MHB No Max for Functions: Examples & Explanations

[evinda]

Hello! (Wave)

If $f: \mathbb{R}^n \to \mathbb{R} (n \in \mathbb{N})$ and $D \subset \mathbb{R}^n$ then it doesn't generally hold that there exists a $\max_{x \in D} f(x)$, i.e. that there is a $\overline{x} \in D$ such that $f(\overline{x}) \geq f(x)$ for each $x \in D$.

For example if $n=1$ and $f(x)=x$, $D=(0,1)$ then $f$ doesn't have a maximum in $(0,1)$. Give examples for $n \geq 2$.

Hint: $f(x_1,x_2)=x_1+x_2$ and find a space $D \subset \mathbb{R}^2$ so that $f$ has't maximum in $D$.How can we find a $D$ such that $f(x_1,x_2)=x_1+x_2$ doesn't have a maximum in $D$ ?Would it be right as follows? $$$$

We consider $D=((1,1), (2,2))$.

If the maximum is in $((1,1),(2,2))$, it holds that $\max_{x \in D} f(x)\subset (2,2)$, say it is $(2- \epsilon_1, 2-\epsilon_2)$. From the density of rationals, we have that there are $q_1, q_2 \in \mathbb{Q}$ such that $2- \epsilon_1< q_1< 2$ and $2- \epsilon_2< q_2< 2 $.

But then $(q_1, q_2 ) \in ((1,1), (2,2))$ and $ (2- \epsilon_1, 2-\epsilon_2) \subset (q_1, q_2)$. Thus $(2- \epsilon_1, 2-\epsilon_2)$ cannot be the maximum.

$$$$

In general, can we find an open and bounded interval $D$ such that a function $f$ has a maximum in it?

 

[I like Serena]

Hello! (Wave)

If $f: \mathbb{R}^n \to \mathbb{R} (n \in \mathbb{N})$ and $D \subset \mathbb{R}^n$ then it doesn't generally hold that there exists a $\max_{x \in D} f(x)$, i.e. that there is a $\overline{x} \in D$ such that $f(\overline{x}) \geq f(x)$ for each $x \in D$.

For example if $n=1$ and $f(x)=x$, $D=(0,1)$ then $f$ doesn't have a maximum in $(0,1)$. Give examples for $n \geq 2$.

Hint: $f(x_1,x_2)=x_1+x_2$ and find a space $D \subset \mathbb{R}^2$ so that $f$ has't maximum in $D$.How can we find a $D$ such that $f(x_1,x_2)=x_1+x_2$ doesn't have a maximum in $D$ ?

Hi evinda! (Mmm)

Suppose we pick $D$ to be an open ball.
How would that work out? (Wondering)

Would it be right as follows? $$$$

We consider $D=((1,1), (2,2))$.

If the maximum is in $((1,1),(2,2))$, it holds that $\max_{x \in D} f(x)\subset (2,2)$

Wouldn't the maximum be some value in $\mathbb R$ instead of a point (or a subset)? (Wondering)

For the $D$ you suggest, we would only have 2 values for $f$.
Let's call them $f_1 = f(1,1)$ and $f_2 = f(2,2)$.
Whichever of them matches the maximum, the corresponding $\overline x$ will be in $D$. (Worried)

 

[evinda]

Suppose we pick $D$ to be an open ball.
How would that work out? (Wondering)

What open ball could we pick for example? (Thinking)

Wouldn't the maximum be some value in $\mathbb R$ instead of a point (or a subset)? (Wondering)

For the $D$ you suggest, we would only have 2 values for $f$.
Let's call them $f_1 = f(1,1)$ and $f_2 = f(2,2)$.
Whichever of them matches the maximum, the corresponding $\overline x$ will be in $D$. (Worried)

So could we compute $f_1 = f(1,1)$ and $f_2 = f(2,2)$ although the intervals of the space $D=((1,1), (2,2))$ are open?

 

[I like Serena]

What open ball could we pick for example? (Thinking)

Say the open unit ball around the origin:
$$D=\{(x,y) : x^2+y^2 < 1 \}$$

So could we compute $f_1 = f(1,1)$ and $f_2 = f(2,2)$ although the intervals of the space $D=((1,1), (2,2))$ are open?

Perhaps I misunderstood.
What did you mean by $D=((1,1), (2,2))$? (Wondering)
If it represents the straight line between the points $(1,1)$ and $(2,2)$ without its endpoints, that would also do just fine. (Nod)
Then the range of the given $f$ would be the interval $(2,4)$, so its maximum would be $4$, except that is not included in the range.

