Can a Function and Its Inverse Prove an Integral Inequality?

[evinda]

Hey again! :p

Let $f:[0,+ \infty) \to \mathbb{R}$ strictly increasing and continuous at $[0,+\infty), f(0)=a$(I am not sure,if it is $a$,it could also be $0$ (Blush)) and let $\lim_{x \to +\infty} f(x)=+\infty$.The range of $f$ is $[0,+\infty)$ and the inverse function $f^{-1}:[0,+\infty) \to [0,+\infty)$ is stricty increasing and continuous at $[0,+\infty)$.Prove that $ab \leq \int_a^bf(x)dx + \int_a^b f^{-1}dx (1) ,\forall a,b>0$ and that the equality just stands for $f(a)=b$.

First,we suppose that $f(a) \geq b$.If $b=f(y)$,then $y \leq a$,because $f$ is strictly increasing.

Then,using the sentence:
Let $f:[a,b] \to \mathbb{R}$ strictly increasing and continuous at $[a,b]$.The range of $f$ is $[f(a),f(b)]$ and the inverse function $f^{-1}:[f(a),f(b)] \to \mathbb{R}$ is stricty increasing and continuous at $[f(a),f(b)]$.Then $$\int_a^b f(x)dx +\int_{f(a)}^{f(b)} f^{-1}(x) dx=bf(b)-af(a)$$

we get that: $$\int_0^y f(x)dx+ \int_a^b f^{-1}(x)dx=yf(y)=by \Rightarrow \int_a^b f^{-1}(x)dx=by-\int_0^y f(x)dx (2)$$

Substituting this in the relation (1),we get $ab \leq \int_a^b f(x)dx+ by-\int_0^y f(x)dx $
To get there,I supposed that the condition is $f(0)=a$.

In my notes,substituting the relation $(2)$ in $(1)$,the result is $ab \leq \int_0^a f(x)dx+yb- \int_0^y f(x) dx$

Could you explain me how I can get this relation?

 

[chisigma]

Hey again! :p

Let $f:[0,+ \infty) \to \mathbb{R}$ strictly increasing and continuous at $[0,+\infty), f(0)=a$(I am not sure,if it is $a$,it could also be $0$ (Blush)) and let $\lim_{x \to +\infty} f(x)=+\infty$.The range of $f$ is $[0,+\infty)$ and the inverse function $f^{-1}:[0,+\infty) \to [0,+\infty)$ is stricty increasing and continuous at $[0,+\infty)$.Prove that $ab \leq \int_a^bf(x)dx + \int_a^b f^{-1}dx (1) ,\forall a,b>0$ and that the equality just stands for $f(a)=b$.

First,we suppose that $f(a) \geq b$.If $b=f(y)$,then $y \leq a$,because $f$ is strictly increasing.

Then,using the sentence:
Let $f:[a,b] \to \mathbb{R}$ strictly increasing and continuous at $[a,b]$.The range of $f$ is $[f(a),f(b)]$ and the inverse function $f^{-1}:[f(a),f(b)] \to \mathbb{R}$ is stricty increasing and continuous at $[f(a),f(b)]$.Then $$\int_a^b f(x)dx +\int_{f(a)}^{f(b)} f^{-1}(x) dx=bf(b)-af(a)$$

we get that: $$\int_0^y f(x)dx+ \int_a^b f^{-1}(x)dx=yf(y)=by \Rightarrow \int_a^b f^{-1}(x)dx=by-\int_0^y f(x)dx (2)$$

Substituting this in the relation (1),we get $ab \leq \int_a^b f(x)dx+ by-\int_0^y f(x)dx $
To get there,I supposed that the condition is $f(0)=a$.

In my notes,substituting the relation $(2)$ in $(1)$,the result is $ab \leq \int_0^a f(x)dx+yb- \int_0^y f(x) dx$

Could you explain me how I can get this relation?

I think that the following pitcure will clarify all!...

Kind regards

$\chi$ $\sigma$

 

[evinda]

I haven't really understood it.. (Thinking)

 

[I like Serena]

I haven't really understood it.. (Thinking)

The substitution is incorrect as you probably already know. (Crying)

Can you give some extra context?
Where is it supposed to lead?
Do you perhaps have a picture in your notes?

 

[evinda]

The substitution is incorrect as you probably already know. (Crying)

Can you give some extra context?
Where is it supposed to lead?
Do you perhaps have a picture in your notes?

