Linked Questions

4
votes
3answers
3k views

How can I prove that a continuous injective function is increasing/decreasing? [duplicate]

If I have a continuous, injective function mapping the real numbers, then it is either increasing or decreasing. This seems intuitively obvious but I can't come up with a neat proof for it.
0
votes
2answers
289 views

Let $f$ be an injective and continous function. Prove that $f$ is monotone. [duplicate]

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be an injective and continous function. Prove that $f$ is monotone. My proof-trying Since $f$ is injective for all $x_{1},x_{2}\in\mathbb{R}$ there is a $y$ ...
1
vote
0answers
356 views

A continuous function $f:\Bbb R\to \Bbb R$ is injective if and only if it is strictly increasing or strictly decreasing [duplicate]

Is the following statement true? A continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ is injective if and only if $f$ is strictly increasing or strictly decreasing. If it's true give a ...
0
votes
2answers
102 views

Hint for real analysis question [duplicate]

Suppose that $f$ is one-to-one and continuous on [$a,b$]. Prove that $f$ is either strictly increasing or strictly decreasing on [$a,b$].
18
votes
3answers
853 views

Solution of functional equation $f(x/f(x)) = 1/f(x)$?

I've been trying to add math rigor to a solution of the functional equation in [1], eq. (22). It is: $$ f\left(\frac{x}{f(x)}\right) = \frac{1}{f(x)}\,, $$ where you know that $f(0)=1$ and $f(-x) = f(...
3
votes
2answers
2k views

When is a continuous function also a bijective function?

To begin with, I would like to set forth a property of continuous functions: There doesn't exist a continuous function $f$ on $\mathbb{R}$ such that $f|_{\mathbb{R}\setminus \mathbb{Q}} : \mathbb{...
0
votes
1answer
4k views

Prove that a monotonically increasing continuous function is invertible.

Let $f:[a,b]\rightarrow[f(a),f(b)]$ be strictly increasing continuous function (i.e $x>y \implies f(x)>f(y)$). Prove that f is invertible. Proving that the function is one-to-one was simple ...
2
votes
1answer
2k views

Inequality and inverse function

Is it possible to affirm that if $B\subseteq \mathbb{R}$ $$f: \mathbb{R} \rightarrow B $$ and $$g: \mathbb{R} \rightarrow B$$ are invertible functions and $$f(x) \leq g(x), \forall x \in \mathbb{R} ...
4
votes
2answers
1k views

$f:U\rightarrow \mathbb{R}$ continuous, injective- prove $f(U)$ is open.

If we have that $f:U\rightarrow \mathbb{R}$ where $U$ is open and $U\subset\mathbb{R}$ and $f$ is continuous and injective I want to show that $f(U)$ is open. So I have considered the function ...
3
votes
1answer
128 views

A continuous function composed with itself [duplicate]

Is there a continuous function $f\colon \mathbb{R}\to \mathbb{R}$ such that $f(f(a)) = -a$ for every $a \in \mathbb{R}$?
0
votes
3answers
75 views

How can I prove that some functions are injective?

Definition of an injective function $$ f(x_1) = f(x_2) => x_1 = x_2 $$ Well I have two functions that I can't prove... $$ f(x) = x^3 + x $$ $$ f(x) = \frac{x}{1-log(x)} $$ log(x) denotes the ...
1
vote
1answer
243 views

Is it true that any injective function $f: \mathbb{R} \to \mathbb{R}$ is strictly monotone?

Is it true that any injective function $f: \mathbb{R} \to \mathbb{R}$ is strictly monotone? If yes, how do I prove it? If not, are there any examples of functions that disprove this statement. I ...
0
votes
2answers
50 views

continuouness of functions

i'm having trouble with the following questions: 1)let $f$ be a continuous function in $\mathbb R$. prove that if $|f|$ is monotonous and rising in $\mathbb R$, then $f$ is monotonous in $\mathbb R$. ...
0
votes
1answer
47 views

Find the integral

Let the function $f(x)$ be thrice differentiable and satisfies $f(f(x))=1-x$ for all $x\in[0,1]$. Then find the value of $$\int_0^1f(x)dx$$ given that $f''(\frac{4}{5})=0$. MY APPROACH: I ...
1
vote
1answer
51 views

Show that function is injective and surjective using given condition on $f$

Consider $f$ a continuous function from $\mathbb{R}$ to $\mathbb{R}$ It is given that $$|f(x)-f(y)|\geq \frac{1}{2}|x-y|\tag{1}$$ for all $x,y$ in $\mathbb{R}$ I want to show that $f$ is one-one and ...