Why should the map $-\Delta^{-1}$ be continuous?

Question

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).

Answer 1

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.