#$L^2$ condition for $f\in H^2(\mathbb{R}^2)$

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hallow arch
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I'm tasked w/ showing that $f\in H^2$ iff $f\in L^2$ and $\Delta f\in L^2$ (Laplacian). I've shown that $f,\Delta f\in L^2\implies f\in H^2$, but going backwards I'm stuck.

I know that the $\norm{f}{H^2}$ is equivalent to $\norm{(1+\abs{\xi}^2)\hat{f}}{L^2}$, but then ik I can't conclude in general that $f+g\in L^2\implies f,g\in L^2$.

Hint is appreciated.

worthy egretBOT
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worthy egretBOT
earnest onyxBOT
hallow arch
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Specifically, I know $f\in H^2\implies (1+\abs{\xi}^2)\hat{f}(\xi)\in L^2$, and so $f+\Delta f\in L^2$ since the fourier transform is invertible on $L^2$

earnest onyxBOT
tulip atlas
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more specifically you are interested in showing that F(f) has a finite L2 norm

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Which would suffice because then (f + Δf) - f would also be in L²

hallow arch
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+close

worthy egretBOT
# hallow arch +close
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worthy egretBOT
# worthy egret

Thank you for your feedback! Cleanse general by deleting it has been awarded 1 helper_points. They now have 121 helper_points. They have 3 helper_points daily left for today.