# Show inequality using the mean value theorem

• MHB
• mathmari
In summary, the conversation involves discussing how to use the mean value theorem to show that for all $x,y \in D$, the absolute value of the difference between the function values of $x$ and $y$ is less than the euclidean norm of $x-y$. The conversation also touches on using the gradient instead of the derivative, finding an upper bound for the norm of the gradient, and analyzing the expression $|f(y)-f(x)|$.
mathmari
Gold Member
MHB
Hey!

Let $D=\left \{x=(x_1, x_2)\in \mathbb{R}^2: x_2>\frac{1}{x_1}, \ x_1>0\right \}$.

We have the function $f: D\rightarrow \left (0,\frac{\pi}{2}\right )$ with $f(x)=\arctan \left (\frac{x_2}{x_1}\right )$.

I want to show using the mean value theorem in $\mathbb{R}^2$ that for all $x,y\in D$ it holds that $$|f(x)-f(y)|<\frac{1}{\sqrt{2}}\|x-y\|$$ where $\|\cdot \|$ is the euclidean norm in $\mathbb{R}^2$.
Do we have to do the following?

We take an interval $(y,x)\in D$ and then from the mean value theorem we get that $$f'(c)=\frac{f(x)-f(y)}{x-y}$$ where $c\in (x,y)$. Then we have to find an upper bound for $f'(c)$.

(Wondering)

Hey mathmari! (Smile)

Shouldn't we use the version of the mean value theorem for functions of more than 1 variable? (Wondering)

I like Serena said:
Shouldn't we use the version of the mean value theorem for functions of more than 1 variable? (Wondering)

Oh yes. So, the difference is that we use here the gradient instead of the derivative at $c$ or not? (Wondering)

Then do we take the norm of $$\nabla f(c)=\frac{f(x)-f(y)}{x-y}$$ and since $f$ is in $\mathbb{R}$ the norm is equal to the absolute vaule? (Wondering)

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mathmari said:
Oh yes. So, the difference is that we use here the gradient instead of the derivative at $c$ or not? (Wondering)

Then do we take the norm of $$\nabla f(c)=\frac{f(x)-f(y)}{x-y}$$ and since $f$ is in $\mathbb{R}$ the norm is equal to the absolute vaule? (Wondering)

Shouldn't it be:
$$f(y)-f(x)=\nabla f((1-c)x+cy)\cdot (y-x)$$
(Wondering)

I like Serena said:
Shouldn't it be:
$$f(y)-f(x)=\nabla f((1-c)x+y)\cdot (y-x)$$
(Wondering)

Ahh ok! So, we have the following: $$\|f(y)-f(x)\|=\|\nabla f((1-c)x+y)\cdot (y-x)\|=\|\nabla f((1-c)x+y)\|\cdot \|(y-x)\|$$ Since the values of the function are in $\mathbb{R}$, we have that the norm is equal to the absolute value:
$$|f(y)-f(x)|=\|\nabla f((1-c)x+y)\|\cdot \|(y-x)\|$$

The gradient of $f$ is the following: $$\nabla f(x_1, x_2)=\begin{pmatrix}f_{x_1} \\ f_{x_2}\end{pmatrix}=\begin{pmatrix}-\frac{x_2}{x_2^2+x_1^2}\\ \frac{x_1}{x_1^2+x_2^2}\end{pmatrix}$$

The norm of the gradient is equal to \begin{align*}\|\nabla f((1-c)x+y)\|&=\|\nabla f((1-c)(x_1, x_2)+(y_1, y_2))\|\\ & =\|\nabla f((1-c)x_1+y_1, (1-c)x_2+y_2)\|\\ & =\sqrt{\left (-\frac{(1-c)x_2+y_2}{[(1-c)x_2+y_2]^2+[(1-c)x_1+y_1]^2}\right )^2+\left (\frac{(1-c)x_1+y_1}{[(1-c)x_1+y_1]^2+[(1-c)x_2+y_2]^2}\right )^2} \\ & = \sqrt{\frac{[(1-c)x_2+y_2]^2+[(1-c)x_1+y_1]^2}{\left ([(1-c)x_2+y_2]^2+[(1-c)x_1+y_1]^2\right )^2}}\\ & =\sqrt{\frac{1}{[(1-c)x_2+y_2]^2+[(1-c)x_1+y_1]^2}} \end{align*}

