This seems quite trivial, however I have never used this before so I am very cautious. Allow me to give some context. I want to bound,
\begin{align} \int_{\Omega}(-\textbf{a}-\textbf{b})v\cdot\nabla vdx, \end{align}
from above where, $\textbf{a},\textbf{b}\in L^{\infty}(\Omega)^{N}$, $v\in H^{1}(\Omega)$ and $\Omega\subset\mathbb{R}^{N}$ is bounded and smooth. The way I do this is by using $-\textbf{a}\leq\|\textbf{a}\|_{\infty}$ and $-\textbf{b}\leq\|\textbf{b}\|_{\infty}$. Now, this seems fine considering
\begin{align} \|f\|_{\infty}=\text{ess}\sup_{\Omega}|f|=\text{ess}\sup_{\Omega}|-f|=\|-f\|_{\infty}\geq-f. \end{align}
Have I done something incorrect here or does everything work fine?
So if $\|\textbf{a}\|_{\infty}=\sup_{x\in\Omega}|\textbf{a}(x)|$ then $|\textbf{a}(x)|\leq\|\textbf{a}\|_{\infty}$ hence, \begin{align} \int_{\Omega}(-\textbf{a}-\textbf{b})v\cdot\nabla v dx&\leq\int_{\Omega}|-(\textbf{a}+\textbf{b})||v\cdot\nabla v|dx\\ &=\int_{\Omega}|\textbf{a}+\textbf{b}||v\cdot\nabla v|dx\\ &\leq\int_{\Omega}(\|\textbf{a}\|_{\infty}+\|\textbf{b}\|_{\infty})|v\cdot\nabla v|dx\\ &\leq(\|\textbf{a}\|_{\infty}+\|\textbf{b}\|_{\infty})\|v\|_{2}\|\nabla v\|_{2}. \end{align}
Does this look more appropriate?
For $a \in L^\infty(\Omega)^N$ I would bound it like follows:
\begin{align*} \int_{\Omega}-\textbf{a}v\cdot\nabla v \, \mathrm dx &\leq \left\lvert \int_{\Omega}-\textbf{a}v\cdot\nabla v \, \mathrm dx \right \vert \\ &= \left \lvert \int_{\Omega} \sum_{k = 1}^N \textbf{a}_k(x) v(x) \partial_k v(x) \, \mathrm dx \right \rvert\\ &\leq \sum_{k = 1}^N \int_{\Omega} \lvert \textbf{a}_k(x) \rvert \lvert v(x) \partial_k v(x) \rvert \, \mathrm dx \\ &\leq \max_{k = 1, \dots, N} \lVert \mathbf a_k \rVert_{L^\infty(\Omega)} \cdot \sum_{k = 1}^N \int_{\Omega} \lvert v(x) \partial_k v(x) \rvert \, \mathrm dx \\ &\leq \max_{k = 1, \dots, N} \lVert \mathbf a_k \rVert_{L^\infty(\Omega)} \cdot \sum_{k = 1}^N \lVert v \rVert_{L^2(\Omega)}\cdot \lVert \partial_k v \rVert_{L^2(\Omega)} \quad \text{(Cauchy-Schwarz)}\\ &\leq \lVert \mathbf a \rVert_{L^\infty(\Omega)^N} \cdot \lVert v \rVert_{L^2(\Omega)} \cdot \lVert v \rVert_{H^1(\Omega)}, \end{align*} where $\lVert v \rVert_{H^1(\Omega)} := \lVert v \rVert_{L^2(\Omega)} + \sum_{k = 1}^N \lVert \partial_k v \rVert_{L^2(\Omega)}$ and $\lVert \mathbf a \rVert_{L^\infty(\Omega)^N} := \max_{k = 1, \dots, N} \lVert \mathbf a_k \rVert_{L^\infty(\Omega)}.$
Finally observe, that $a + b \in L^\infty(\Omega)^N$.