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Given two continuous and positive semi-definite functions $f,g: \mathbb{R}^n \mapsto \mathbb{R}$ is it possible to find the maximal distance between the functions over a bounded subset of $\mathbb{R}^n$ (i.e., $\max\limits_{\vec{x} \in D} \left|f(\vec{x}) - g(\vec{x})\right|$ where $\forall x \in D \subset \mathbb{R}^n: x_L \leq x \leq x_U$)? If not, is it possible to find a (tight) over-approximation?

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    Why does $f$ and $g$ have to be quadratic forms? – 2017-02-07
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    @user400479 What's you definition of positive semi-definition? is it that for all $\vec x$ you have $ f(\vec x)\geq 0$ ? – 2017-02-07

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As a counterexample, consider $g(\vec x)=\alpha f(\vec x)$ for some $\alpha\geq 0$ and $f(\vec x)=|\vec x|$, then you see that there is no upper bound on that max. In fact $$ \max_{\vec x}|f(\vec x)-g(\vec x)|=\alpha\max_{\vec x}|\vec x| $$ which is not bounded from above.

Does it answer you question?

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    Good point, I'll edit the question to only consider $\vec{x}$ in some bounded subset of the domain – 2017-02-07
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    Provided that with $|D|<\infty$ you mean that the diagonal of $D$ is bounded, you can just take $f(\vec x)=\frac{1}{|\vec x -\vec x_0|}$ with $\vec x_0\in D$. – 2017-02-07
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    I meant that $D$ is a bounded subset of $\mathbb{R}^n$ (i.e. a finite set of vectors each of length $n$), this implies that there would be a maximal element in $D$ and hence $\max\limits_\vec{x} |\vec{x}|$ would be bounded. – 2017-02-08
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    $x_0$ can't be in the domain of $f(\vec{x}) = \frac{1}{|\vec{x} - \vec{x_0}|}$ given that the function is undefined at this point – 2017-02-08
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    If bounded means finite for you, then the answer is much more easy. You can just list all the values and use for example the triangular inequalities. Anyway, I would say that your problem requires a more formal statement... As it is now, it is misleading, a bounded set is not usually defined as a finite set, even if it can be finite. – 2017-02-08
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    Can you be more specific? I've altered the question slightly to make $D$ bounded in the sense that it has a well-defined minimal and maximal value. – 2017-02-10
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    I am sorry @user400479, your question is now even more ambiguous, what are $x_U$ and $x_L$? are them vectors? how can you define the relations $\geq,\leq$ for vectors? Sincerely, I suggest you either to ask colleague of yours to some clarification about you problem or to report the exercise where you question come from... I am trying, but I cant really help you in these conditions. Sorry for saying you this – 2017-02-11