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:

[![enter image description here](https://i.stack.imgur.com/9TFnG.png)](https://i.stack.imgur.com/9TFnG.png)

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.