Open your mind before you answer it

Calculus Level 4

Suppose $f:[0,1]\rightarrow[0,1]$ is a continuous non-decreasing function with $f(0)=0$ and $f(1)=1$ . Define $g(y) = \min \{ x \in [0,1] | f(x) \geq y \}$ , then:

g

is non-decreasing. B: If

g

is continuous then

f

is strictly increasing. both A and B neither A nor B

1 solution

Prasun Biswas
Jun 25, 2014

Since it is already said that $f$ is continuous and non-decreasing, so it is increasing (whether $g$ is continuous or not).

According to given def. of $g$ , the value of $g(a)$ is the min value of the set $\{x\in [0,1]| f(x)\geq a\}$ which will increase with increase in $a$ . That is, $g(a)$ will increase if $a$ increases.

So, g is non-decreasing

P.S. -- I haven't actually studied this part in Calculus till now, so feel free to correct me if I'm wrong... :)

[I realize that this solution is quite messed up, I'll edit it shortly. You might want to check Dan Lawson's comment below (message added on 04-11-2015)]

I have two concerns with this solution:

I don't think non-decreasing and continuous implies strictly increasing. For example, if $f(x)$ is defined to be $2x$ for $0 \leq x \leq \frac{1}{2}$ and $1$ for $\frac{1}{2} \leq x \leq 1$ , then $f$ is both continuous and non-decreasing but is not strictly increasing.
Even if the function $f$ was strictly increasing by the conditions of the problem, then B would be vacuously true, not false.

It is the case that B is false, however. Using the example above, $g$ is continuous from $[0,1]$ to $[0, \frac{1}{2}]$ but $f$ is not strictly increasing.

Dan Lawson - 6 years, 11 months ago

That's a nice solution.

Pranav Rao - 5 years, 7 months ago

Open your mind before you answer it

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