Continuity of bilinear form

Question

Let $X$ be a Banach space and $b(\cdot,\cdot):[0,C] \times X \to \mathbb{R}$ be a form that is continuous wrt. the first argument, linear wrt. the second argument and satisfies $$b(t,x) \leq K\lVert x \rVert_X$$ a boundedness condition for all $x$ and all $t \in [0,C]$. Suppose $x_h \to x$ in $X$ as $h \to 0.$ Does $$b(t+h, x_h) \to b(t,x)$$ as $h \to 0?$ How do I prove this?

Answer 1

Fix $\epsilon >0$.

If $b$ is continuous wrt the first argument, then for any $x\in X$ and $\epsilon>0$, $\exists \delta>0$ such that $|h|<\delta$ implies $|b(t+h,x)-b(t,x)|\leq \dfrac{\varepsilon}{2}$.

Now, since $x_h\rightarrow x$, it means that there exists also $\eta>0$ such that $|h|<\eta\Rightarrow \|x_h-x\|<\dfrac{\varepsilon}{2C}$.

When $|h|<\min(\delta,\eta)$, $$\begin{align*}|b(t+h,x_h)-b(t,x)| & =|b(t+h,x_h+x-x)-b(t,x)| \\ &=|b(t+h,x_h-x)+b(t+h,x)-b(t,x)|\end{align*}$$ by linear w.r.t the second argument, so $$\begin{align*}|b(t+h,x_h)-b(t,x)| & \leq |b(t+h,x_h-x)|+|b(t+h,x)-b(t,x)| \\ & \leq |b(t+h,x_h-x)|+\dfrac{\varepsilon}{2} \\ & \leq C\|x_h-x\| + \dfrac{\varepsilon}{2} \\ & \leq \varepsilon\end{align*}$$