Help w/ 'given function f which maps an open interval S...'

[SophieToft]

This is a bit embarassing but I just became doubtful about math question.

Given a function f which maps an open interval S to R^n

\(\displaystyle f:S \rightarrow \mathbb{R}^n\)

The meaning of this, isn't that that the values of S becomes a subset of \(\displaystyle \mathbb{R}^n\)???

Sorry for making such a stupid question, but I just had to be sure?

 

[stapel]

No; this means that the function f does something to the elements of S so that the results are elements of Rⁿ.

Eliz.

 

[SophieToft]

No; this means that the function f does something to the elements of S so that the results are elements of Rⁿ.

Eliz.

Hi Eliz,

f is continious function.

Does that make a diference?

my problem is working on normalized vector spaces.

S an an open interval and \(\displaystyle f: S -> \mathbb{R}^n\) is a continuous function.

\(\displaystyle || \cdot ||\) be norm on \(\displaystyle \mathbb{R}^n\). show

1) There exist a K > 0 such that \(\displaystyle ||y|| \leq K||y||_1 ; \ y \in \mathbb{R}^n, ||y||_1 = \sum_{i=1} ^n |y_i|\).

My Solution:

According to the definition the norm of a vector y in R^n is the non-negative scalar \(\displaystyle ||y|| = \sqrt{y_1^{2} + y_2^{2} + \cdots y_n^2}\)

The L1-norm can be written as \(\displaystyle ||y||_1 = |y_1| + |y_2| + \cdots + |y_n|\)

Expanding the inequality:

\(\displaystyle \sqrt{y_1^{2} + y_2^{2} + \cdots y_n^2} \leq K|y_1| + K|y_2| + \cdots + K|y_n|\)

Is it then concludable from this that since K>0, then the right side of the inequality sign will always be larger than the left side?

Does this sound right? Or am I missing something?

Sincerely Yours and God bless

Sophie Toft

 

[pka]

All of that is hopelessly confusing to me.
Can you simply tell us what is given and what you are to prove?
Also, what in the world does the continuous function f have to do with it?

 

[SophieToft]

Hello my friends,

Sorry for the confusing statement:

I'm trying to prove the following:

Let S an an open interval and \(\displaystyle f: S -> \mathbb{R}^n\) is a continuous function.

\(\displaystyle || \cdot ||\) be norm on \(\displaystyle \mathbb{R}^n\). show

a) There exist a K > 0 such that \(\displaystyle ||y|| \leq K||y||_1 ; \ y \in \mathbb{R}^n, ||y||_1 = \sum_{i=1} ^n |y_i|\).


Sincerley and God bless

Sophie Toft

All of that is hopelessly confusing to me.
Can you simply tell us what is given and what you are to prove?
Also, what in the world does the continuous function f have to do with it?

 

[pka]

Are we to assume that f is a surjection (i.e. onto)?
If not, and I suspect not, what again does f have to do with it?

 

[SophieToft]

Yes we are suppose to assume that

Sincerely Yours and God bless

Sophie T

Are we to assume that f is a surjection (i.e. onto)?
If not, and I suspect not, what again does f have to do with it?

 

[pka]

Sorry to say, but I don’t think I can help you.
You still have explained what f has to do with this problem
I think that it has nothing to do with it.

There is a well-known theorem that states: “On any finite dimensional vector space all norms are equivalent.” Now what you have been told to prove is one-half of the definition of ‘equivalent norms’. So your question may be a part of a larger theorem. Because we have no way of knowing the order of your text material, anymore than that would be just a guess. But, if this is about equivalent norms a continuous function has nothing to do with the proof. Maybe you can lookup equivalent norms and get some guidance from your text.

 

[stapel]

f is continious function. Does that make a diference?

No.

For instance, let f0,1) -> R² be defined as \(\displaystyle f(x)\, =\, \left(\frac{1}{x},\, \frac{1}{x\, +\, 1}\right)\). This function maps S = (0, 1) (an open interval in R) into R², but that does not make S somehow a subset of R².

Eliz.

 

[SophieToft]

Its because there is a second question sorry.

b) Show that the mapping \(\displaystyle S \ni u \rightarrow ||f(u)|| \in \mathbb{R}\)

This means since f takes elements from S (does something to them). where u is an element of S, then the values normalized function f belongs to the set of all real numbers ???

Sincerely Yours
Sophie Toft

 

[pka]

Now, I see come real confusion on your part.
If for any \(\displaystyle f:S \mapsto R^n ,\), continuous or otherwise, \(\displaystyle \left\| {f(u)} \right\| \in R.\)

By the very definition of norm we have \(\displaystyle \left\| {f(u)} \right\|\) is a real number!

 

[SophieToft]

The text formulation is as follows:

Let S an an open interval and \(\displaystyle f: S -> \mathbb{R}^n\) is a continuous function.

Let \(\displaystyle || \cdot ||\) be norm on \(\displaystyle \mathbb{R}^n\).

Show the following

1) There exist a K > 0 such that \(\displaystyle ||y|| \leq K||y||_1 ; \ y \in \mathbb{R}^n, ||y||_1 = \sum_{i=1} ^n |y_i|\).

2) Show that the mapping \(\displaystyle S \ni u \rightarrow ||f(u)|| \in \mathbb{R}\) is continous.

Sincerley Sophie

 

[pka]

Well, why did you not tell us to begin with?
Lets prove \(\displaystyle x\left\| {f(x)} \right\|\) is continuous at u.

Using the continuity of f, there exists \(\displaystyle d > 0\) such that if \(\displaystyle \left\| {u - v} \right\| < d\) then \(\displaystyle \left| {f(v)} \right| \le 1 + \left| {f(u)} \right|.\)

If \(\displaystyle c > 0\) then let \(\displaystyle e = \min \left\{ {\frac{c}{{2\left( {1 + \left| {f(u)} \right|} \right)}},d,\frac{c}{{\left\| u \right\| + 2}}} \right\}.\)


If \(\displaystyle \left\| {u - v} \right\| < e\) then
\(\displaystyle \begin{array}{rcl}
\left\| {uf(u) - vf(v)} \right\| & \le & \left\| {uf(u) - uf(v)} \right\| + \left\| {uf(v) - vf(v)} \right\| \\
& \le & \left\| u \right\|\left| {f(u) - f(v)} \right| + \left\| {u - v} \right\|\left| {f(v)} \right| \\
& < & \frac{c}{2} + \frac{c}{2} = c \\
\end{array}.\)