Find the best upper bound of $\varepsilon$ such that $g(x) = 1 + \varepsilon f(x) > 0$, $f(x)$ bounded on a closed set?

Question

Consider a smooth function $f : [0, 1] \to \mathbb{R}$. Moreover, consider:

$$g(x) = 1 + \varepsilon f(x),$$

for $x \in [0, 1]$ and some $\varepsilon > 0$.

Which are the conditions on $\varepsilon$ which guarantee that $g(x) > 0$? I guess the answer is something in the form:

$$0 < \varepsilon < M,$$

for some $M > 0$.

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My attempt

Let $[0, 1] = I^- \cup I^+$, where:

• $f(x) < 0 ~\forall x \in I^-$;
• $f(x) \geq 0 ~\forall x \in I^+$.

For $x \in I^+$, then $g(x) > 0$.

For $x \in I^-$, since function $f$ is (smooth, then) bounded, then:

$$\exists F > 0 : f(x)> -F ~\forall x \in I^-.$$

Therefore:

$$g(x) = 1 + \varepsilon f(x) > 1 - \varepsilon F ~\forall x \in I^-.$$

If $\varepsilon < \frac{1}{F}$, then:

$$1 - \varepsilon F > 0,$$

and hence

$$g(x) > 0 ~\forall x \in [0, 1].$$

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Questions

Is there some easies way to achieve this kind of results?

Is there any better upper bound for $\varepsilon$?

Answer 1

It's easier than you think. You just need that your function is continuous. Indeed, since $f$ is continuous, for the extreme value theorem $f$ must attain a maximum $M$ and a minimum $m$.
Then (SPOILER)

$$ g(x)\geq 0 \quad \forall x\in [0,1] \Leftrightarrow 1+\varepsilon f(x)\geq 0 \Leftrightarrow 1+\varepsilon m\geq 0 $$ If $m=0$, it's obvious that $g(x)>0$, otherwise we select $\varepsilon >0$ such that $\varepsilon\geq \frac{-1}{m}$

Answer 2

If $f(x) \ge 0$, then $1+\epsilon f(x) > 0$ for all values of $\epsilon$.

If $f(x) < 0$, then

\begin{align} g(x) > 0 &\iff 1 + \epsilon f(x) > 0 \\ &\iff \epsilon f(x) > -1 \\ &\iff \epsilon < -\dfrac{1}{f(x)} \end{align}

So $\displaystyle \epsilon = -\max_{f(x)<0} \dfrac{1}{f(x)}$