A (Poincare?) inequality on $L^1$ space In my lecture notes, I was tasked to prove the following result:

Let $\Omega$ be the horizontal strip $\mathbb R\times (-b,b)$ for some $b>0$ and let $u$ be a $C^1(\Omega)$ function with boundary value $u(x,\pm b)=0$. Show that
$$\int_{\Omega}|u|dx dy\leq 2b\int_{\Omega} |\nabla u|dx dy.$$

So far, the best result I achieved (though it could be wrong) is
$$\int_{\Omega}|u|dx dy\leq\int_{\Omega} |\nabla u|dx dy,$$
so when $b<\dfrac12$, I don't have the desired inequality.
My current approach is summarised as below:

*

*prove $$|u(x,y)|\leq \int_{-\infty}^\infty |\nabla u(x,y)| dx,\ \ \ |u(x,y)|\leq \int_{-b}^b |\nabla u(x,y)| dy$$
for all $x,y\in\Omega$ by Fundamental theorem of calculus;

*prove $$|u|^2\leq \int_{-\infty}^\infty |\nabla u(x,y)| dx\cdot\int_{-b}^b |\nabla u(x,y)| dy\implies\int_\Omega|u|^2dx dy\leq \left(\int_{\Omega} |\nabla u|dx dy\right)^2,$$as $\int_{-\infty}^\infty |\nabla u(x,y)| dx$ is a constant with respect to $x$ and $\int_{-b}^b |\nabla u(x,y)| dy$ is a constant with respect to $y$;

*by Cauchy-Schwarz's inequality, prove
$$\left(\int_{\Omega}|u|dxdy\right)^2\leq \int_\Omega|u|^2dx dy\leq \left(\int_{\Omega} |\nabla u|dx dy\right)^2,$$
which is the result I got.

Yet, throughout my proof, nothing related to $b$ appears, so I guess there is something wrong in my proof which I failed to figure out.
 A: You cannot use the fundamental theorem that easily on the $x$ variable as the function is not $0$ for $x\to \pm \infty$. By the fundamental theorem in $y$,
$$
|u(x,y')| \leq \int_{-b}^{y'} |\partial_yu(x,y)|\,\mathrm d y \leq \int_{-b}^{b} |\partial_yu(x,y)|\,\mathrm d y.
$$
Since the right-hand side is independent of $y'$,
$$
\int_{-b}^b |u(x,y')|\,\mathrm d y' \leq \int_{-b}^b \mathrm d y'\int_{-b}^{b} |\partial_yu(x,y)|\,\mathrm d y = 2\,b\int_{-b}^{b} |\partial_yu(x,y)|\,\mathrm d y.
$$
and so integrating in $x$ gives the result.

Remarks. Only the fact that $u(x,-b)=0$ was used. In the case when additionally $u(x,b)=0$, one can get a better estimate. If $y'<0$, one uses the same estimate to get
$$
|u(x,y')| \leq \int_{-b}^{0} |\partial_yu(x,y)|\,\mathrm d y.
$$
while when $y>0$, similarly,
$$
|u(x,y')| \leq \int_{0}^{b} |\partial_yu(x,y)|\,\mathrm d y.
$$
Hence,
$$
\int_{-b}^b |u(x,y')|\,\mathrm d y' = \int_{-b}^0 |u(x,y')|\,\mathrm d y' + \int_0^b u(x,y')\,\mathrm d y'
\\
\leq b \int_{-b}^{0} |\partial_yu(x,y)|\,\mathrm d y + b \int_{0}^{b} |\partial_yu(x,y)|\,\mathrm d y
$$
giving after integration in $x$,
$$
\iint_{\Omega}|u(x,y)|\,\mathrm d y \leq b \iint_{\Omega}|\partial_yu(x,y)|\,\mathrm d y \leq b \iint_{\Omega}|\nabla u(x,y)|\,\mathrm d y.
$$
