prove that Opeartor $\partial _x^2:{L^2}(0,1) \to {H^{ - 2}}(0,1)$ is invertible

Question

we can easily show that $$\partial _x^2:{L^2}(0,1) \to {H^{ - 2}}(0,1)$$ is continuous, How we can shiw that it is invertible ? thanks.

Answer 1

We can't show it's invertible. If $f(t)=t$ or $f(t)=1$ then $\partial_x^2 f=0$.

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It appears that the OP was led to the question by an assertion regarding $(I-\partial^2)^{-1}$ in the literature. We should note that $I-\partial^2\ne\partial^2$. But there's a more subtle point that should perhaps be clarified.

When people talk about these things they don't always mean what they appear to mean. Leaving aside Sobolev spaces since I know very little about them, an example. One might easily find someone asserting that $$(I-\partial^2)^{-1}:L^2(\mathbb R)\to L^2(\mathbb R)$$is bounded. When someone says this they do not mean that there is an operator $I-\partial^2:L^2\to L^2$ with a bounded inverse!

The fact is that $I-\partial^2$ is an invertible operator on, for example, the Schwarz space. This is easily verified via the Fourier transform; the operator just multiplies $\hat f$ by $1+\xi^2$. The inverse mutiplies $\hat f$ by $(1+\xi^2)^{-1}$, and that inverse extends to a bounded operator on $L^2$ (as one easily sees again by looking at the Fourier transform.) That's what's meant by saying that $(I-\partial^2)^{-1}$ is bounded on $L^2$.

Is $I-\partial^2:L^2(\Omega)\to H^{-2}(\Omega)$ actually invertible? Seems to me that this would depend on $\Omega$. But the invertibility of this operator (from that space to that space) is really not part of what people mean when they talk about the boundedness of the (formal) inverse.

Irrelevant comment re "Depends on $\Omega$": Not that it matters, but $I-\partial^2:L^2(\Omega)\to H^{-2}(\Omega)$ is certainly not invertible for $\Omega=(0,1)$, since $e^t$ is in the kernel. But it is invertible for $\Omega=\mathbb R$; this is trivial from the characterization of Sobolev spaces in terms of the Fourier transform. Note again that this question is less relevant than you might think to the question of boundedness of the operator that people denote $(I-\partial^2)^{-1}$.

Or so it seems to me - if someone who actually knows the difference between a differential operator and a hole in the ground has any corrections or clarifications that would be great.