I want to prove that there is no $L^2(\mathbb{R}^N)$-solution of the equation $−\Delta \phi =\lambda \phi$ for every $\lambda \in \mathbb{R}$. I know that the Pohozaev identity asserts for $N\geq 3$, $\Omega \subset \mathbb{R}^N$ a non-empty, bounded and open subset with smooth boundary $\delta\Omega$ that if $ \phi \in \mathcal{C}^2(\Omega) \cap \mathcal{C}_0(\Omega)$ is a solution of $−\Delta \phi =\lambda \phi$ on $Ω$, then $$∫_Ω2NF(ϕ)−(N−2)ϕf(ϕ)\ dx=∫_{∂Ω}(\frac{∂ϕ}{∂ν})^2(x⋅ν) \ dσ$$
where $F(ϕ)=∫_0^\phi f(τ) dτ=λϕ$ and $f(ϕ)=λϕ$.
I wonder if this is still true for the case $Ω=\mathbb{R}^N$. If it is, then one would have $$∫_Ω2NF(ϕ)−(N−2)ϕf(ϕ) \ dx=2λ∫_{\Omega = \mathbb{R}^N}ϕ^2(x) dx=0\ \ \ \ \ ∀λ∈R$$ and from this, it follows that $\Vert ϕ\Vert ^2_{L^2}=0$ and so $ϕ=0 $ a.e. on $R^N$, but since $\phi $ is continuous then $\phi = 0$ everywhere on $\mathbb{R} ^N$. Overall, the only $L^2(\mathbb{R}^N)$-solution of the equation $−Δϕ=λϕ$ is the function zero.
Is my reasoning correct? Thank you in advance.