{R}$ and the function should be vanish on the boundary.
I also saw a lot of the examples with $u(x) \in H^1_0(\Omega)$ with $u$ is even not continuous.
Could you provide me any idea?
The example I gave in my last comment can be simplified a lot, you can just take $$ \Large f(x) = \frac{1}{(1+x^2)^{\frac14}}. $$ It is not square integrable because we have $$ \Large \frac{1}{(1+x^2)^{\frac12}} \geq \frac{1}{(2x^2)^{\frac12}} = \frac{1}{\sqrt{2}}\frac{1}{x} $$ for $x \geq 1$ and its derivative $$ \Large f'(x) = -\frac{x}{2(1+x^2)^{\frac54}} $$ is square integrable because $$ \Large \frac{x^2}{4(1+x^2)^{\frac52}} \leq \frac{1+x^2}{(1+x^2)^{\frac52}} \leq \frac{1}{1+x^2} $$ for all $x \in \mathbb{R}$.