Exercise 46, Chapter 1 from Real Analysis, Carothers

Question

I have this homework exercise and I need some quidelines in order to solve it.

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be continuous.

(a) If $f(0) > 0$, show that $f(x) > 0$ for all $x$ in some open interval $(-a , a )$.

(b) If $f(x ) \geq 0$ for every rational $x$, show that $f(x ) \geq 0$ for all real $x$ . Will this result hold with $\geq 0$ replaced by $>0$? Explain.

Answer 1

1. Suppose not then for every $\epsilon_n=1/n \exists x_n\in (-1/n,1/n)$ with $f(x_n)=0$. We have $x_n\rightarrow0$ and since f is continuous $f(0)=lim_{n\rightarrow \infty } f(x_n)=0$ thus a contradiction
2. For $f:R\rightarrow R$ given by $f(x)=(x-\sqrt 2)^2,$ $f(r)>0 \forall r\in Q$ but $f(\sqrt2)=0$. However for a function satisfying the condition we have given $x\in R$ $\exists$ a sequence $r_n$ of rationals such that $r_n\rightarrow x$. Then we have $f(r_n)>0 \forall n$ thus $f(x)=lim_{n\rightarrow \infty } f(r_n)≥0$

Answer 2

It should be inutitively clear that these statements are true. The trick (as in any first chapter of a math text) is formalizing that intuitition.

Hints: For the first question, what does the $\epsilon$-$\delta$ definition of continuity tell you about $f(x)$ at $x = 0$ with $\epsilon = f(0)$? For the second question, remember that for any $x \in \Bbb R$, there is a sequence of rational numbers $x_n$ with $x_n \to x$. What's the connection between continuous functions and sequences?

Answer 3

Recall the definition of continuity at the point $a$:

$\forall$ $e>0$, $\exists$ $b>0$ s.t $\forall$ $x$ s.t $|x-a|<b$, then $|f(x)-f(a)|<e$.

Now, for part a), take $e<\frac{1}{2}f(0)$ and $a=0$. Then we have:

$\exists$ $b>0$ s.t $\forall$ $x$ s.t $|x|<b$, then $|f(x)-f(0)|<e<\frac{1}{2}f(0)$. But then we'd have

$-\frac{1}{2}f(0) < f(x) -f(0) < \frac{1}{2}f(0)$ which implies by looking at the first inequality that

$f(x)>\frac{1}{2}f(0) > 0$

$QED$

Try part b) by yourself but notice that we can apply part a) for every rational point and find a small neighbourhood around every rational point where the function is positive. Also, use that $\forall x \in \Re$, there is a sequence of rationals converging to $x$.

And you probably should edit it to make it $f(x)\geq0$ $\forall$ $x \in \Re$

Answer 4

First one is precisely the Sign Preserving Property, about which you can read about here : http://www.cut-the-knot.org/fta/brodie.shtml

Second one : (There may be other methods too, but I say one of them) : You can take any $r \in \mathbb{R}$ and argue that you have a sequence $\{x_n \}_{n=1}^{\infty}$ that converges to $r$ .