Let $X$ and $Y$ be real Hilbert spaces and let $b : X\times Y \to \mathbb R$ be a bilinear form which satisfies
$$ M := \sup_{x \in X \setminus \{0\}} \sup_{y \in Y \setminus \{0\}} \frac{|b(x,y)|}{\| x \|_X \| y \|_Y} < \infty $$
$$ \beta := \inf_{x \in X \setminus \{0\}} \sup_{y \in Y \setminus \{0\}} \frac{b(x,y)}{\| x\|_X \| y \|_Y} \in (0,\infty).$$
The first identity means that we can associate a bounded linear operator $B \in \mathcal L(X, Y')$ to $b$ which satisfies $$ \langle Bx, y \rangle_{Y' \times Y} = b(x,y) \quad \forall x \in X, y \in X.$$ The second identity says that it has a bounded inverse $B^{-1} \in \mathcal L(Y', X)$. In particular, we have for all $x \in X$: $$ \sup_{y \in Y \setminus \{0\}} \frac{b(x,y)}{\| y \|_Y} \ge \beta \| x\|_X. $$ If we assume that $X$ is dense and continuously embedded in $Y'$, then we can consider $B$ as an (unbounded) linear operator $B \colon X \subseteq Y' \to Y'$. My question is if we can somehow deduce that this operator is accretive, i.e., if $$ \langle Bx, x \rangle_{Y'} \ge 0 \quad \forall x \in X? $$ It seems related to the second-to-last inequality, but I do not know if it actually follows.
For a concrete example to illustrate where this problem arises: Consider a Gelfand triple $V \hookrightarrow H \hookrightarrow V'$ where $H$ and $V$ are some real separable Hilbert spaces. Suppose that the bilinear form $a \colon V \times V \to \mathbb R$ is bounded and coercive. Then $A \in \mathcal L(V, V')$ with bounded inverse. Now set $X = \{ u \in L^2(0,T; V) : u' \in L^2(0,T; V'), u(0) = 0 \}$, $Y = L^2(0,T;V)$ and $$ b(x,y) = \int_J \langle x'(t) + Ax(t), y \rangle_{V' \times V} dt $$ Then the conditions for $M$ and $\beta$ above can be verified (see for instance Section 71.1.2 from Finite Elements by Ern and Guermond). Thus, we can say $B := \partial_t + A \in \mathcal L(X,Y')$.