here is an exercise from axler's book: linear algebra done right 3ed
Suppose Q is a polynomial on $\mathbb{R^n}$. Prove that there exists a harmonic polynomial P on $\mathbb{R^n}$ such that P(x)=Q(x) for every x $\in$ $\mathbb{R^n}$ with $ \parallel x\parallel=1$. [The only fact about harmonic functions that you need for this exercise is that if P is a harmonic function on $\mathbb{R^n}$ and $P(x)=0$ for every x $\in$ $\mathbb{R^n}$ with $ \parallel x\parallel=1$ then $P=0$
Axler gives a hint that the desired harmonic polynomial P is of the form Q+$ (1-\parallel x\parallel^2)$R for some polynomial R. and he ask to prove the existence of such polynomial R on $\mathbb{R^n}$ that verifies $Q+ (1-\parallel x\parallel^2)R$ is harmonic by defining an operator T on a suitable vector space by Tr=$\Delta((1-\parallel x\parallel^2)r$
and then showing that T is injective and hence surjective. My attempt: first I prove that T is injective (T is defined on the set of polynomial of $\mathbb{R^n}$ to itself) :
assume $\Delta((1-\parallel x\parallel^2)r=0$ so $(1-\parallel x\parallel^2)r$ is harmonic polynomial and since this polynomial equal zero for every $\parallel x\parallel=1$ I conclude using the hint that $(1-\parallel x\parallel^2)r=0$ therefore $r=0$ so T is injectif.The problem is that I can't prove that T is surjective (I can't deduce from injectivity because it's not a finite dimentional vector space).
if I succeed at proving that T is surjective then I can conclude that T is a isomorphisim and since $-\Delta Q \in $ (the image of T) I deduced the existence of a polynomial r st $\Delta((1-\parallel x\parallel^2)r)=-\Delta Q$ which is equivalent to say $\Delta(Q+(1-\parallel x\parallel^2)r)=0$. is my logic right? or did I need to change the vector space to a finite one? note: this is an exerice from inner product chapter
Restrict the map to polynomials of degree less than equal to degree of q. Note that here degree means highest sum of the indices of the monomials with nonzero coefficients.