So you mean that we don't conclude at this relation: $ab \leq \int_0^a f(x)dx+yb- \int_0^y f(x) dx$ but at this $ab \leq \int_a^b f(x)dx+ by-\int_0^y f(x)dx $ ?

No,there is no picture in my notes.. (Shake) I have to show this inequality:$ab \leq \int_a^bf(x)dx + \int_a^b f^{-1}dx ,\forall a,b>0$.
Using the sentence,that I have written in the first post, we find an inequality which we substistute at the relation we want to prove. Doing calculations we conclude to a relation, so if we show that this relations stands then the initial inequality we are supposed to proved stands too.

 

[I like Serena]

So you mean that we don't conclude at this relation: $ab \leq \int_0^a f(x)dx+yb- \int_0^y f(x) dx$ but at this $ab \leq \int_a^b f(x)dx+ by-\int_0^y f(x)dx $ ?

No,there is no picture in my notes.. (Shake) I have to show this inequality:$ab \leq \int_a^bf(x)dx + \int_a^b f^{-1}dx ,\forall a,b>0$.
Using the sentence,that I have written in the first post, we find an inequality which we substistute at the relation we want to prove. Doing calculations we conclude to a relation, so if we show that this relations stands then the initial inequality we are supposed to proved stands too.

Your inequality is not true.

Pick for instance $a=b$.
Then we get:
$$a^2 \le \int_a^a f(x)dx + \int_a^a f^{-1}(x)dx = 0$$
This is false for any $a>0$.

 

[evinda]

Your inequality is not true.

Pick for instance $a=b$.
Then we get:
$$a^2 \le \int_a^a f(x)dx + \int_a^a f^{-1}(x)dx = 0$$
This is false for any $a>0$.

Now I found the exercise in my textbook.. (Wait) I have to show that this inequality is true: $ab \leq \int_0^a f(x) dx + \int_0^b f^{-1} (x) dx, \forall a,b >0 (1) $ and that the equality is true iff $f(a)=b$.

First,we suppose that $f(a) \geq b$. If $b=f(y)$,then $y \leq a$.
From the sentence,we get that: $\int_0^y f(x) dx+ \int_0^b f^{-1}(x) dx=yb \Rightarrow \int_0^b f^{-1}(x) dx=yb-\int_0^b f(x)dx$

$(1) \Rightarrow ab \leq \int_0^a f(x) dx+yb-\int_0^y f(x)dx \Rightarrow b(a-y) \leq \int_y^a f(x)dx \leq \int_y^a f(a) dx =f(a)(a-y) \Rightarrow f(a) \geq b$

But..why do we suppose at the beginning that $b=f(y)$ ?

 

[I like Serena]

Now I found the exercise in my textbook.. (Wait) I have to show that this inequality is true: $ab \leq \int_0^a f(x) dx + \int_0^b f^{-1} (x) dx, \forall a,b >0 (1) $ and that the equality is true iff $f(a)=b$.

First,we suppose that $f(a) \geq b$. If $b=f(y)$,then $y \leq a$.
From the sentence,we get that: $\int_0^y f(x) dx+ \int_0^b f^{-1}(x) dx=yb \Rightarrow \int_0^b f^{-1}(x) dx=yb-\int_0^b f(x)dx$

$(1) \Rightarrow ab \leq \int_0^a f(x) dx+yb-\int_0^y f(x)dx \Rightarrow b(a-y) \leq \int_y^a f(x)dx \leq \int_y^a f(a) dx =f(a)(a-y) \Rightarrow f(a) \geq b$

That makes much more sense! (Happy)

But..why do we suppose at the beginning that $b=f(y)$ ?

It is not an assumption, it is a definition of $y = f^{-1}(b)$, so we can refer to it with a single symbol.

 

[evinda]

It is not an assumption, it is a definition of $y = f^{-1}(b)$, so we can refer to it with a single symbol.

Why do we know that there is a $y$ such that $y = f^{-1}(b)$ ? (Thinking)

 

[I like Serena]

Why do we know that there is a $y$ such that $y = f^{-1}(b)$ ? (Thinking)

Both $f$ and $f^{-1}$ are continuous on $[0, \infty)$.
Furthermore the second integral in your inequality ends at $b$. Therefore $f^{-1}(b)$ must exist.

 

[evinda]

Both $f$ and $f^{-1}$ are continuous on $[0, \infty)$.
Furthermore the second integral in your inequality ends at $b$. Therefore $f^{-1}(b)$ must exist.

I understand!Thank you very much! (Clapping)