Proving Triangle Inequality for L-Normed Vector Space

[Mathman23]

Hi

I'm given the following assignment which deals with to looks like an L-normed vectorspace:

Prove that,

$$|f(y_1) - f(y_2)| \leq || y_1 - y_2||$$

To prove this do I approach the above as a triangle inequality or as a cauchy-swartz inequality?

Best Regards,

Fred

 

[benorin]

What do you know about f ? How is the norm defined?

 

[Mathman23]

Hello and thank You for Your reply,

f is defined as follows:

$$f: \mathbb{R}^n \rightarrow \mathbb{R}$$ and

$$y1, y2 \in \mathbb{R}^n$$

Best Regards,

Fred

What do you know about f ? How is the norm defined?

 

[benorin]

More must be given, (or I am just that tired) since

$$|f(y_1) - f(y_2)| \leq || y_1 - y_2||$$

does not, in general (under those conditions), hold for arbitrary f:R^n->R, like, say, put $$f(y)=2 ||y||$$ with y2=0 and y1=y gives

$$|f(y) - f(0)| = 2||y|| \not\leq || y - 0||$$

 

[Mathman23]

Hello again,

According to my textbook the first step is to show that

$$f(y_2) \leq || y_1 - y_2|| + f(y_1)$$

The is a sub-problem of a problem which deals distance from a point to a set.

f is defined as the distance from a point in $$\mathbb{R}^n$$ to a subset S of $$\mathbb{R}^n$$

and finally I'm suppose to conclude that f is Uniform continuites on $$\mathbb{R}^n$$

Any idears?

Best Regards,

Fred

More must be given, (or I am just that tired) since

$$|f(y_1) - f(y_2)| \leq || y_1 - y_2||$$

does not, in general (under those conditions), hold for arbitrary f:R^n->R, like, say, put $$f(y)=2 ||y||$$ with y2=0 and y1=y gives

$$|f(y) - f(0)| = 2||y|| \not\leq || y - 0||$$

 

[benorin]

Hello again,

According to my textbook the first step is to show that

$$f(y_2) \leq || y_1 - y_2|| + f(y_1)$$

The is a sub-problem of a problem which deals distance from a point to a set.

f is defined as the distance from a point in $$\mathbb{R}^n$$ to a subset S of $$\mathbb{R}^n$$

and finally I'm suppose to conclude that f is Uniform continuites on $$\mathbb{R}^n$$

Any idears?

Best Regards,

Fred

Since f is defined as the distance from a point in $$\mathbb{R}^n$$ to a subset S of $$\mathbb{R}^n$$, we should have

$$f_S(y)=\mbox{inf}\left\{ \|x-y\| : x\in S\right\}$$ for $$y\in\mathbb{R}^n$$

I put a subscript of S on f since f depends on S, but consider S fixed and supress the subscript from here on.

$$f(y_2) \leq || y_1 - y_2|| + f(y_1)$$ is the statement that

$$\mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\}$$,

how could you prove that?

 

[Mathman23]

Since f is defined as the distance from a point in $$\mathbb{R}^n$$ to a subset S of $$\mathbb{R}^n$$, we should have

$$f_S(y)=\mbox{inf}\left\{ \|x-y\| : x\in S\right\}$$ for $$y\in\mathbb{R}^n$$

I put a subscript of S on f since f depends on S, but consider S fixed and supress the subscript from here on.

$$f(y_2) \leq || y_1 - y_2|| + f(y_1)$$ is the statement that

$$\mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\}$$,

how could you prove that?

By showing that normed distance from x to y1 and from x to y2 are equal? Cause they belong to the same subset?

Best Regards
Fred

 

[benorin]

By the triangle inequality, for all $$x,y_1,y_2\in\mathbb{R}^n$$, we have

$$\|x-y_2\| \leq \| y_1 - y_2\| +\|x-y_1\|$$,

and since $$x\in S\mbox{ and }S\subset\mathbb{R}^n$$, we may restrict x to be in S and inf over x in S to obtain

$$\mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \mbox{inf}\left\{\| y_1 - y_2\| + \|x-y_1\| : x\in S\right\} = \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\}$$

 

[Mathman23]

Hello again and thank You for Your answer,

And this shows, that

$$|f(y_1) - f(y_2)| \leq || y_1 - y_2||$$

is true?

Best Regards

Fred

By the triangle inequality, for all $$x,y_1,y_2\in\mathbb{R}^n$$, we have

$$\|x-y_2\| \leq \| y_1 - y_2\| +\|x-y_1\|$$,

and since $$x\in S\mbox{ and }S\subset\mathbb{R}^n$$, we may restrict x to be in S and inf over x in S to obtain

$$\mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \mbox{inf}\left\{\| y_1 - y_2\| + \|x-y_1\| : x\in S\right\} = \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\}$$

 

[benorin]

Swap y1 and y2, see what you get.

 

[Mathman23]

Hello again,


then

$$|f(y_1) - f(y_2)| \leq || y_1 - y_2||$$

since

$$\mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} - \| y_1 - y_2\| \leq \mbox{inf}\left\{\| y_1 - y_2\| + \|x-y_1\| : x\in S\right\}$$ ?

