I am dealing with the following problem. Given the nonlinearity $|u|^2u$ where where $u \in H_0^1(\mathbb{R},\mathbb{C}) \cap H^2(\mathbb{R},\mathbb{C}$).
I need the inequality
$||u|u|^2 ||_{L^2}\leq (||u||_{L^2}||) (||u||_{L^2})(||u||_{L^2})$ to be true; I could also use the $H_0^1$ norm.
Apparently, it can be satisfied with Cauchy-Schwarts - if so, somebody please tell me how!
The paper that I am reading satisfies it automatically.
I am trying to prove this statement for my purposes (because my operator has compact resolvent, and is bounded!).
If an operator in $X$ with compact resolvent is bounded, $X$ must be finite-dimensional.
which is described as a problem in Kato's book. Then, I feel I could use the fact that all norms are equivalent.
Please help with either problem!
Much love
CD
The statement you are looking for cannot hold. Maybe you forgot to mention a crucial piece of information though. Maybe one has a chance with some assumption on the $H^2$ norm of $u$.
If you want a proof, see below.
One clearly has that $\|u |u|^2\|_{L^2} = \|u\|_{L^6}^3$. Moreover, $(\|u\|_{L^2}) (\|u\|_{L^2}) (\|u\|_{L^2}) = (\|u\|_{L^2})^3$ yields that your statement is equivalent to proving that $$ \|u\|_{L^6} \leq \|u \|_{L^2}. $$ for all $u \in H^1_0 \cap H^2$.
It is relatively clear that the result above cannot hold but, for completeness, I give an ad hoc proof below.
Let $u^* \in L^2 \setminus L^6$. By the density of $C^{\infty}_c$ in $L^2$ we have that there exists a sequence $(u_j)_{j\in \mathbb{N}}$ such that $u_j \to u^*$, with $u_j \in C^{\infty}_c$.
Assume towards a contradiction that $\|u\|_{L^6} \leq \|u \|_{L^2}$ holds for all $u \in H^1_0 \cap H^2$, so in particular, since $C^{\infty}_c \subset H^1_0 \cap H^2$ we have that $$ \|u_j- u_k\|_{L^6} \leq \|u_j-u_k\|_{L^2} \text{ for all }j,k \in \mathbb{N}. $$ Hence, $(u_j)_{j\in \mathbb{N}}$ is a Cauchy sequence in $L^6$ and by the completeness of $L^6$ it's limit is $u^{**} \in L^6$, i.e. $u^{**} \neq u^*$. Standard arguments show that this cannot be true. For example, one can choose a pointwise convergent subsequence of $(u_j)_{j\in \mathbb{N}}$ which produces a contradiction.