Why should the map $-\Delta^{-1}$ be continuous? I'm reading an article by Wei-Ming Ni about the existence of solutions for the elliptic problem $$\Delta u +|x|^\lambda |u|^\tau =0,$$ 
in the unit ball $\Omega$ in dimension $>2$. I'm looking for solutions in $E=\tilde{H}^1_0$, which means this functions are also radially symmetric and in $H^1_0$. The scalar product is $$\langle u,v \rangle_E = \int_\Omega \nabla u \cdot \nabla v $$
The last step of the proof is based on the definition of a functional $T:E\to E$ such that 
$$\langle Tu,v \rangle_E = \int_\Omega |x|^\lambda |u|^\tau v.$$
How do I know such an operator is well-defined? 
Moreover, it says it can be written in the form $Tu = -\Delta^{-1} (|x|^\lambda |u|^\tau),$ and it goes on saying that the map $T_1 : H^{-1} \to E$ such that $|x|^\lambda|u|^\tau \mapsto  -\Delta^{-1} (|x|^\lambda |u|^\tau)$ is continuous. I can't get why this should be well-defined (and continuous).
 A: First of all, there should be a constraint on the exponent $\tau$ to guarantee certain integrability  conditions. Define
$$n^*:=\frac{2(n+\lambda)}{n-2}, $$
where $n$ denotes the dimension of the ball. Then $E$ embeds continuously in the Banach space $ X:=L^{n^*}(0,1; r^{\lambda+n-1} dr)$. (For a proof of this I refer to this old paper of mine.)
Define $\tau^*:=n^*-1$. (This is the critical Sobolev exponent.) Assume
$$\tau \leq \tau^*. $$
Set $r:=|x|$, $x\in \mathbb{R}^n$. Use  Holder's inequality  for the conjugate exponents $n^*$ and $\frac{n^*}{\tau^*}$  and the Sobolev embedding $E\to X$ to verify that
$$\left\vert\int_\Omega r^\lambda |u|^\tau v dx\right\vert \leq C\Vert u\Vert_E\cdot \Vert v\Vert_E. $$
This proves that $Tu$ defines a linear functional on $E$. The dual of $E$ is $H^{-1}$.
The Laplacian  defines an isomorphism $\Delta E\to H^{-1}$ with inverse  $\Delta^{-1}$. The continuity statement   is proved by observing that the map
$$ E\ni u\mapsto |u|^\tau \in L^{\frac{n^*}{\tau^*}}(0,1, r^{\lambda+n-1} dr). $$
is continuous.
