I have $u,v \in H^1(0,1)$ with $v(0) = 0$. I would like to get a $C >0$ such that
$$|u(1)v(1)| \le C \|u\|_{H^1} \|v\|_{H^1}$$
What I thought about was writing
$$u(1)v(1) = (uv)(1) = \int_0^1 (uv)' + (uv)(0) = \int_0^1 (uv)'$$
Thus
$$|u(1)v(1)| = \left| \int_0^1 (uv)' \right| \le \|(uv)'\|_{L^2} \le \|uv\|_{H^1}$$
At this point I'm wondering if the $H^1$ norm is multiplicative, i.e. if $\|uv\|_{H^1} \le \|u\|_{H^1} \|v\|_{H^1}$, OR at least if there is some $L>0$ such that for all $H^1(0,1)$ functions $w,r$, we have $\|wr\|_{H^1} \le L\|w\|_{H1} \|r\|_{H^1}$ (which seems like a weird thing, though anyway..)