Let $f,g:\mathbb{R}^n \to \mathbb{R}$ be two continuous functions where $f(\tilde{x}) < g(\tilde{x})$ for some $\tilde{x} \in \mathbb{R}^n$. According to this post, there exist a neighborhood $B(\tilde{x}, \delta)$ such that $f(x) < g(x)$ for all $x \in B(\tilde{x}, \delta)$. While we proof this statement we end up with $g(\tilde{x}) -f(\tilde{x}) -2 \epsilon < g(x) - f(x)$ where $\epsilon =\frac{g(\tilde{x}) -f(\tilde{x})}{2} $ gets the job done.
My question:
I think as long as $\epsilon \le \frac{g(\tilde{x}) -f(\tilde{x})}{2} $, we are fine and $f(x) < g(x)$. Can we use different $\epsilon$ let us say $\epsilon =\frac{g(\tilde{x}) -f(\tilde{x})}{4} $ and conclude $0<\frac{g(\tilde{x}) -f(\tilde{x})}{2} < g(x) - f(x)$? If so, what is the meaning of this? What is the intuition behind this? Is it the right way of finding a lower bound on the difference of two perturbed function?
Let $h(x)=g(x)-f(x)$. Then $h(\tilde{x})>0$. Then, there exists a neighborhood where $ h(\tilde{x})/2 < h(x)$. Since $h$ is continuous, for every $\epsilon>0$ there is $B(\tilde{x}, \delta)$ with $\delta>0$ such that $|h(x)-h(\tilde{x})|<\epsilon$. One can write $-\epsilon < h(x)-h(\tilde{x})<\epsilon$. Use $-\epsilon < h(x)-h(\tilde{x})$ and let your epsilon be $h(\tilde{x})/2$.