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Given $a\in \mathbb R$ and a function $f: \mathbb R \to \mathbb R$, prove that if $f$ is continuous at $a$ and $f(a)>0$, then $\exists \delta >0 $ s.t. $f(x)>0$, $\forall x \in (a-\delta,a+ \delta )$.

I tried it as follows-

To prove - ($f$ is continuous at $a$) and $f(a)>0 \implies \exists \delta>0$, $f(x)>0$, $0<|x-a|<\delta$

Assume $f$ is continuous at $a$ and $f(a)>0$

Then $$\forall \epsilon >0, \exists \delta>0,\text{ s.t. }|x-a|< \delta \implies |f(x)-f(a)| <\epsilon$$

$$|f(x)|-|f(a)|= |f(x)|-f(a) \le |f(x)-f(a)| <\epsilon$$

$$|f(x)|<\epsilon +f(a) $$

$$-\epsilon -f(a)<f(x)<\epsilon +f(a) $$

I can't think of a way to continue this to prove that $f(x)>0$

egreg
  • 238,574
S.Dan
  • 1,115

2 Answers2

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Given that $f$ is continuous at $a$ and $f(a)>0$, then by continuity at $a$, there exists a $\delta>0$ for $\varepsilon=\dfrac{f(a)}{2}>0$ such that for all $x\in \left(x-\delta,x+\delta\right) \implies |f(x)-f(a)|<\dfrac{f(a)}{2}\implies \color{blue}{0}< \dfrac{f(a)}{2}<\color{blue}{f(x)}<\dfrac{3f(a)}{2} \qquad \square$

The following is true. Let $f(x)$ be continuous at $a$, and $f(a)\neq 0$, then there is some neighborhood of $a$ such that $f(x)$ and $f(a)$ has same sign. The proof for the case where $f(a)<0$ is also similar, take $\varepsilon=-\dfrac{f(a)}{2}$.

hrkrshnn
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Let $f(a)=b$ with $0< b$. Let $\epsilon=\frac{b}{2}$ then since $f(x)\rightarrow b$ as $x \rightarrow a$ there is a $\delta >0$ such that

$$|x-a|< \delta \rightarrow |f(x)-b| < \frac{b}{2}$$

this latter condition says

$$-\frac{b}{2} < f(x)-b$$ which gives

$$0< \frac{b}{2} < f(x)$$

Rene Schipperus
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