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Let $\lambda>0$ and $n\geq 1$. Prove that the operator

$$-\Delta+\lambda I:H^2(\mathbb{R}^n)\to L^2(\mathbb{R}^n)$$

is invertible and find the norm

$$\left|\left|\left(-\Delta+\lambda I\right)^{-1}\right|\right|_{L^2(\mathbb{R}^n)\to L^2(\mathbb{R}^n)}.$$

Futhermore show that

$$\left(-\Delta+\lambda I\right)^{-1}:H^s(\mathbb{R}^n)\to H^{s+2}(\mathbb{R}^n)\qquad s\in\mathbb{R}$$

is bounded and find the norm.

I don't know how to find norm of inverse when $\lambda$ is not $0$ and I'm not sure if invertibility can be proved in same way.

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1 Answer 1

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Consider $g(x) = \|x\|^2 + \lambda$. Then $$(-\Delta + \lambda I)u = F^{-1}M_gFu$$ where $F$ is the Fourier transform and $M_g$ is the operator defined by $M_gu = gu$.

Since $g(x) \geq \lambda$ for all $x \in \mathbb{R}^n$, $1/g$ exists and is bounded by $1/\lambda$. Define $$Au = F^{-1}M_{1/g}Fu$$

Then since $M_g^{-1} = M_{1/g}$ $$(-\Delta + \lambda)A = F^{-1}M_gFF^{-1}M_{1/g}F = I$$ and similarly $$A(-\Delta + \lambda) = I$$

so that $A$ is the inverse of $-\Delta + \lambda$.

Now we have that $$\|Au\|_2 = \|F^{-1}M_{1/g}Fu\|_2 = \|M_{1/g}Fu\|_2 \le \|1/g\|_\infty\|Fu\|_2 = 1/\lambda\|u\|_2$$

so that $\|(-\Delta + \lambda)^{-1}\| \le 1/\lambda$.

To see that the above inequality is optimal, choose $u_n$ such that $Fu_n$ is the characteristic function of the ball of radius $n$ around $0$ and consider $\|Au_n\|_2$ as $n \to \infty$.

To deal with the case $(-\Delta + \lambda)^{-1}: H^s \to H^{s+2}$ recall that

$$\|u\|_{H^s} = \|F^{-1}(1 + \|x\|^2)^{s/2}Fu\|_2$$ and proceed similarly.

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Thanks!! I didn't used Fourier but is easier your approach =). –  Gastón Burrull Jun 11 '12 at 4:38
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