Is everything correct so far? (Wondering)

Now we have to find an upper bound of that norm, or not? (Wondering)

mathmari said:
Ahh ok! So, we have the following: $$\|f(y)-f(x)\|=\|\nabla f((1-c)x+y)\cdot (y-x)\|=\|\nabla f((1-c)x+y)\|\cdot \|(y-x)\|$$

I'm afraid I forgot a $c$ in front of $y$. (Blush)
We should have:
$$\|f(y)-f(x)\|=\|\nabla f((1-c)x+cy)\cdot (y-x)\|$$

And we get an inequality don't we?
That is, by Cauchy:
$$\|f(y)-f(x)\|\le\|\nabla f((1-c)x+cy)\|\cdot \|(y-x)\|$$
(Thinking)

mathmari said:
Since the values of the function are in $\mathbb{R}$, we have that the norm is equal to the absolute value:
$$|f(y)-f(x)|=\|\nabla f((1-c)x+y)\|\cdot \|(y-x)\|$$

The gradient of $f$ is the following: $$\nabla f(x_1, x_2)=\begin{pmatrix}f_{x_1} \\ f_{x_2}\end{pmatrix}=\begin{pmatrix}-\frac{x_2}{x_2^2+x_1^2}\\ \frac{x_1}{x_1^2+x_2^2}\end{pmatrix}$$

The norm of the gradient is equal to \begin{align*}\|\nabla f((1-c)x+y)\|&=\|\nabla f((1-c)(x_1, x_2)+(y_1, y_2))\|\\ & =\|\nabla f((1-c)x_1+y_1, (1-c)x_2+y_2)\|\\ & =\sqrt{\left (-\frac{(1-c)x_2+y_2}{[(1-c)x_2+y_2]^2+[(1-c)x_1+y_1]^2}\right )^2+\left (\frac{(1-c)x_1+y_1}{[(1-c)x_1+y_1]^2+[(1-c)x_2+y_2]^2}\right )^2} \\ & = \sqrt{\frac{[(1-c)x_2+y_2]^2+[(1-c)x_1+y_1]^2}{\left ([(1-c)x_2+y_2]^2+[(1-c)x_1+y_1]^2\right )^2}}\\ & =\sqrt{\frac{1}{[(1-c)x_2+y_2]^2+[(1-c)x_1+y_1]^2}} \end{align*}

Is everything correct so far?

Now we have to find an upper bound of that norm, or not?

If we add a $c$ in front of $y$, then I believe it is correct. (Nod)

And yes, we have to find an upper bound in terms of $\|(y-x)\|$. (Thinking)

I like Serena said:
I'm afraid I forgot a $c$ in front of $y$. (Blush)
We should have:
$$\|f(y)-f(x)\|=\|\nabla f((1-c)x+cy)\cdot (y-x)\|$$

And we get an inequality don't we?
That is, by Cauchy:
$$\|f(y)-f(x)\|\le\|\nabla f((1-c)x+cy)\|\cdot \|(y-x)\|$$
(Thinking)
If we add a $c$ in front of $y$, then I believe it is correct. (Nod)

And yes, we have to find an upper bound in terms of $\|(y-x)\|$. (Thinking)
Does it hold that $\|f(y)-f(x)\|=|f(y)-f(x)|$ because $f$ is in $\mathbb{R}$ ? So, we have the following: $$\|f(y)-f(x)\|=\|\nabla f((1-c)x+cy)\cdot (y-x)\|\leq \|\nabla f((1-c)x+cy)\|\cdot \|y-x\|$$

