You can conclude that $A - B$ is a harmonic polynomial: By Weyl's Lemma, $A-B$ is a regular distribution that can be expressed by some $u\in\mathcal C^\infty(\mathbb R^n)$. If you take the Fourier transform, you see that $|\xi|^2 (\hat{A} - \hat{B}) = 0$, so $\hat{A} - \hat{B}$ is supported at the origin. Therefore, $A - B$ is a polynomial. (This required the classification of distributions supported at a single point -- see Hormander, Vol. 1.)

If you know beforehand that, say, $A - B \in L^2$, then you can conclude that $A = B$.

-Dallas

You can conclude that $A - B$ is a harmonic polynomial: If you take the Fourier transform, you see that $|\xi|^2 (\hat{A} - \hat{B}) = 0$, so $\hat{A} - \hat{B}$ is supported at the origin. (All of this makes sense at the level of tempered distributions.) Therefore, $A - B$ is a polynomial. (This required the classification of distributions supported at a single point -- see Hormander, Vol. 1.)

If you know beforehand that, say, $A - B \in L^2$, then you can conclude that $A = B$.

-Dallas

You can conclude that $A - B$ is a harmonic polynomial. If you take Fourier transform, you see that $|\xi|^2 (\hat{A} - \hat{B}) = 0$, so $\hat{A} - \hat{B}$ is supported at the origin. Therefore, $A - B$ is a polynomial. (This required the classification of distributions supported at a single point -- see Hormander, Vol. 1.) If you know beforehand that, say, $A - B \in L^2$, then you can conclude that $A = B$.

-Dallas

You can conclude that $A - B$ is a harmonic polynomial: By Weyl's Lemma, $A-B$ is a regular distribution that can be expressed by some $u\in\mathcal C^\infty(\mathbb R^n)$. If you take the Fourier transform, you see that $|\xi|^2 (\hat{A} - \hat{B}) = 0$, so $\hat{A} - \hat{B}$ is supported at the origin. Therefore, $A - B$ is a polynomial. (This required the classification of distributions supported at a single point -- see Hormander, Vol. 1.)

If you know beforehand that, say, $A - B \in L^2$, then you can conclude that $A = B$.

-Dallas