Show That if F and G Are Continuous at X0 Then F G is Continuous at X0 as Well

evinda

Hello!!! (Talking)

I am given this exercise:

Let $f,g:[a,b] \to \mathbb{R}$ continuous functions such that $\int_a^b f(x) dx=\int_a^b g(x) dx$. Show that $\exists x_0 \in [a,b]$ such that $f(x_0)=g(x_0)$ .

That's what I thought:

We consider the function $h(x)=f(x)-g(x)$. So we have to show that if $h:[a,b] \to \mathbb{R}$ continuous and $\int_a^b h(x) dx=0$,then $\exists x_0$ such that $h(x_0)=0.$

Let's suppose that $h(x) \neq 0, \forall x \in [a,b]$.
Then $h(x)>0 \text{ or } h(x)<0$ in $[a,b]$.
Let $h(x)>0 , \forall x \in [a,b]$
Then $f(x)-g(x)>0 \Rightarrow f(x)>g(x) \Rightarrow \int_a^b f(x) dx > \int_a^b g(x) dx \Rightarrow \int_a^b (f(x)-g(x)) dx>0 \Rightarrow \int_a^b h(x) dx>0$,that can't be true,since $\int_a^b h(x) dx=0$.
So, $\exists x_0$ such that $h(x_0)=0$ .

Could you tell me if it is right? (Blush) (Thinking)

stainburg

A whatever theorem claims that:

If $\displaystyle h$ is continuous on $\displaystyle [a,b]$, there is a point $\displaystyle x_0\in [a,b]$, such that

$\displaystyle \int_a^b h(x)dx=h(x_0)(b-a)$.

Now let $\displaystyle h=f-g$. Can you begin with this?

evinda

A whatever theorem claims that:
If $\displaystyle h$ is continuous on $\displaystyle [a,b]$, there is a point $\displaystyle x_0\in [a,b]$, such that

$\displaystyle \int_a^b h(x)dx=h(x_0)(b-a)$.

Now let $\displaystyle h=f-g$. Can you begin with this?

So,you mean that we suppose that $h(x_0) \neq 0$ and we also know that $b-a \neq 0$,and then we conclude that $\int_a^b h(x)dx \neq 0$,that is a contradiction??

Is that what I have done wrong?? (Thinking)

stainburg

So,you mean that we suppose that $h(x_0) \neq 0$ and we also know that $b-a \neq 0$,and then we conclude that $\int_a^b h(x)dx \neq 0$,that is a contradiction??
Is that what I have done wrong?? (Thinking)

If you doubt the whatever theorem, I will give some hint.

Clearly, $\displaystyle h\in \mathcal{R}[a,b]$.

If $\displaystyle m\leq h(x)\leq M$ at each $\displaystyle x\in [a,b]$, then

$\displaystyle m(b-a) \leq \int_a^b h(x)dx \leq M(b-a)$.

Hence, there is a number $\displaystyle \mu \in [m,M]$ such that

$\displaystyle \int_a^b h(x)dx=\mu (b-a)$.

Since $\displaystyle h$ is continuous on $\displaystyle [a,b]$, there must be an $\displaystyle x_0\in [a,b]$ such that $\displaystyle h(x_0)=\mu$.

evinda

If you doubt the whatever theorem, I will give some hint.
Clearly, $\displaystyle h\in \mathcal{R}[a,b]$.

If $\displaystyle m\leq h(x)\leq M$ at each $\displaystyle x\in [a,b]$, then

$\displaystyle m(b-a) \leq \int_a^b h(x)dx \leq M(b-a)$.

Hence, there is a number $\displaystyle \mu \in [m,M]$ such that

$\displaystyle \int_a^b h(x)dx=\mu (b-a)$.

Since $\displaystyle h$ is continuous on $\displaystyle [a,b]$, there must be an $\displaystyle x_0\in [a,b]$ such that $\displaystyle h(x_0)=\mu$.

Ok,I understand the theorem!! But could you also tell me if that what I tried is right or if I have done something wrong? (Thinking)

stainburg

Ok,I understand the theorem!! But could you also tell me if that what I tried is right or if I have done something wrong? (Thinking)

The statement

Let $\displaystyle h(x)>0$ , $\displaystyle \forall x \in [a,b]$

is wrong, because you were not discussing under the condition, $\displaystyle \int_a^b h(x)dx=0$.

By the way, you cannot forget the continuity condition.

evinda

The statement
is wrong, because you were not discussing under the condition, $\displaystyle \int_a^b h(x)dx=0$.

By the way, you cannot forget the continuity condition.

Why is it wrong?? (Wondering) I suppose,for contradiction that $h(x) \neq 0, \forall x \in [a,b]$,so it is either $h(x)>0 \text{ or } h(x)<0, \forall x \in [a,b]$.So,I take the case $h(x)>0, \forall x \in [a,b]$ ,and I want to find a relation that is not true,to have the wanted contradiction.

stainburg

Why is it wrong?? (Wondering) I suppose,for contradiction that $h(x) \neq 0, \forall x \in [a,b]$,so it is either $h(x)>0 \text{ or } h(x)<0, \forall x \in [a,b]$.So,I take the case $h(x)>0, \forall x \in [a,b]$ ,and I want to find a relation that is not true,to have the wanted contradiction.

OK, it's my fault. I really meant that it cannot lead to the conclusion. Continuity is the key. If not (continuous), the conclusion is not true.

stainburg

Why is it wrong?? (Wondering) I suppose,for contradiction that $h(x) \neq 0, \forall x \in [a,b]$,so it is either $h(x)>0 \text{ or } h(x)<0, \forall x \in [a,b]$.So,I take the case $h(x)>0, \forall x \in [a,b]$ ,and I want to find a relation that is not true,to have the wanted contradiction.

