Theorem : Let $M_0=(x_1,...,x_m) \in D(f) \subset R^m$ be the point of continuity of a function $f(M)$. If $f(M_0)\neq 0$ and $f(M_0)>0$, then $\exists B_{\delta}(M_0) \subset D(f)$ open ball with radius $\delta$, s.t. $\forall M \in D(f) \cap B_\delta(M_0)$ holds $f(M)f(M_0)>0$.

I understand the meaning of the theorem, but I'm looking for an obvious strategy to prove it. My try so far, I'm not quite sure if its correct, could you please check it?

If the function is continous at point $M_0$, we can say $\forall \epsilon>0$ $\exists \delta_0>0$ $\forall M \in D(f)$, the distance $\rho(M,M_0)<\delta_0$ in other words $M \in B_{\delta_0}(M_0) => |f(M)-f(M_0)|<\epsilon$

$0 < f(M_0)-\epsilon < f(M) < f(M_0)+\epsilon$

Here, I suppose, I need to give a proof by contradiction.

$\forall \delta>0$, no matter how small $\delta$ is, $ \exists M' \in B_\delta(M_0) \cap D(f)$ s.t. $f(M')\leq 0$

let us fix $\delta=\frac{1}{n}$. $\exists N(\delta)\in \mathbf{N}$, that $\forall n>N$, $\exists \{ {M_n}^{'} \}_N^{\infty}{} \in D(f)$ sequence, s.t. $\rho(\{{M_n}^{'}\},M_0)<\delta=\frac{1}{n}$ and $f(\{{M_n}^{'}\})\leq 0$
because of continuity at $M_0$, we are getting $f(\{{M_n}^{'}\}) \rightarrow f(M_0)\leq 0$ which is a contradiction, because we have $f(M_0)>0$.QED.
Thank you for your time.


Suppose $f(M_0) >0$, then there exists an $\epsilon >0$ such that $$f(M_0) \geq \epsilon.$$ Now let us work with $\epsilon/2$, from continuity, we know there exists a $\delta>0$ such that for each $M\in D(f) \cap B_\delta(M_0)$ we have $$|f(M_0) - f(M)|\leq \epsilon/2.$$ Now because $f(M_0) \geq \epsilon$, we get $$f(M) \geq \epsilon/2 >0.$$ The product of two positive numbers is positive, we have for each $M\in D(f) \cap B_\delta(M_0)$, $f(M_0)f(M) >0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.