Continuity of a step function

Question

I found the following question in GRE 1268 practice test. I was wondering if you could tell me how to approach this problem since everytime I encounter a similar problem I have troubles solving it. Thanks for your help!

The function $f:\mathbb{R} \rightarrow \mathbb{R}$ is defined as follows,

$$f(x) = 3x^2 \text{ if } x \in \mathbb{Q}$$ and $$f(x) = -5x^2 \text{ if } x \notin \mathbb{Q}$$

Which of the following is true?

A) f is discontinuous at all $x \in \mathbb{R}$

B)f is continuous only at $x = 0$ and differentiable only at $x = 0$

C) f is continuous only at $x = 0$ and nondifferentiable at all $x \in \mathbb{R}$

D) f is continuous at all $x \in \mathbb{Q}$ and nondifferentiable at all $x \in \mathbb{R}$

E) f is continuous at all $x \notin \mathbb{Q}$ and nondifferentiable at all $x \in \mathbb{R}$.

Start with the definition. Where do you get stuck? – user251257 Oct 27 '16 at 16:56 — user251257, Oct 27 '16 at 16:56
http://math.stackexchange.com/questions/1922417/ – BCLC Oct 27 '16 at 17:44 — BCLC, Oct 27 '16 at 17:44

score 1 · Answer 1 · answered Oct 27 '16 at 22:13

1

Briefly: Any function on $\mathbb R$ that satisfies $|f(x)| \le Cx^2$ is differentiable at $0.$ Just go back to the definition of the derivative (note that $f(0)=0$). It follows that such a function is continuous at $0.$ We have this in our problem with the given $f(x),$ with $C = 5.$ Let's go through the GRE drill: It follows that A is false, and so are C, D, E. That leaves only B.

answered Oct 27 '16 at 22:13

zhw.

107,943

Re first sentence, is that because $$0 \le \frac{|f(x) - f(0)|}{|x - 0|} \le \frac{Cx^2}{|x|} = Cx \ ?$$ Kind of like 38 III here? – BCLC Oct 28 '16 at 04:53
Also do we need any condition on $f(0)$? – BCLC Oct 28 '16 at 04:54
1

Yes, good proof, and $f(0)=0$ follows from the given inequality. – zhw. Oct 28 '16 at 05:24
zhw. Thanks!... – BCLC Oct 28 '16 at 05:42
1

Oh one thing in your proof. At the end it should be $C|x|.$ – zhw. Oct 28 '16 at 05:50

score 0 · Accepted Answer · answered Oct 27 '16 at 17:36

$f(x)$ is continuous at $x=a$ iff for every sequence $x_n$ that converges to $a$, the sequence $f(x_n)$ converges to $f(a)$

That's either the definition or a property from the standard $\varepsilon-\delta$ definition.

So consider an irrational sequence $i_n$ and a rational sequence $r_n$ that both converge to some number $a$. Then we have

$$\lim f(i_n) = \lim -5i_n^2 = -5a^2$$

$$\lim f(r_n) = \lim 3r_n^2 = 3a^2$$

We have $-5a^2 = 3a^2 \iff a=0$

Hence the $f(x)$ is continuous at $x=0$ and nowhere else.

So what about differentiability?

$f$ is not differentiable for any $x \ne 0$ since differentiability implies continuity. So let's see if $f$ is differentiable at $x=0$:

$$f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$$

$$= \lim_{h \to 0} \frac{f(h)}{h} \tag{*}$$

$(*)$ is supposed* to be equal to both of the following:

$$\lim_{n \to \infty} \frac{f(ih_n)}{ih_n} \tag{1}$$

$$\lim_{n \to \infty} \frac{f(rh_n)}{rh_n} \tag{2}$$

where $ih_n$ is an irrational sequence, $rh_n$ is a rational sequence and both converge to $0$

Compute $(1)$:

$$\lim_{ih_n \to 0} \frac{f(ih_n)}{ih_n} = \lim_{ih_n \to 0} \frac{-5ih_n^2}{ih_n} = \lim_{ih_n \to 0} \frac{-5ih_n}{1} = 0$$

Compute $(2)$:

$$\lim_{rh_n \to 0} \frac{f(rh_n)}{rh_n} = \lim_{rh_n \to 0} \frac{3rh_n^2}{rh_n} = \lim_{rh_n \to 0} \frac{3rh_n}{1} = 0$$

Hence, $f'(0)$ exists and namely is zero.

Therefore, $f$ is continuous and differentiable at $x=0$ and nowhere else.

*I think there's some rule like

$$\lim_{x \to a} f(x) = L \iff$$

for all sequences $x_n$ that converge to $a$

$$\lim_{n \to \infty} f(x_n) = L$$

Continuity of a step function

2 Answers2