Suppose $A$ is an $n\times n$ positive definite matrix, and let $f:\mathbb{R}^n \to \mathbb{R}$ be defined as $f(x) = x^TAx$.
Then suppose we know that for two vectors $x$ and $y$, we have $|x_i-y_i| \leq r$, and $0 \leq x_i,y_i \leq 1$ for all $i \in \{1,2,\ldots,n\}$, then can we bound |f(x)-f(y)| in terms of $r$?
Some hints: Let $A=P'DP$ where D is diagonal. (More work is needed for the case where the case where the decomposition is not possible).
Then $f(x)=x'P'DPx = x'Dx$ where $\bar{x}=Px$
$$f(x)-f(y)=\bar{x}'D\bar{x}-\bar{y}'D\bar{y}=(\bar{x}+\bar{y})'D(\bar{x}-\bar{y})\leftarrow \mbox{here is where the decomposition is useful}\\ =\overline{(x+y)}'D\overline{(x-y)}=(x+y)'A(x-y)$$
Since $\forall i, |x_i-y_i| \leq r$ and $|x_i+y_i| \leq 2$, $$|f(x)-f(y)|=|(x+y)'A(x-y)|\leq 2r \sum_{i=1}^n \sum_{j=1}^n |a_{ij}|$$