Conditions for Kernel coercivity on a bilinear form

Question

Let $H = L^2([0,1])$, $K \in L^{\infty}([0,1]^2)$. I have the following bilinear form: $$ a(u,v)=\int_{[0,1]^2}{K(x,y) u(x) v(y) dx dy} \;\;\; \forall u,v \in H $$ What conditions do I have to set for $ K $ so that $ a $ is coercive (i.e. $ \exists \, a_0 > 0 \; s.t. \; a(u,u) \ge a_0 \|u\|_H^2 $)? I'm suspecting both $ K $ symmetric and positive definite, but I am still not sure of how to formally prove it.

Answer 1

The bilinear form $ a $ is coercive as long as we make the following assumptions regarding the Kernel function $ K $:

• $ K(x,x) \in C([0,1]) $ ;
• $ \inf_{x \in [0,1]} K(x,x) = a_0 > 0 $ ;
• $ K(x,y) = 0 \;\;\; \forall \; x \ne y $ .

Using these three conditions, $ \forall u \in H $ we get

$$ \begin{aligned} a(u,u) & = \int_{[0,1]^2} K(x,y) \: u(x) \: u(y) \: dx \, dy \\ & = \int_{0}^{1} K(x,x) \: u(x)^2 \: dx \\ & \ge a_0 \int_{0}^{1} u(x)^2 dx = a_0 \| u \|_{H}^{2} \end{aligned} $$

And thus, we get coercivity.

Answer 2

I'll try to answer, but since I'm not a matematician I would be glad to have some feedback on the correctness of this answer.

I think your bilinear form is never coercive. Indeed, we can rewrite $a(u,u)$ as

$$\int_0^1 u(x)Tu(x)\ dx \qquad Tu(x):=\int_0^1 K(x,y)u(y)\ dy$$

Then, since $K$ is bounded, $T$ is an Hilber-Shmidt operator on $H$, so is compact in particular. Thus, by the properties of compact operators, $$\forall \delta>0\ \exists u\in H:\qquad \|Tu\|_H\le \delta\|u\|_H$$

so that, taking this $u$, we get

$a(u,u)=\int_0^1 u(x)Tu(x)\ dx\le \|u\|_H\|Tu\|_H\le \delta \|u\|_H^2$

since this holds for every $\delta>0$, the bilinear form cannot be coercive.