Problem Statement: Let $H := H^1(0,2) := W^{1,2}(0,2)$ and define the function $B : H^2 \to \mathbb{R}$ by
$$B(f,g) := \int_0^2\!\! f'g' + \int_0^1\!\! f \int_0^1\!\! g$$
Then $B$ is a bilinear form; I wish to show it is continuous, in the sense that $\exists C \ge 0$ such that, for any $f,g \in H$, we have
$$|B(f,g)| \le C \cdot \|f\|_H\cdot \|g\|_H$$
Update: It has occurred to me that this arises as part of Exercise $8.25$ in Haim Brezis' Functional Analysis, Sobolev Spaces, and Partial Differential Equations.
$ \newcommand{\ip}[2]{\left\langle #1 , #2 \right\rangle} \newcommand{\n}[1]{\|#1\|} \newcommand{\norm}[1]{\|#1\|} \newcommand{\nc}{\newcommand} \nc{\ipw}[2]{\ip{#1}{#2}_{W^{1,2}(0,2)}} \nc{\nw}[1]{\norm{#1}_{W^{1,2}(0,2)}} \nc{\ipl}[2]{\ip{#1}{#2}_{L^2(0,2)}} \nc{\nl}[1]{\norm{#1}_{L^2(0,2)}} \nc{\al}[1]{\begin{align*}#1\end{align*}} \nc{\abs}[1]{\left|#1\right|} \nc{\para}[1]{\left(#1\right)} $ My Attempts: As I understand it, the relevant norms and inner products come in from $L^2$ as so:
$$\al{ \ip{f}{g}_{L^2(0,2)} &:= \int_0^2 \!\! fg \\ \norm{f}_{L^2(0,2)} &:= \sqrt{ \int_0^2 \!\!f^2 } \\ \ipw{f}{g} &:= \ip{f}{g}_{L^2(0,2)} + \ip{f'}{g'}_{L^2(0,2)} \\ &= \int_0^2\!\!fg + \int_0^2 \!\!f'g' \\ \nw{f} &:= \sqrt{ \ipw{f}{f} } \\ &= \sqrt{\ip{f}{f}_{L^2(0,2)} + \ip{f'}{f'}_{L^2(0,2)} } \\ &= \sqrt{ \int_0^2 \!\!f^2 + \int_0^2\!\! (f')^2 } }$$
Hence we want to show $\exists C \ge 0$ such that for all $f,g \in W^{1,2}(0,2)$,
$$\abs{B(f,g)} \le C \sqrt{ \para{ \int_0^2\!\! f^2 + \int_0^2 \!\!(f')^2 }\para{ \int_0^2 \!\!g^2 + \int_0^2\!\! (g')^2 } }$$
I've tried working this in a few ways, but the product of the integrals proves to be the peskiest bit. For instance, running through some basic ideas (triangle inequality, Cauchy-Schwarz) in a "forward" sense, I found
$$ \al{ \abs{B(f,g)} &= \abs{ \int_0^2 \!\!f'g' + \para{ \int_0^1\!\! f } \para{ \int_0^1\!\! g } } \\ &\le \abs{ \int_0^2 \!\!f'g' }+ \abs{ \int_0^1\!\! f } \abs{ \int_0^1\!\! g } \\ &\le \int_0^2\!\!\abs{ f'g' }+ \para{ \int_0^1\!\! \abs{f }}\para{ \int_0^1 \!\!\abs{ g }} \\ &\le \int_0^2\!\!\abs{ f'g' }+ \para{ \int_0^2\!\! \abs{f }}\para{ \int_0^2 \!\!\abs{ g }} \tag{$\ast$} \\ &= \ipl{ \abs{f'} }{\abs{g'}} + \ipl{ \abs{f} }{1} \cdot \ipl{\abs{g}}{1} \\ &\le \nl{ \abs{f'} }\cdot \nl{\abs{g'}} + \nl{ \abs{f} } \cdot \nl{1} \cdot \nl{\abs{g}} \cdot \nl{1} \\ &= \nl{ \abs{f'} }\cdot \nl{\abs{g'}} + 2 \cdot \nl{ \abs{f} } \cdot \nl{\abs{g}} \\ &= \sqrt{ \para{ \int_0^2 \!\!(f')^2 } \para{ \int_0^2\!\! (g')^2 } } + 2 \sqrt{ \para{ \int_0^2\!\! f^2 } \para{ \int_0^2\!\! g^2 } } }$$
...and from here, nothing really comes to mind.
