Negative ground state energy in two dimensions

Question

Let $\epsilon > 0$ and $V\in L^{1+\epsilon}(\mathbb R^2) \cap L^\infty(\mathbb R^2)$ real-valued, $V \leq 0, V \neq 0$ and let $$\mathcal E(\psi) := \int_{\mathbb R^2} \lvert \nabla \psi(x) \rvert^2 \, dx + \int_{\mathbb R^2} V(x) \lvert \psi(x) \rvert^2 \, dx, \quad \psi \in H^1(\mathbb R^2) = W^{1,2}(\mathbb R^2).$$ Consider the ground state energy $$E_0 := \inf\{\mathcal E(\psi) : \psi \in H^1(\mathbb R^2), \lVert \psi \rVert _2 = 1\}.$$ which can be shown to be bounded from below. I want to show that $E_0 < 0$. However, I did not know how to approach this problem. The usual approach taking $f \in C_c^\infty(\mathbb R^2)$ and letting $$f_n = nf(nx)$$ does not seem to work since both terms scale the same way under this dilation. Can anyone help me come up with a better trial state?

Answer 1

I got a solution now - in case someone is interested in the future, I'll leave it here. We need a differently scaling trial state, so the logarithm comes in handy in dimension $d=2$. Let $f\in C^\infty(\mathbb R, [0,1]), f(x) = 1$ if $x < 1$ and $f(x) = 0$ if $x> 2$. Let $R>0$ and define $\chi_R \colon \mathbb R^2 \to [0,1]$, $\chi_R(x) := f(\frac{1}{R}\log \lvert x \rvert )$. Then $\chi_R \in C_c^\infty(\mathbb R^2), \chi_R(x) = 1$ if $\lvert x \rvert < e^R$ and $\chi_R(x) = 0$ if $\lvert x \rvert > e^{2R}$. We then have $$\nabla \chi_R(x) = f'(\frac{1}{R} \log \lvert x \rvert) \frac{1}{R} \frac{1}{\lvert x \rvert} \frac{x}{\lvert x \rvert},$$ so $$\int_{\mathbb R^2} \lvert \nabla \chi_R(x) \rvert^2 \, dx = \frac{1}{R^2}\int_{\mathbb e^R < \lvert x \rvert < e^{2R}} \left \lvert f'(\frac{1}{R} \log \lvert x \vert) \right \rvert^2 \frac{1}{\lvert x \rvert^2} \, dx \leq \frac{\lVert f' \rVert_\infty \, \omega_d}{R^2} \int_{e^R}^{e^{2R}} \frac{1}{r} \, dr = \frac{\lVert f' \rVert_\infty \, \omega_d}{R} \stackrel{R\to \infty}{\longrightarrow} 0.$$ On the other hand, $$\int_{\mathbb R^2} V(x) \lvert \chi_R(x) \rvert^2 \, dx \leq \int_{\mathbb R^2} V(x) \mathbb 1_{ \{ \lvert x \rvert \leq e^{2R} \} }(x) \, dx \stackrel{R\to \infty}{\longrightarrow} \int_{\mathbb R^2} V(x) \, dx < 0.$$ Hence for sufficiently large $R$, $\mathcal E(\chi_R) < 0.$ Choosing $\tilde{\chi}_R := \frac{\chi_R}{\lVert \chi_R \rVert_2}$, we still have $\mathcal E(\tilde{\chi}_R) < 0, \lVert \tilde{\chi}_R \rVert_2 = 1$ and hence $$E_0 \leq \mathcal E(\tilde{\chi}_R) < 0,$$ as desired.