L-normed vector space

Mathman23

Hi

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

Prove that,

[tex]|f(y_1) - f(y_2)| \leq || y_1 - y_2||[/tex]

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:

[tex]f: \mathbb{R}^n \rightarrow \mathbb{R}[/tex] and

[tex]y1, y2 \in \mathbb{R}^n[/tex]

Best Regards,

Fred

benorin

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

[tex]|f(y_1) - f(y_2)| \leq || y_1 - y_2||[/tex]

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

[tex]|f(y) - f(0)| = 2||y|| \not\leq || y - 0||[/tex]

Mathman23

Hello again,

According to my textbook the first step is to show that

[tex] f(y_2) \leq || y_1 - y_2|| + f(y_1)[/tex]

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 [tex] \mathbb{R}^n [/tex] to a subset S of [tex]\mathbb{R}^n[/tex]

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

Any idears?

Best Regards,

Fred

benorin

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

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

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

[tex] f(y_2) \leq || y_1 - y_2|| + f(y_1)[/tex] is the statement that

[tex] \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\}[/tex],

how could you prove that?

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

benorin

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

[tex]\|x-y_2\| \leq \| y_1 - y_2\| +\|x-y_1\|[/tex],

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

[tex] \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\}[/tex]

Mathman23

Hello again and thank You for Your answer,

And this shows, that

[tex]|f(y_1) - f(y_2)| \leq || y_1 - y_2||[/tex]

is true?

Best Regards

Fred

benorin

Swap y1 and y2, see what you get.

Mathman23

Hello again,


then

[tex]|f(y_1) - f(y_2)| \leq || y_1 - y_2||[/tex]

since

[tex] \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\} [/tex] ?

Best Regards

Fred

benorin

Of this

[tex] \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\}[/tex]

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

[tex] \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\}[/tex]

in this, swap y1 and y2, to get

[tex] \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\}[/tex]

and since [tex] \| y_2 - y_1\| = \| y_1 - y_2\| ,[/tex] we have this pair of inequalities

[tex]\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} } [/tex]

thinking of theses as being of the form

[tex]A\leq k+B[/tex]
[tex]B\leq k+A[/tex]

we then have

[tex]A-B\leq k[/tex] and
[tex]B-A\leq k[/tex]

thus [tex]|B-A| \leq k[/tex] i.e. [tex]|f(y_1) - f(y_2)| \leq || y_1 - y_2||[/tex]

Mathman23

Hello again, and thank You for Your answers,

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

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

Best Regards,

Fred

benorin

What is the definition of a uniformly continuous function?

Mathman23

Definition:uniformly continuous function

S is a subset of [tex]\mathbb{R}^n[/tex] and [tex]f: S \rightarrow \mathbb{R}[/tex] 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

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.

Mathman23

regarding uniform continuity

then according to the definition of uniformly continious functions then
[tex]|x - y| = |x| - |y|[/tex]

From here I'm a bit unsure

Do I do the following y = x + [tex]\delta[/tex] then [tex]x \geq 0 [/tex]

[tex]|x - (x+ \delta)| = \delta[/tex]

then [tex]|x - (x+ \delta)| < \epsilon[/tex]

futermore since [tex]\delta > 0[/tex] according to the definition:

[tex]\delta < \epsilon[/tex]

therefore f is uniformly continious.

Am I on the right path here?

Best Regards
Fred