This problem is from a mathematical analysis exam
Define this bilinear form on $L^2([0,1])$: $$ a(u,v)=\int_{[0,1]^2}{k(x,y) u(x) v(y) dx dy} \qquad \forall u,v \in L^2([0,1]) $$
Can this bilinear form be coercive on $L^2$ for some choice of $k \in L^{\infty}([0,1]^2)$?
I am quite concerned by the "official" solution, which is
Yes, 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 L^2([0,1]) $ we get
$$ \begin{align} a(u,u) & = \int_{[0,1]^2} k(x,y) \: u(x) \: u(y) \: dx \, dy \\ & = \int_{0}^{1} k(x,x) \: u(x)^2 \: dx \\ & \ge a_0 \int_{0}^{1} u(x)^2 dx = a_0 \| u \|_2^{2} \end{align} $$
My point is that in the previous example, $k$ is almost everywhere zero on $[0,1]^2$, so, the bilinear form should be identically zero. Instead, in order to make this passages, one should take $k(x,y)=\delta_x(y)$, which of course is not in $L^\infty$