A continuous function that is negative at some point will be negative around that point

Question

Proposition:

$f: [0, \infty) \to \mathbb{R}$ be a continuous function and suppose that $f(0) < 0$, then $\exists \alpha > 0, \text{s.t. } f(t)<0, \forall t \in [0, \alpha] .$

Sketch:

Proof attempt:

• Since $f$ is continuous, therefore for all $t, t^\prime \in \mathbb{R}_{\geq 0}$, $\forall \epsilon >0, \exists \delta >0$ such that $|t^\prime - t| < \delta \implies |f(t^\prime) - f(t)| < \epsilon$.

  Then take $t^\prime = 0$, we have for all $t$, $|t| < \delta \implies |f(0) - f(t)| < \epsilon$.

  Take $\delta = \alpha$, then $|f(0) - f(t)| < \epsilon, \forall t \in (-\alpha, \alpha)$.

  Since $|f(0) - f(t)| < \epsilon \Longleftrightarrow -\epsilon < f(0) - f(t) < \epsilon$, therefore $f(t) < f(0) + \epsilon$.

  Thus for any $0< \epsilon < |f(0)|$, $f(t) < 0$ for all $t \in (0, \alpha)$.

There are couple problems in the proof:

1. how do I argue that $f(t) <0$ for a closed interval as in the proposition?

2. how do I go from $|t| < \alpha \Longleftrightarrow -\alpha < t < \alpha$ to $t \in (0, \alpha)$?

Please let me know if there are any other problems in the proof, feel free to post yours if you can construct a more concise one.

Answer 1

Your two questions can be simply addressed by just observing that you can shrink the range that $t$ can be in however you want. For instance, if you know that $f(t)<0$ for all $t$ in $(-\alpha,\alpha)$, you also know that $f(t)<0$ for all $t\in[0,\alpha)$, since any $t\in[0,\alpha)$ is also in $(-\alpha,\alpha)$. (Actually, note that you can only use values of $t$ which are in the domain of the function in the first place, so you actually only know that $f(t)<0$ for $t\in [0,\alpha)$, since for $t<0$ it is undefined.) You can also use this idea to get a closed interval: can you think of a closed interval $[0,\beta]$ that would be contained in $[0,\alpha)$? The answer is hidden below.

Just take any $\beta$ such that $0<\beta<\alpha$, and then $[0,\beta]$ is a subset of $[0,\alpha)$. So whenever $t\in[0,\beta]$, $t\in [0,\alpha)$ as well, so $f(t)<0$. If you want a specific value of $\beta$, you could take $\beta=\alpha/2$.

There are some other issues in your proof. First, your definition of continuity is incorrect. Your definition would be true of any function at all, since you could just choose $\delta$ to be less than $|t-t'|$ so that the implication $|t^\prime - t| < \delta \implies |f(t^\prime) - f(t)| < \epsilon$ is vacuously true. The "for all $t$" in the statement needs to come after the "there exists $\delta$" so that $\delta$ can't depend on $t$.

Also, beware that your choice of $\delta$ (and therefore $\alpha$) depends on $\epsilon$. So you shouldn't really be talking about $\delta$ until you've chosen the $\epsilon$ you're going to use, as you do when you say $0<\epsilon<|f(0)|$. Or, if you want to leave the choice of $\epsilon$ until the end, you should state explicitly that you are using an $\epsilon$ that will be chosen later.

Answer 2

Since the function is only defined for nonnegative values of $t$, continuity at zero in this context is: for every $\epsilon > 0$, there exists $\delta > 0$ such that $| f(0) - f(t) | < \epsilon$ whenever $t \in [0, \delta)$ (your usage of continuity talks about values of $t < 0$, which doesn't make sense in this context).

Now, let $\epsilon = |f(0)|$. This gives you a corresponding $\delta > 0$ such that $f$ is negative on $[0, \delta)$. Now, take $\alpha = \delta / 2$ to get that $f$ is negative on $[0, \alpha]$, since $[0,\alpha]\subset [0,\delta)$.

Answer 3

1. you already spoiled $\alpha$, so I define $\beta = \frac{1}{2}\alpha$ It is clear, that $[ -\beta, \beta ] \subset (-\alpha, \alpha)$. Hence if some statement is true for $ x \in (-\alpha, \alpha)$ it is true for $ x \in [-\beta, \beta ]$ as well.

2. same reasoning: $[0, \beta] \subset [-\beta, \beta]$

Answer 4

With $\epsilon=-\frac {f (0)}{2} $ the continuity at $0$ insures that $$\exists \eta>0\; : \forall x\in (0,\eta) $$

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