If $u_m \to u$ and $v_m \to v$, does $b(u_m,v_m) \to b(u,v)$?

Question

In a Hilbert space $H$, if $u_m \to u$ and $v_m \to v$, does $b(u_m,v_m) \to b(u,v)$ if $b$ is a bounded bilinear form on $H$?

Answer 1

Since $b$ is bounded bilinear form on $H$ then $$|b(u,v)|\leq M||u||||v||$$ hence \begin{align}|b(u_m,v-m)-b(u,v)|&\leq|b(u_m,v_m)-b(u_m,v)|+|b(u_m,v)-b(u,v)|\\ &\leq M||u_m||||v_m-v||+M||v||||u_m-u||\end{align}

and since $(u_m)$ is convergent then it's bounded: $||u_m||\leq C$ so \begin{align}|b(u_m,v-m)-b(u,v)|&\leq M(C||v_m-v||+||u_m-u||)\to0\end{align} and we can conclude.