Furthermore, I understand what you mean. You want to make sure that both $\displaystyle g(x)\leq f(x)$ and $\displaystyle g(x)\geq f(x)$ hold on some subsets of [a,b]. This cannot lead to the conclusion. For example,

$\displaystyle f(x)=\begin{cases} 1,&&0\leq x<1/2\\1/20,&&x=1/2\\0, &&1/2< x\leq 1\end{cases}$ and $\displaystyle g(x)=1/2$, $\displaystyle x\in [0,1]$. Clearly, $\displaystyle \int_0^1f(x)dx=\int_0^1g(x)dx$ but there is no such $\displaystyle x_0$

evinda

If we want to show that $\exists x_0 \in [a,b]$ such that $h(x_0)=0$,by finding a contradiction, don't we have to suppose that $h(x) \neq 0 , \forall x \in [a,b]$? So, $h(x)>0 \forall x \in [a,b] \text{ or } h(x)<0 \forall x \in [a,b]$.Because, if $h$ would take both positive and negative values in $[a,b]$,according to a theorem,it would exist a point at which it would be equal to $0$.

stainburg

If we want to show that $\exists x_0 \in [a,b]$ such that $h(x_0)=0$,by finding a contradiction, don't we have to suppose that $h(x) \neq 0 , \forall x \in [a,b]$? So, $h(x)>0 \forall x \in [a,b] \text{ or } h(x)<0 \forall x \in [a,b]$.Because, if $h$ would take both positive and negative values in $[a,b]$,according to a theorem,it would exist a point at which it would be equal to $0$.

Then you should use the continuity condition. That's all.

evinda

So you mean that this is wrong:
Then $f(x)-g(x)>0 \Rightarrow f(x)>g(x) \Rightarrow \int_a^b f(x) dx > \int_a^b g(x) dx \Rightarrow \int_a^b (f(x)-g(x)) dx>0 \Rightarrow \int_a^b h(x) dx>0$,that can't be true,since $\int_a^b h(x) dx=0$.
So, $\exists x_0$ such that $h(x_0)=0$ ??

Do I have to do it in an other way?? (Wondering)

stainburg

So you mean that this is wrong:
Then $f(x)-g(x)>0 \Rightarrow f(x)>g(x) \Rightarrow \int_a^b f(x) dx > \int_a^b g(x) dx \Rightarrow \int_a^b (f(x)-g(x)) dx>0 \Rightarrow \int_a^b h(x) dx>0$,that can't be true,since $\int_a^b h(x) dx=0$.
So, $\exists x_0$ such that $h(x_0)=0$ ??
Do I have to do it in an other way?? (Wondering)

I think you did it in a complicated way...After all, you have to follow my way...

evinda

I think you did it in a complicated way...After all, you have to follow my way...

I have understood your way.... But would my way be also right?? (Wondering)

stainburg

I have understood your way.... But would my way be also right?? (Wondering)

I have shown that if you ignore continuity condition you get nothing.

evinda

I have shown that if you ignore continuity condition you get nothing.

So,do I have to do it like that?

$h$ is continuous,so it has a positive minimum value at $y$, let $h(y)$.So, $h(x) \geq h(y) \Rightarrow \int_a^b h(x) \geq h(y)(b-a)>0 $,that is a contradiction.

stainburg

So,do I have to do it like that?
$h$ is continuous,so it has a positive minimum value at $y$, let $h(y)$.So, $h(x) \geq h(y) \Rightarrow \int_a^b h(x) \geq h(y)(b-a)>0 $,that is a contradiction.

Yeah, it has something to do with the continuity... but why do you like contradiction so much?

evinda

Yeah, it has something to do with the continuity... but why do you like contradiction so much?

I usually use contradiction to solve such type of problems.. (Nerd)

Opalg

Hello!!! (Talking)
I am given this exercise:

Let $f,g:[a,b] \to \mathbb{R}$ continuous functions such that $\int_a^b f(x) dx=\int_a^b g(x) dx$. Show that $\exists x_0 \in [a,b]$ such that $f(x_0)=g(x_0)$ .

That's what I thought:

We consider the function $h(x)=f(x)-g(x)$. So we have to show that if $h:[a,b] \to \mathbb{R}$ continuous and $\int_a^b h(x) dx=0$,then $\exists x_0$ such that $h(x_0)=0.$

Let's suppose that $h(x) \neq 0, \forall x \in [a,b]$.
Then $h(x)>0 \text{ or } h(x)<0$ in $[a,b]$.
Let $h(x)>0 , \forall x \in [a,b]$
Then $f(x)-g(x)>0 \Rightarrow f(x)>g(x) \Rightarrow \int_a^b f(x) dx > \int_a^b g(x) dx \Rightarrow \int_a^b (f(x)-g(x)) dx>0 \Rightarrow \int_a^b h(x) dx>0$,that can't be true,since $\int_a^b h(x) dx=0$.
So, $\exists x_0$ such that $h(x_0)=0$ .

Could you tell me if it is right? (Blush) (Thinking)

Your method is correct. In it, you need to use continuity twice. First, you need the intermediate value theorem to justify the assertion that if $h$ takes both positive and negative values, then it must also take the value $0$ somewhere. Second, you need continuity to justify the statement that a strictly positive function has a strictly positive integral.

evinda

Your method is correct. In it, you need to use continuity twice. First, you need the intermediate value theorem to justify the assertion that if $h$ takes both positive and negative values, then it must also take the value $0$ somewhere. Second, you need continuity to justify the statement that a strictly positive function has a strictly positive integral.

Great!!!Thank you very much!!! (Party)

Show That if F and G Are Continuous at X0 Then F G is Continuous at X0 as Well

Show that f(x0)=g(x0)

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