If $l_i$ is the first weak eigenvalue of $-\Delta$ in a domain $G_i$ and $G_1\subseteq G_2$, then $l_1\ge l_2$ and equality is possible

Question

Let

• $\Omega_i\subseteq\mathbb{R}^n$ be a domain
• $\lambda_i$ be the first weak eigenvalue of $-\Delta$ in $\Omega_i$

It's easy to verify that $\Omega_1\subseteq\Omega_2$ implies $\lambda_1\ge \lambda_2$, by considering the corresponding Rayleigh quotients.

However, while I'm quite sure that we don't need to have strict inequality, i.e. $\lambda_1>\lambda_2$, I failed to find an elegant counter example.

Answer 1

It does has strict inequality: Let $f_1$ satisfies $-\Delta f_1 = \lambda f_1$ and $\|f_1\|_{L^2(\Omega_1)} = 1$ on $\Omega_1$. If $\Omega_1 \subset \Omega_2$, then the function

$$g(x) = \begin{cases} f_1(x) & \text{ if } x\in \Omega_1 \\ 0 & \text{ if not.}\end{cases}$$

is a $W^{1, 2}_0(\Omega)$ function, $\|g\|_{L^2(\Omega_2)} = \|f_1\|_{L^2(\Omega_1)}=1$ and

$$\lambda_1= \frac{\int_{\Omega_2} \|\nabla g\|^2}{\|g\|_{L^2(\Omega_2)}}=\int_{\Omega_2} \|\nabla g\|^2 = \int_{\Omega_1} \|\nabla f_1\|^2.$$

If $\lambda_2 = \lambda_1$, then the Rayleigh quotients is attained by $g$, which implies that $g$ is also the first eigenfunction. But it is impossible as the first eigenfunction must be strictly positive (Note $g$ is not strictly positive).