The norm of the gradient is equal to \begin{align*}\|\nabla f((1-c)x+cy)\|&=\|\nabla f((1-c)(x_1, x_2)+c(y_1, y_2))\|\\ & =\|\nabla f((1-c)x_1+cy_1, (1-c)x_2+cy_2)\|\\ & =\sqrt{\left (-\frac{(1-c)x_2+cy_2}{[(1-c)x_2+cy_2]^2+[(1-c)x_1+cy_1]^2}\right )^2+\left (\frac{(1-c)x_1+cy_1}{[(1-c)x_1+cy_1]^2+[(1-c)x_2+cy_2]^2}\right )^2} \\ & = \sqrt{\frac{[(1-c)x_2+cy_2]^2+[(1-c)x_1+cy_1]^2}{\left ([(1-c)x_2+cy_2]^2+[(1-c)x_1+cy_1]^2\right )^2}}\\ & =\sqrt{\frac{1}{[(1-c)x_2+cy_2]^2+[(1-c)x_1+cy_1]^2}} \\ & = \frac{1}{\sqrt{[(1-c)x_2+cy_2]^2+[(1-c)x_1+cy_1]^2}} \end{align*} We have that $x_2>\frac{1}{x_1}$ and $y_2>\frac{1}{y_1}$.

So \begin{align*}[(1-c)x_2+cy_2]^2+[(1-c)x_1+cy_1]^2&>\left [\frac{(1-c)}{x_1}+\frac{c}{y_1}\right ]^2+[(1-c)x_1+cy_1]^2\\ & =\left [\frac{(1-c)y_1+cx_1}{x_1y_1}\right ]^2+[(1-c)x_1+cy_1]^2\end{align*} Right? What can we do next? (Wondering)

We have to bound $[(1-c)x_2+cy_2]^2+[(1-c)x_1+cy_1]^2$ from below from $2$, right? But how? (Wondering)

mathmari said:
We have to bound $[(1-c)x_2+cy_2]^2+[(1-c)x_1+cy_1]^2$ from below from $2$, right? But how? (Wondering)

Yep. (Nod)

So we want to show that:
$$\| (1-c)x+cy) \|^2 \overset ?\ge 2$$
yes?
This is the squared length of a vector between $x$ and $y$ isn't it?

Suppose we start with $\|x\|^2 \overset ?\ge 2$.
Is it true? (Wondering)Oh, and yes, since $f$ is a scalar function, we have: $\| f(y)-f(x) \| = |f(y)-f(x)|$.

I like Serena said:
So we want to show that:
$$\| (1-c)x+cy) \|^2 \overset ?\ge 2$$
yes?
This is the squared length of a vector between $x$ and $y$ isn't it?

Yes! (Nod)

I like Serena said:
Suppose we start with $\|x\|^2 \overset ?\ge 2$.
Is it true? (Wondering)

Why do we have to suppose that? (Wondering)
I like Serena said:
Oh, and yes, since $f$ is a scalar function, we have: $\| f(y)-f(x) \| = |f(y)-f(x)|$.

Ah ok!

mathmari said:
Why do we have to suppose that? (Wondering)

We want to know if the property holds for any point on the line segment between $x$ and $y$ don't we?
So I propose to check $x$ and $y$ first, which correspond to $c=0$ respectively $c=1$.
And then prove that it holds more generally for any point on the line segment. (Thinking)

Is the line segment contained in $D$? (Wondering)

Ahh ok.

We have that $$\|x\|=\sqrt{x_1^2+x_2^2}>\sqrt{x_1^2+\frac{1}{x_1^2}}$$ How do we know that this is bigger than 2? I got stuck right now (Wondering)

mathmari said:
Ahh ok.

We have that $$\|x\|=\sqrt{x_1^2+x_2^2}>\sqrt{x_1^2+\frac{1}{x_1^2}}$$ How do we know that this is bigger than 2? I got stuck right now (Wondering)

What happens for $x_1=1$, $x_1 \to 0$, and $x_1 \to \infty$?
There is some symmetry to it, isn't there? (Wondering)

Alternatively, can we find the extrema with $x_1>0$? Wondering)

I like Serena said:
What happens for $x_1=1$, $x_1 \to 0$, and $x_1 \to \infty$?
There is some symmetry to it, isn't there? (Wondering)

For $x_1=1$ we have $\sqrt{1+1}=\sqrt{2}$. For $x_1 \to 0$ and for $x_1 \to \infty$ we have that $\sqrt{x_1^2+\frac{1}{x_1^2}}\to \infty$, right? (Wondering)

