How to show if $f(c) > 0$ there exist a neighborhood over which $f(x)>0$?

549 Views Asked by At

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous at $c$ and let $f(c)>0$. Show that there exists a neighborhood $V_{\delta}(c)$ of $c$ such that if $x \in V_{\delta}(c)$ then $f(x)>0$.

The above statement is taken as a fact in Wolfe conditions proof so I thought it might be useful to have the proof.

I tried the continuity definition at $c$ to show it

$$\forall \epsilon > 0, \exists \gamma > 0 ; \\ \forall x \in \mathbb{R}, |x-c| < \gamma \Rightarrow |f(x)-f(c)| < \epsilon$$

For the case $f(x) > f(c)$ it works. I do not know how to proceed for the other way around?

1

There are 1 best solutions below

6
On BEST ANSWER

Recall that, in general, for $c > 0$,

$\vert b - a \vert = \vert a - b \vert < c \Longrightarrow a -c < b < a + c; \tag 1$

now with $f(x)$ continuous and

$f(c) > 0, \tag 2$

we may choose

$0 < \epsilon < f(c) \tag 3$

and then there exists a $0 < \delta \in \Bbb R$ such that

$\vert x - c \vert < \delta \Longrightarrow \vert f(x) - f(c) \vert < \epsilon; \tag 4$

but in the light of (1) this implies

$0 < f(c) - \epsilon < f(x) < f(c) + \epsilon, \tag 5$

which says that

$\vert x - c \vert < \delta \Longrightarrow f(x) > 0. \tag 6$