Let $\epsilon \in [0, \infty[$. Consider the following operator on $L^2(\mathbb{R})$: 
\begin{equation}
H(\epsilon) = -\frac{d^2}{dx^2} + x^2 + \epsilon |x|.
\end{equation} 
How does one show that the lowest eigenvalue of $H(\epsilon)$, denoted by $\lambda\_1(\epsilon)$ satisfies: 



> 
> \begin{equation}
> \epsilon \mapsto \lambda\_1(\epsilon) \text{ is increasing on } [0, \infty[
> \end{equation} 
> 
> 
> 


and furthermore that 



> 
> \begin{equation}
> \lim\_{\epsilon \rightarrow \infty} \lambda\_1(\epsilon) = +\infty?
> \end{equation}
> 
> 
> 


I had the following idea. Define the operator \begin{equation} \mathcal{U}: u \mapsto u \big(x+\frac{\epsilon}{2} \big)
\end{equation}
on $L^2(\mathbb{R})$. $\mathcal{U}$ is a unitary operator with inverse / adjoint $\mathcal{U}^{-1}: v \mapsto v(x-\frac{\epsilon}{2})$. By using $\mathcal{U}$ and observing that 
\begin{equation}
x^2 + \epsilon |x| = \big(|x| + \frac{\epsilon}{2} \big)^2 - \frac{\epsilon^2}{4},
\end{equation}
I tried showing that that $H(\epsilon)$ is unitarily-equivalent to another operator with known eigenvalues (which would ideally satisfy the conditions in the boxes). However I fail to obtain such an operator. All feedback is welcome.



Limit (at infinity) for the lowest eigenvalue of a perturbed harmonic oscillator
Consider two operators $L\_1w=-w''+U(x)w$ with eigenvalues $\lambda\_k$ and $L\_2w=-w''+V(x)w$
with eigenvalues $\mu\_k$. If $U\geq V$ then $\lambda\_k\geq \mu\_k$.
To prove this consider the Rayleigh ratio:
$$R\_j(w)=\frac{\int \overline{w}L\_jw}{\int |w|^2}.$$
The smallest eigenvalue is the minimum of the Rayleigh ratio.
(Higher eigenvalues are obtained from a Maximin problem for the Rayleigh ratio,
so they also increase when the potential increases).


To prove that the eigenvalues tend to infinity when $\epsilon\to\infty$, compare
with the operator $-w''+\epsilon|x|w$. Let $\lambda\_0(\epsilon)$
be the smallest eigenvalue, and $w$ the corresponding eigenfunction:
$$-w''+\epsilon|x|w=\lambda\_0(\epsilon)w.$$
Set $t=\epsilon^{1/3}x,$ $w(x)=y(\epsilon^{1/3}x).$ Then
$$-y''+|t|y=\lambda\_0(\epsilon)\epsilon^{-2/3}y.$$
Therefore $\lambda\_0(\epsilon)=\epsilon^{2/3}\lambda\_0(1).$ This tends to $\infty$
when $\epsilon\to\infty$.