 

[evinda]

Say the open unit ball around the origin:
$$D=\{(x,y) : x^2+y^2 < 1 \}$$

How do we know that it is an open ball? (Thinking)

Perhaps I misunderstood.
What did you mean by $D=((1,1), (2,2))$? (Wondering)
If it represents the straight line between the points $(1,1)$ and $(2,2)$ without its endpoints, that would also do just fine. (Nod)
Then the range of the given $f$ would be the interval $(2,4)$, so its maximum would be $4$, except that is not included in the range.

I don't know.. I am a little confused right now. Should we find a space $D \subset \mathbb{R}^2$ of the form $D=((x_1, x_2))$ ? (Sweating)

 

[I like Serena]

How do we know that it is an open ball? (Thinking)

It's the definition of an open ball. Quoting you in your other thread:

$B_d(x, \epsilon) := \{ y \in \mathbb{R}^n: d(x,y)< \epsilon\}$ -> open ball with center $x$ and radius $\epsilon$.

To see that the ball is indeed open, we can check if we can find a mini-ball around each point in the ball that is completely inside the ball... (Thinking)

I don't know.. I am a little confused right now. Should we find a space $D \subset \mathbb{R}^2$ of the form $D=((x_1, x_2))$ ? (Sweating)

Let's start with your notation $D=((x_1, x_2))$... what does it mean? (Wondering)

 

[evinda]

It's the definition of an open ball. Quoting you in your other thread:To see that it the ball is indeed open, we can check if we can find a mini-ball around each point in the ball that is completely inside the ball... (Thinking)

How can we determine if we can find a mini-ball around each point in the ball that is completely inside the ball?

Which points $(x,y)$ do we check to see if they satisfy the relation $x^2+y^2<1$ , picking $D=\{ (x,y): x^2+y^2<1\} $? (Thinking)

Let's start with your notation $D=((x_1, x_2))$... what does it mean? (Wondering)

That the values that the function $f$ can take belong to the interval $(x_1,x_2)$ ? Or am I wrong? (Thinking)

 

[I like Serena]

How can we determine if we can find a mini-ball around each point in the ball that is completely inside the ball?

Which points $(x,y)$ do we check to see if they satisfy the relation $x^2+y^2<1$ , picking $D=\{ (x,y): x^2+y^2<1\} $? (Thinking)

Let's pick a point $(x_1,y_1)$ that is an element of the ball $D$.
That is, it satisfies $x_1^2+y_1^2<1$.
Now the question is, can we find an $\epsilon > 0$ so small that the disk around the point $(x_1,y_1)$ with radius $\epsilon$ fits completely inside $D$?
Which $\epsilon$ could be pick to achieve that? (Wondering)

That the values that the function $f$ can take belong to the interval $(x_1,x_2)$ ? Or am I wrong? (Thinking)

I don't think so. (Worried)

We're running into the ambiguity what $(x_1,x_2)$ means. Is it a point in $\mathbb R^2$ or is it an open interval from $x_1$ to $x_2$?
When this happens, we often need to deduce from the context what was intended.
In this case $D$ is a subset of $\mathbb R^2$, so I'd interpret $(x_1,x_2)$ as a point unless it is specifically mentioned that it is an interval. Moreover, from your opening post we can also deduce that $(x_1,x_2)$ must be a point.
Ideally, we always state explicitly what is intended. (Nerd)

Either way, the extra set of parentheses in $D = ((x_1,x_2))$ has no known meaning to me.
I can only guess what it might mean. (Doh)

 

[evinda]

Let's pick a point $(x_1,y_1)$ that is an element of the ball $D$.
That is, it satisfies $x_1^2+y_1^2<1$.
Now the question is, can we find an $\epsilon > 0$ so small that the disk around the point $(x_1,y_1)$ with radius $\epsilon$ fits completely inside $D$?
Which $\epsilon$ could be pick to achieve that? (Wondering)

So do we have to check if there is an $\epsilon>0$ such that $(x_1- \epsilon)^2+(y_1-\epsilon)^2<1$?

So if we pick $D=\{ (x,y): x^2+y^2<1 \}$ in order to show that $f(x_1,x_2)=x_1+x_2$ doesn't have a maximum in $D$ how can we proceed?
Can we assume that $f$ has a maximum in $D$ at the point $(\overline{x_1}, \overline{x_2})$?
If so, then $\overline{x_1}+\overline{x_2}$ is the greatest value for which $\overline{x_1}+\overline{x_2}<1$.
How could we continue?

 

[I like Serena]

So do we have to check if there is an $\epsilon>0$ such that $(x_1- \epsilon)^2+(y_1-\epsilon)^2<1$?