Reaching the fourth line, or something like it, seems to be essential so I can "extend" those integrals to being over $(0,2)$ in lieu of $(0,1)$, making them mesh more amicably with the Sobolev norms. The "loosest" formulation of the work up to that point would have in particular
$$\abs{B(f,g)} \le\abs{ \int_0^2\!\! f'g' }+ \para{ \int_0^2 \!\!\abs{f }}\para{ \int_0^2 \!\!\abs{ g }}$$
by applying triangle inequality later, but it doesn't seem to "get" me anything. I certainly can't extend the bounds without having the absolute value of the inputs either, as otherwise I could actually lose, not gain, value.
So instead I tried working backwards (i.e. assuming $C$ exists and working from the right- to the left-hand sides, in the hope somehow I might reach an intermediate step:
$$\al{ &C \cdot \nw{f} \cdot \nw{g} \\ &= C \sqrt{ \para{ \int_0^2\!\! f^2 + \int_0^2\!\! (f')^2 }\para{ \int_0^2 \!\!g^2 + \int_0^2 \!\!(g')^2 } } \\ &= C \sqrt{ \int_0^2\!\! f^2 \int_0^2 \!\!g^2 + \int_0^2 \!\!f^2 \int_0^2 \!\!(g')^2 + \int_0^2\!\! (f')^2 \int_0^2\!\! g^2 + \int_0^2\!\! (f')^2 \int_0^2 \!\!(g')^2 } \\ &\ge C \para{ \sqrt{ \int_0^2\!\! f^2 \int_0^2\!\!g^2 } + \sqrt{\int_0^2\!\! f^2 \int_0^2 \!\!(g')^2 }+ \sqrt{\int_0^2 \!\!(f')^2 \int_0^2\!\! g^2} +\sqrt{ \int_0^2 \!\!(f')^2 \int_0^2\!\! (g')^2 } }\\ &= C \para{ \nl{f} \nl{g} + \nl{f} \nl{g'} + \nl{f'} \nl{g} + \nl{f'} \nl{g'} } \\ &\ge C \para{ \abs{ \ipl{f}{g} } + \abs{ \ipl{f}{g'} } + \abs{ \ipl{f'}{g} } + \abs{ \ipl{f'}{g'} } } \\ &\ge C \para{ \abs{ \ipl{f}{g} } + \abs{ \ipl{f'}{g'} } } \\ &\ge C \abs{ \ipl{f}{g} + \ipl{f'}{g'} } \\ &= C \abs{ \ipw f g } \\ &= C \abs{ \int_0^2 \!\!fg + \int_0^2\!\! f'g' } }$$
Sadly, none of this seemed to pan out either, and didn't get me any real insights, though I feel like it's close somehow.
Does anyone have any ideas as to how I might proceed in either sense, or perhaps an approach or angle I'm overlooking?
You were very close to the end: we have $\|f\|_{L^2},\|f'\|_{L^2} \leq \|f\|_{W^{1,2}}$ (same for $g$), therefore $$\|f'\|_{L^2}\|g'\|_{L^2} + 2\|f\|_{L^2}\|g\|_{L^2} \leq 3 \|f\|_{W^{1,2}}\|g\|_{W^{1,2}}.$$ The expression on the left-hand side appeared in your first attempt. By the way, you can get a better constant if you use Cauchy--Schwarz on the integral over $(0,1)$ and then extend the interval.