L-normed vector space

1. Feb 8, 2006

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

2. Feb 8, 2006

benorin

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

3. Feb 8, 2006

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

4. Feb 8, 2006

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

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

5. Feb 8, 2006

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

6. Feb 8, 2006

benorin

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?

7. Feb 8, 2006

Mathman23

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

Last edited: Feb 8, 2006
8. Feb 8, 2006

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

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

9. Feb 8, 2006

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

Last edited: Feb 8, 2006
10. Feb 8, 2006

benorin

Swap y1 and y2, see what you get.

11. Feb 8, 2006

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

12. Feb 8, 2006

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

13. Feb 8, 2006

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

14. Feb 8, 2006

benorin

What is the definition of a uniformly continuous function?

15. Feb 8, 2006

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

16. Feb 8, 2006

benorin

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.

17. Feb 8, 2006

Mathman23

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

Last edited: Feb 8, 2006
18. Feb 8, 2006

benorin

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

Last edited: Feb 8, 2006
19. Feb 8, 2006

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

20. Feb 8, 2006

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) ?