Yup. (Thinking)

So if we pick $D=\{ (x,y): x^2+y^2<1 \}$ in order to show that $f(x_1,x_2)=x_1+x_2$ doesn't have a maximum in $D$ how can we proceed?
Can we assume that $f$ has a maximum in $D$ at the point $(\overline{x_1}, \overline{x_2})$?
If so, then $\overline{x_1}+\overline{x_2}$ is the greatest value for which $\overline{x_1}+\overline{x_2}<1$.
How could we continue?

The maximum on the closure of $D$ is at $P = (\frac 12 \sqrt 2, \frac 12 \sqrt 2)$ and has the value $\sqrt 2$.
However, that point is on the boundary and does not belong to $D$ itself.
Any point in $D$ will always have a value that is slightly less, and whichever point we pick, we can always pick a point closer to $P$, which has a higher value.
In other words, $D$ does not have a maximum.
It does have a supremum, which is $\sqrt 2$.

 

[evinda]

Yup. (Thinking)

Could we pick for example $\epsilon=x_1$ ? (Thinking)

The maximum on the closure of $D$ is at $P = (\frac 12 \sqrt 2, \frac 12 \sqrt 2)$ and has the value $\sqrt 2$.

How do we deduce this?

Any point in $D$ will always have a value that is slightly less, and whichever point we pick, we can always pick a point closer to $P$, which has a higher value.

Could you explain it further to me?

 

[I like Serena]

Could we pick for example $\epsilon=x_1$ ? (Thinking)

Not generally.
Suppose $(x_1,y_1) = (0.9, 0)$, then if we pick $\epsilon=0.9$ the resulting disk won't fit inside $D$. (Worried)

How do we deduce this?

Could you explain it further to me?

How would you find the maximum of a regular function, say $f: x \mapsto x^2$, on the interval $[-1,2]$? (Wondering)

 

[evinda]

Not generally.
Suppose $(x_1,y_1) = (0.9, 0)$, then if we pick $\epsilon=0.9$ the resulting disk won't fit inside $D$. (Worried)

So can't we find an $\epsilon$ for the general case? (Thinking)

How would you find the maximum of a regular function, say $f: x \mapsto x^2$, on the interval $[-1,2]$? (Wondering)

We find that $f$ is decreasing for $x<0$ and increasing for $x>0$, so it achieves its maximum for $x=2$.

 

[I like Serena]

So can't we find an $\epsilon$ for the general case? (Thinking)

What is the distance of $(x_1, y_1)$ to the edge of $D$? (Wondering)
Perhaps we can use that as $\epsilon$.

We find that $f$ is decreasing for $x<0$ and increasing for $x>0$, so it achieves its maximum for $x=2$.

Yep!
And what is the maximum if the interval is $(-1,2)$? (Wondering)

 

[evinda]

What is the distance of $(x_1, y_1)$ to the edge of $D$? (Wondering)
Perhaps we can use that as $\epsilon$.

With edge of $D$ do you mean the circle with center $(0,0)$ and radius $1$?

Yep!
And what is the maximum if the interval is $(-1,2)$? (Wondering)

There is no maximum, only a supremum at the point $x=2$.
So does it hold that any function does not have a maximum on an open and bounded interval, but just a supremum? (Thinking)

 

[I like Serena]

With edge of $D$ do you mean the circle with center $(0,0)$ and radius $1$?

Yes. (Sweating)

There is no maximum, only a supremum at the point $x=2$.
So does it hold that any function does not have a maximum on an open and bounded interval, but just a supremum? (Thinking)

How about $f: x \mapsto 2-x^2$. Does it have a maximum on $(-1,2)$? (Wondering)

 

[evinda]

Yes. (Sweating)

How can we find the distance? (Sweating)

How about $f: x \mapsto 2-x^2$. Does it have a maximum on $(-1,2)$? (Wondering)

No, $f$ has just a supremum at the point $x=-1$.

 

[I like Serena]

How can we find the distance? (Sweating)

The distance of $(x_1,y_1)$ to the origin is $\sqrt{x_1^2+y_1^2}$.
The distance of the origin to any point on the boundary of $D$ is $1$. (It's a unit disk.)
Therefore the distance of $(x_1,y_1)$ to the boundary is $1-\sqrt{x_1^2+y_1^2}$.
If we pick that as $\epsilon$ we're good to go.

No, $f$ has just a supremum at the point $x=-1$.

Let's see... (Thinking)

$f(-1)=2-(-1)^2=1$, but $f(0)=2-0^2=2$. Isn't that bigger? (Wondering)

 

[evinda]

The distance of $(x_1,y_1)$ to the origin is $\sqrt{x_1^2+y_1^2}$.
The distance of the origin to any point on the boundary of $D$ is $1$. (It's a unit disk.)
Therefore the distance of $(x_1,y_1)$ to the boundary is $1-\sqrt{x_1^2+y_1^2}$.
If we pick that as $\epsilon$ we're good to go.