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I like Serena said:
Alternatively, can we find the extrema with $x_1>0$? Wondering)

\begin{align*}[(1-c)x_2+cy_2]^2+[(1-c)x_1+cy_1]^2&>\left [(1-c)\cdot \frac{1}{x_1}+c\cdot \frac{1}{y_1}\right ]^2+[(1-c)x_1+cy_1]^2\\ & >\left (\frac{1-c}{x_1}\right )^2+\left (\frac{c}{y_1}\right )^2+\left ((1-c)x_1\right )^2+\left (cy_1\right )^2 \\ & = \frac{(1-c)^2}{x_1^2}+\frac{c^2}{y_1^2}+(1-c)^2x_1^2+c^2y_1^2 \\ & =(1-c)^2\cdot \left (\frac{1}{x_1^2}+x_1^2\right )+c^2\cdot \left(\frac{1}{y_1^2}+y_1^2\right )\end{align*}

Let $g(x)=\frac{1}{x^2}+x^2$. We have that \begin{align*}&g'(x)=-\frac{2}{x^3}+2x=\frac{2(-1+x^4)}{x^3}\Rightarrow g'(x)=0\Rightarrow x=\pm 1\overset{x>0}{\Longrightarrow } x=1\\ &g''(x)=\frac{6}{x^4}+2>0 \Rightarrow g''(1)>0\end{align*} This means that the function $g$ has a minimum at $x=1$. So, we have \begin{equation*}g(x)>g(1)\Rightarrow \frac{1}{x^2}+x^2>\frac{1}{1^2}+1^2=1+1=2\end{equation*}

Is this correct? (Wondering)

mathmari said:
For $x_1=1$ we have $\sqrt{1+1}=\sqrt{2}$. For $x_1 \to 0$ and for $x_1 \to \infty$ we have that $\sqrt{x_1^2+\frac{1}{x_1^2}}\to \infty$, right? (Wondering)
mathmari said:
Let $g(x)=\frac{1}{x^2}+x^2$. We have that \begin{align*}&g'(x)=-\frac{2}{x^3}+2x=\frac{2(-1+x^4)}{x^3}\Rightarrow g'(x)=0\Rightarrow x=\pm 1\overset{x>0}{\Longrightarrow } x=1\\ &g''(x)=\frac{6}{x^4}+2>0 \Rightarrow g''(1)>0\end{align*} This means that the function $g$ has a minimum at $x=1$. So, we have \begin{equation*}g(x)>g(1)\Rightarrow \frac{1}{x^2}+x^2>\frac{1}{1^2}+1^2=1+1=2\end{equation*}

Is this correct? (Wondering)
Yep. (Nod)

I like Serena said:
Yep. (Nod)

But then we have \begin{align*}[(1-c)x_2+cy_2]^2+[(1-c)x_1+cy_1]^2& >(1-c)^2\cdot \left (\frac{1}{x_1^2}+x_1^2\right )+c^2\cdot \left(\frac{1}{y_1^2}+y_1^2\right ) \\ & > (1-c)^2\cdot 2+c^2\cdot 2 \end{align*}
If the (1-c) and c would not be squared, the result would be equal to 2. What do we have to do in this case? (Wondering)

mathmari said:
But then we have \begin{align*}[(1-c)x_2+cy_2]^2+[(1-c)x_1+cy_1]^2& >(1-c)^2\cdot \left (\frac{1}{x_1^2}+x_1^2\right )+c^2\cdot \left(\frac{1}{y_1^2}+y_1^2\right ) \\ & > (1-c)^2\cdot 2+c^2\cdot 2 \end{align*}
If the (1-c) and c would not be squared, the result would be equal to 2. What do we have to do in this case? (Wondering)

How about we look at it slightly differently.
The points $(1-c)x+cy$ with $0<c<1$ form the line segment between $x$ and $y$.
We want to prove that any such point $(1-c)x+cy$ on the line segment has $\|(1-c)x+cy\|^2 > 2$, don't we?
And we already know that this is the case for any point $u=(u_1,u_2)$ in the area defined by $u_2>\frac 1{u_1}$ don't we?
Doesn't it suffice then if we can tell if the whole line segment is in that area? (Wondering)

For the record, you are effectively trying to prove that the line segment is in a convex area.
We could, but we already have a theorem for that, which is not entirely trivial to prove.
Instead I suggest to observe that the area defined by $x_2>\frac 1{x_1}$ with $x_1>0$ is a convex area. (Nerd)
(Can you tell why that is? (Wondering))
Consequently the line segment between $x$ and $y$ is fully contained in the area.
Therefore every point on the line segment has the the desired property.