Best Regards

Fred

Swap y1 and y2, see what you get.

 

[benorin]

Of this

$$\mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \mbox{inf}\left\{\| y_1 - y_2\| + \|x-y_1\| : x\in S\right\} = \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\}$$

you need only first and last bits (including the <= sign), namely

$$\mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\}$$

in this, swap y1 and y2, to get

$$\mbox{inf}\left\{ \|x-y_1\| : x\in S\right\} \leq \| y_2 - y_1\| + \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\}$$

and since $$\| y_2 - y_1\| = \| y_1 - y_2\| ,$$ we have this pair of inequalities

$$\boxed{\begin{array}{cc} \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\} \\ \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\} \leq \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \end{array} }$$

thinking of theses as being of the form

$$A\leq k+B$$
$$B\leq k+A$$

we then have

$$A-B\leq k$$ and
$$B-A\leq k$$

thus $$|B-A| \leq k$$ i.e. $$|f(y_1) - f(y_2)| \leq || y_1 - y_2||$$

 

[Mathman23]

Hello again, and thank You for Your answers,

I need to conclude that the function f is is uniformly continius on $$\mathbb{R}^n$$.

In order to show this by proving that definition of uniformly continious functions applies to my specific f function?

Best Regards,

Fred

Of this

$$\mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \mbox{inf}\left\{\| y_1 - y_2\| + \|x-y_1\| : x\in S\right\} = \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\}$$

you need only first and last bits (including the <= sign), namely

$$\mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\}$$

in this, swap y1 and y2, to get

$$\mbox{inf}\left\{ \|x-y_1\| : x\in S\right\} \leq \| y_2 - y_1\| + \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\}$$

and since $$\| y_2 - y_1\| = \| y_1 - y_2\| ,$$ we have this pair of inequalities

$$\boxed{\begin{array}{cc} \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \leq \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\} \\ \mbox{inf}\left\{ \|x-y_1\| : x\in S\right\} \leq \| y_1 - y_2\| + \mbox{inf}\left\{ \|x-y_2\| : x\in S\right\} \end{array} }$$

thinking of theses as being of the form

$$A\leq k+B$$
$$B\leq k+A$$

we then have

$$A-B\leq k$$ and
$$B-A\leq k$$

thus $$|B-A| \leq k$$ i.e. $$|f(y_1) - f(y_2)| \leq || y_1 - y_2||$$

 

[benorin]

What is the definition of a uniformly continuous function?

 

[Mathman23]

Definition:uniformly continuous function

S is a subset of $$\mathbb{R}^n$$ and $$f: S \rightarrow \mathbb{R}$$ be a real valued function. The function f is said to be uniformly continuous, if there for every epsilon > 0, exists a delta > 0 such that,

|f(x) - f(y) | < epsilon

for all x,y \in S, which statisfies that ||x - y|| < delta

Do I then apply this definition to my given f(x) ? and use it to show how f(x) = 0, if x belongs to Y ?

Best Regards
Fred

What is the definition of a uniformly continuous function?

 

[benorin]

Do I then apply this definition to my given f(x) ? and use it to show how f(x) = 0, if x belongs to Y ?
Fred

Yes, if you wish to prove f is a uniformly continuous function do just that: the first bit is done, so just choose delta = epsilon and it is done.

And do U mean x belongs to S? Indeed it is a simple matter that f(x)=0 for every x in S, since (for fixed x) the shortest distance between x and y when y is allowed to be x is 0.

 

[Mathman23]

Yes, if you wish to prove f is a uniformly continuous function do just that: the first bit is done, so just choose delta = epsilon and it is done.

And do U mean x belongs to S? Indeed it is a simple matter that f(x)=0 for every x in S, since (for fixed x) the shortest distance between x and y when y is allowed to be x is 0.

regarding uniform continuity

then according to the definition of uniformly continious functions then
$$|x - y| = |x| - |y|$$

From here I'm a bit unsure

Do I do the following y = x + $$\delta$$ then $$x \geq 0$$

$$|x - (x+ \delta)| = \delta$$

then $$|x - (x+ \delta)| < \epsilon$$

futermore since $$\delta > 0$$ according to the definition:

$$\delta < \epsilon$$

therefore f is uniformly continious.

Am I on the right path here?

Best Regards
Fred

 

[benorin]

Definition:uniformly continuous function

S is a subset of $$\mathbb{R}^n$$ and $$f: S \rightarrow \mathbb{R}$$ be a real valued function. The function f is said to be uniformly continuous, if there for every epsilon > 0, exists a delta > 0 such that,

|f(x) - f(y) | < epsilon

for all x,y \in S, which statisfies that ||x - y|| < delta

Recall that we just proved that

$$|f(x) - f(y) | \leq \| x-y\|$$ for all $$x,y\in\mathbb{R}^n$$

we now wish to show that

$$\mbox{ For every } \epsilon >0, \mbox{ there exists a } \delta >0\mbox{ such that }\| x-y\| < \delta \Rightarrow |f(x)-f(y)| <\epsilon$$

So fix $$\epsilon >0$$. Choose $$\delta = \epsilon$$ so that

$$\| x-y\| < \delta \Rightarrow |f(x)-f(y)| \leq \| x-y\| < \delta =\epsilon$$

where we have used that bit of which I wrote "Recall that we just proved ..." to get the $$|f(x)-f(y)| \leq \| x-y\|$$ part of the above line.