I haven't really understood why the distance of $(x_1,y_1)$ to the boundary is $1-\sqrt{x_1^2+y_1^2}$. (Sweating)
Could you explain it further to me?

Let's see... (Thinking)

$f(-1)=2-(-1)^2=1$, but $f(0)=2-0^2=2$. Isn't that bigger? (Wondering)

Yes, so $f$ achieves its maximum at the point $x=0$.

 

[I like Serena]

I haven't really understood why the distance of $(x_1,y_1)$ to the boundary is $1-\sqrt{x_1^2+y_1^2}$. (Sweating)
Could you explain it further to me?

Let's consider the real line and suppose we have some $0\le x<1$.
The distance from $0$ to $x$ is $|x|$, while the distance from $x$ to $1$ is $1-|x|$. Yes? (Thinking)

The same holds in any number of dimensions: the distance from $\mathbf 0$ to $\mathbf x$ is $||\mathbf x||$, while the distance from $\mathbf x$ to the boundary of the unit ball is $1-||\mathbf x||$. (Thinking)

Yes, so $f$ achieves its maximum at the point $x=0$.

Aha! So not all functions have their maximum or supremum on the boundary! (Smirk)

 

[evinda]

Let's consider the real line and suppose we have some $0\le x<1$.
The distance from $0$ to $x$ is $|x|$, while the distance from $x$ to $1$ is $1-|x|$. Yes? (Thinking)

The same holds in any number of dimensions: the distance from $\mathbf 0$ to $\mathbf x$ is $||\mathbf x||$, while the distance from $\mathbf x$ to the boundary of the unit ball is $1-||\mathbf x||$. (Thinking)

A ok. And in our case do we take the euclidean norm? (Thinking)

Aha! So not all functions have their maximum or supremum on the boundary! (Smirk)

But how do we deduce in our case when we have the function $f(x_1,x_2)=x_1+x_2$ and $D=\{ (x,y): x^2+y^2<1\}$ that the maximum is on the boundary? Because of the fact that the fuction is increasing? (Thinking)

 

[I like Serena]

A ok. And in our case do we take the euclidean norm? (Thinking)

Yes.

But how do we deduce in our case when we have the function $f(x_1,x_2)=x_1+x_2$ and $D=\{ (x,y): x^2+y^2<1\}$ that the maximum is on the boundary? Because of the fact that the fuction is increasing? (Thinking)

It's a linear function.
Linear functions always have their maximum at the boundary. (Nerd)

To find where, we can go 2 ways:
1. Parametrize the boundary, substitute, and find the maximum.
2. Use symmetry considerations to find the maximum.

For approach 1 that means substituting $x=\cos\phi, y=\sin\phi$ in $f(x,y)$, take the derivative with respect to $\phi$, set it to zero, and solve. (Thinking)

For approach 2 that means that we consider that f(x,y) is symmetric in x and y. (Thinking)

 

[evinda]

Yes.
It's a linear function.
Linear functions always have their maximum at the boundary. (Nerd)

To find where, we can go 2 ways:
1. Parametrize the boundary, substitute, and find the maximum.

For approach 1 that means substituting $x=\cos\phi, y=\sin\phi$ in $f(x,y)$, take the derivative with respect to $\phi$, set it to zero, and solve. (Thinking)

So is it like that?

$$f(x,y)=f(\phi)= \cos \phi + \sin \phi$$

$$f'(\phi)=-\sin \phi+ \cos \phi$$

$$f'(\phi)=0 \Rightarrow \sin \phi= \cos \phi \Rightarrow \phi=\frac{\pi}{4} \text{ or } \phi=-\frac{3 \pi }{4}$$
So, $x=y= \frac{\sqrt{2}}{2}$.

Is it a general methodoly to find the maximum of a linear function?

 

[I like Serena]

So is it like that?

$$f(x,y)=f(\phi)= \cos \phi + \sin \phi$$

$$f'(\phi)=-\sin \phi+ \cos \phi$$

$$f'(\phi)=0 \Rightarrow \sin \phi= \cos \phi \Rightarrow \phi=\frac{\pi}{4} \text{ or } \phi=-\frac{3 \pi }{4}$$
So, $x=y= \frac{\sqrt{2}}{2}$.

Is it a general methodoly to find the maximum of a linear function?

Good! (Smile)

It's a general methodology to find the maximum of any function on a boundary.

For a linear function we can make it easier for ourselves if we pick a boundary that consists of line segments, such as the unit square. Then the maximum must be on a corner. (Mmm)