I like Serena said:
For the record, you are effectively trying to prove that the line segment is in a convex area.
We could, but we already have a theorem for that, which is not entirely trivial to prove.
Instead I suggest to observe that the area defined by $x_2>\frac 1{x_1}$ with $x_1>0$ is a convex area. (Nerd)
(Can you tell why that is? (Wondering))
Consequently the line segment between $x$ and $y$ is fully contained in the area.
Therefore every point on the line segment has the the desired property.

This is not relevant to the ongoing discussion. Apologies for my intrusion.

But I have to ask. Is it just me or are you deliberately copying mathmari's style of placing emojis? (Wondering)

caffeinemachine said:
This is not relevant to the ongoing discussion. Apologies for my intrusion.

But I have to ask. Is it just me or are you deliberately copying mathmari's style of placing emojis? (Wondering)

In 2011, before MHB existed, I started posting smiley rich content.
I felt that was more welcoming and motivating than dry mathematical responses.
At the time I was inspired by Tiny TimTM, who had been voted Best Homework Helper.
As I got some positive feedback, I continued the practice and refined it.

Then, in 2013, mathmari and evinda joined MHB.
To be honest, I do not recall who started using smileys. We probably both did.
Now, after so many years, I believe we mutually developed this mode of communication that we are both comfortable with.
Admittedly I generally post less smileys in threads of other members nowadays, since somehow it doesn't feel quite so appropriate any more.

On a different note caffeinemachine, you have grown to quite an impressive mathematical caliber!
Can we perhaps see some more activity from you please?
I'm also curious what you are up to nowadays and I think other members may be interested as well.
After all, you have a long history on MHB.
That probably deserves a separate thread though. You can be sure of a positive response from me (not sure yet if I should add an emoji)!

I like Serena said:
In 2011, before MHB existed, I started posting smiley rich content.
I felt that was more welcoming and motivating than dry mathematical responses.
At the time I was inspired by Tiny TimTM, who had been voted to be the Best Homework Helper.
As I got some positive feedback, I continued the practice and refined it.

Then, in 2013, mathmari and evinda joined MHB.
To be honest, I do not recall who started using smileys. We probably both did.
Now, after so many years, I believe we mutually developed this mode of communication that we are both comfortable with.
Admittedly I generally post less smileys in threads of other members nowadays, since somehow it doesn't feel quite so appropriate any more.

On a different note caffeinemachine, you have grown to quite an impressive mathematical caliber!
Can we perhaps see some more activity from you please?
I'm also curious what you are up to nowadays and I think other members may be interested as well.
After all, you have a long history on MHB.
That probably deserves a separate thread though. You can be sure of a positive response from me (not sure yet if I should add an emoji)!

Thank you for the kind words about my mathematical calibre! I will try to be more active. I have started a thread in the introductions forum. Now I should get out of here before mathmari kills me.

caffeinemachine said:
Thank you for the kind words about my mathematical calibre! I will try to be more active. I have started a thread in the introductions forum. Now I should get out of here before mathmari kills me.

Thanks!
Perhaps mathmari will be a little more forgiving if you could provide some insight in her thread https://mathhelpboards.com/analysis-50/show-f-form-24346.html in Analysis.
I am stuck myself in that thread, but perhaps you wouldn't be? (Wondering)

I like Serena said:
Thanks!
Perhaps mathmari will be a little more forgiving if you could provide some insight in her thread https://mathhelpboards.com/analysis-50/show-f-form-24346.html in Analysis.
I am stuck myself in that thread, but perhaps you wouldn't be? (Wondering)
I am afraid it's not clear to me what is being asked there. Could you please explain the question?

caffeinemachine said:
I am afraid it's not clear to me what is being asked there. Could you please explain the question?

The question is:
mathmari said:
Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be twice continuously differentiable and homogeneous of degree $2$.

To show that the function has its possible local extremas at its roots, do we have show that the first derivative, i.e. the gradient is equal to $0$ if the function is equal to $0$ ?