--Ben

 

[Mathman23]

Thank You again,

If have two small final followup questions:

(a) If S is closed set, and if x \notin S, then f(x) > 0

(b) if Y is closed, then

X = {x \in R^n | f(x) = 0}

I know in (a) that since x is not in S, then the metric distance between x,y is larger than zero, but does that explain f(x) > 0??

Any hints on how to solve (b) ?

Best Regards

Fred

Yes, if you wish to prove f is a uniformly continuous function do just that: the first bit is done, so just choose delta = epsilon and it is done.

And do U mean x belongs to S? Indeed it is a simple matter that f(x)=0 for every x in S, since (for fixed x) the shortest distance between x and y when y is allowed to be x is 0.

 

[benorin]

"I know in (a) that since x is not in S, then the metric distance between x,y is larger than zero, but does that explain f(x) > 0??"

How do you know this is so? If had, say, S=(0,1) (the segment of the real line strictly between 0 and 1) what than is f(0) ?

 

[Mathman23]

"I know in (a) that since x is not in S, then the metric distance between x,y is larger than zero, but does that explain f(x) > 0??"

How do you know this is so? If had, say, S=(0,1) (the segment of the real line strictly between 0 and 1) what than is f(0) ?

If S = (x=1, y = 0) then f(x) > 0 ??

I'm a bit lost here, any hints?

Sincerely
Fred

 

[benorin]

This is a trick question:

How do you know this is so? If had, say, S=(0,1) (the segment of the real line strictly between 0 and 1) what then is f(0) ?

I was trying to ensure you understood why "I know in (a) that since x is not in S, then the metric distance between x,y is larger than zero..."

Understand that if y is the particular point in S whose distance to x is a minimum, then f(x)-||x-y||, that is to say the value of f(x) is that minimum distance between x and y.

A geometric example will help: We're in $$\mathbb{R}^2$$. Let S be the unit square [0,1]x[0,1]. Now fix a point x in $$\mathbb{R}^2$$, I'll pick x=(2,1/2). Now f[(2,1/2)] can be determined as follows:

first, find the point in S (the unit square) which is closest to (2,1/2) (which is x in our example), and denote this "closest point" by y. So what is y? Well clearly y is on the right-hand edge of the unit square S since were it not, we could easily find a point in S that is closer to (2,1/2). Every point on the right-hand edge of S is of the form (1,t), where $$0\leq t\leq 1$$. The distance $$d$$ between (1,t) and (2,1/2) is given by the usual distance formula

$$d=\sqrt{(2-1)^2+(1/2 -t)^2}=\sqrt{1+(1/2 -t)^2}$$

and we seek the value of t with $$0\leq t\leq 1$$ that minimizes $$d$$. We reason that the minimum occurs for t=1/2 and hence determine that y=(1,1/2).

Second, we recall that f(x) is the distance between x and y, i.e. f(x)=||x-y||, and, in this example, the distance is given by the usual formula, so we have

$$f[(2,1/2)] = ||(2,1/2)-(1,1/2)||=\sqrt{(1-2)^2+(1/2 -1/2)^2}=1$$.

Which makes sense, if you think about the diagram (you should draw one). End Geometric Example.

Notice that in the above example S was a closed set. In the so-called trick question I posed at the top of this post, S was (0,1) on the real line (which is not a closed set) and I had asked what then is f(0)? We would first determine the point y in (0,1) whose distance to 0 is a minimum, at least we would if there were such a point, but 0 is a limit point of (0,1) so there are infinitely many points in (0,1) infinitely close to 0, but there is always one closer... strictly speaking, there is no point y in (0,1) of minimum distance to 0, but there is an infimum (the inf) of the set of such distances, which is 0 (this is the value of the distance, not the point y at which it occurs), and thus f(0)=0. But notice that 0 (the point) is not in S, and yet we have f(0)=0 and not f(0)>0. Why?


Answer: S is not closed. In the example where S was the unit square in R^2, S was closed. Can you find a point z in R^2 that is not is S such that f(z)=0? Why not?

 

[Mathman23]

Hi all,

I been thinking regarding:

(a) If S is closed set, and if $$x \notin S$$, then f(x) > 0

(b) if Y is closed, then

$$X = \{x \in R^n | f(x) = 0 \}$$

I know that according to my textbook the definition of a closed set is a as follows:

A subset F of $$\mathbb{R}^n$$ is closed if its complement:

\mathbb{R}^n \mathrm{\} $$F = \{ x \in R^n | x \notin F \}$$ is open.

Do I then need to show that x is an internal point of $$\mathbb{R}^n$$


Sincerely

Fred