Suppose I know that $h(f)$ is bounded by $h(f) \geq (1 -2ab)\Vert f' \Vert^2_{L^2} - \frac{4a}{b} \Vert f \Vert^2_{L^2}$ with $a,b > 0$ and $f \in H^1$. Then for any $b \leq \frac{1}{2a}$ we can find a constant $c > 0$ s.t $h(f) \geq -c \Vert f \Vert^2_{L^2}$. Let $c_\infty$ denote the optimal bound.

My question is: How do I bound $h(f) + (1+c_\infty)\Vert f \Vert^2_{L^2}$ by some $\alpha > 0$ s.t

$\alpha^2 \Vert f \Vert^2_{H^1} \leq h(f) + (1+c_\infty)\Vert f \Vert^2_{L^2}$

With $c = c_\infty$ then I can obviously bound it with the $L^2$-norm, but I dont see how its done with the Sobolev norm, how can I pull out a positive alpha from

$ h(f) + (1+c_\infty)\Vert f \Vert^2_{L^2} \geq (1+c_\infty - \frac{4a}{b} \Vert f \Vert^2_{L^2})+ (1 -2ab)\Vert f' \Vert^2_{L^2} $ ?

Any help is greatly appreciated.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.