Also how can we show that $f$ is in the form $f(x)=\frac{1}{2}x^T\cdot H_f(0)\cdot x$, where $H_f(0)$ is the Hessian Matrix of $f$ at $0$ ? Could you give me a hint?

Homogeneous of degree $2$ means that $f(t\mathbf x)=t^2f(\mathbf x)$ for any $t$ and any $\mathbf x$.
So how can we find that $Df(\mathbf x)=\mathbf 0$ implies that $f(\mathbf x)=0$?
And how can we find that $f(x)=\frac{1}{2}x^T\cdot H_f(0)\cdot x$?
Or put differently, that $f(\mathbf x)=\frac 12 D^2f(0)\mathbf x^2$?

I could find the latter by assuming that $f$ is infinitely differentiable. However, that is not given.
And I still don't know how to prove the first statement.

I like Serena said:
The question isomogeneous of degree $2$ means that $f(t\mathbf x)=t^2f(\mathbf x)$ for any $t$ and any $\mathbf x$.
So how can we find that $Df(\mathbf x)=\mathbf 0$ implies that $f(\mathbf x)=0$?
And how can we find that $f(x)=\frac{1}{2}x^T\cdot H_f(0)\cdot x$?
Or put differently, that $f(\mathbf x)=\frac 12 D^2f(0)\mathbf x^2$?

I could find the latter by assuming that $f$ is infinitely differentiable. However, that is not given.
And I still don't know how to prove the first statement.
Here's a comment on the first one. I think I must be missing something.

We have $f(t\mathbf x)=t^2f(\mathbf x)$ for all $t$ and all $\mathbf x$. Let $\mathbf a\in \mathbf R^n$ be such that $Df_{\mathbf a}=0$. We want to show that $f(\mathbf a)=0$. Am I right?

So we have the equation $f(t\mathbf a)=t^2f(\mathbf a)$ for all $t$. Differentiating both sides at $t=1$ we have
$$Df_{\mathbf a}(\mathbf a) = 2f(\mathbf a)$$
But LHS is $0$. Doesn't that do it?

caffeinemachine said:
Here's a comment on the first one. I think I must be missing something.

We have $f(t\mathbf x)=t^2f(\mathbf x)$ for all $t$ and all $\mathbf x$. Let $\mathbf a\in \mathbf R^n$ be such that $Df_{\mathbf a}=0$. We want to show that $f(\mathbf a)=0$. Am I right?

So we have the equation $f(t\mathbf a)=t^2f(\mathbf a)$ for all $t$. Differentiating both sides at $t=1$ we have
$$Df_{\mathbf a}(\mathbf a) = 2f(\mathbf a)$$
But LHS is $0$. Doesn't that do it?

Yep. I believe that does the job. Thanks!

## 1. What is the mean value theorem?

The mean value theorem is a fundamental theorem in calculus that states that if a function is continuous on a closed interval and differentiable on the open interval, then there exists at least one point within that interval where the slope of the tangent line is equal to the average slope between the endpoints of the interval.

## 2. How is the mean value theorem used to show inequality?

The mean value theorem can be used to show inequality by comparing the slope of the tangent line at the point where it is equal to the average slope, to the slope of the function at other points within the interval. If the slope at the point where it is equal to the average slope is greater than the slope at other points, then the function is increasing and the inequality is satisfied. If the slope at the point is less than the slope at other points, then the function is decreasing and the inequality is not satisfied.

## 3. What type of functions can the mean value theorem be applied to?

The mean value theorem can be applied to any continuous function on a closed interval that is differentiable on the open interval. This includes polynomial, exponential, logarithmic, and trigonometric functions, among others.

## 4. Can the mean value theorem be used to prove all inequalities?

No, the mean value theorem can only be used to show inequalities for functions that satisfy the conditions of continuity and differentiability on the specified interval. It cannot be used to prove inequalities for functions that do not meet these criteria.

## 5. What is the significance of using the mean value theorem to show inequality?

Using the mean value theorem to show inequality allows for a more rigorous and precise mathematical proof compared to other methods. It also helps to illustrate the relationship between the slope of a function and its behavior on a specific interval.

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