Modern Abstract Analysis

BAYMAX

04:07

9 hours ago, by Martin Sleziak

Continuity of norm can be considered as a special case of this for $E=\{0\}$.

9 hours ago, by Martin Sleziak

On page 36 it is mentioned that in a metric space $X$ the function $$d_E(X)=\operatorname{dist}(x,E)$$ is continuous for any non-empty set $E\subseteq X$.

I thought of this as - "Norm of x that is $||x||$, as the distance of a point $x$ in space $X$ from Origin"

so $||x|| = Inf \{ d_{{0}} (x) : x \in X\}$ is continuous on $X$

so are we basically saying this

$||.|| :X \rightarrow \Bbb{R} $ is a continuous function?

also

2)

I think in the above conversation we were trying to show that the mapping $F(x) \rightarrow ||F(x)||$ is continuous

may be can I call this a continuous transformation?

Martin Sleziak

04:53

@BAYMAX Yes, that is what we are claiming.

In fact $\|x\|=d(0,x)$.

@BAYMAX Since norm is continuous and $F$ is continuous, their composition $x\mapsto \|F(x)\|$ is also continuous.

BAYMAX

Nice one there!

is that we call a continuous transformation?

or there is some other terminology for it?

Martin Sleziak

Main site: Is a norm a continuous function? As I mentioned, you may find other posts.

@BAYMAX I guess it would be ok, but continuous function (and perhaps continuous map) seem to be used more often.

IIRC Limaye has chapter on continuous functions metric spaces.

Yes, it starts on page 33. I should have said section rather than chapter.

BAYMAX

ha it's in your language I guess :) strana means page perhaps!

yes section will also help me quick as I have the text with me! and can quickly see the preface section!

Martin Sleziak

Oh, sorry, you mean I should have used this link: books.google.com/books?id=BaEgbIoPTygC&pg=PA33 (Anyway you were posting links ending with .co.in - I guess we both simply copied where google redirected us based of language preferences.)

BAYMAX

No problem..just found a different version! it's ok.

BAYMAX

05:24

also I see you added me to the list of room owners,I am very much thankful for that!,I will try my best to serve the duties of a room owner.

@MartinSleziak

BAYMAX

08:16

.)

$X =C[0,1]$

be the space of all real valued continuous functions

on$[0,1]$ .Let $T: X \rightarrow \mathbb{R}$, be alinear functional defined by $T(f) = f(1)$.Let $X_{1} = (X,||.||_{1})$ and $X_{2} = (X,||.||_{\infty})$.Then $T$ is neither continuous on $X_{1}$ nor on $X_{2}$ ?

I think $||f||_{1} = \int_{0}^{1} f dx$ is this correct?

or it has some other definitions?

and is $||f||_{\infty} = sup(f(x),x \in [0,1])$

?

I thought of using this result that if $f$ is continuous on $X$ then there exists some $\alpha$ such that $||T(f)|| \leq \alpha ||f||$

as per here we have to show $||f(1)|| \leq \alpha ||f||$

also what is the norm associated with space $\Bbb{R}$,I think it is modulus?

Martin Sleziak

In both definitions you are missing absolute value, you should have
\begin{align*}
\|f\|_1 &= \int_0^1 |f(x)| \,dx\\
\|f\|_\infty &= \sup_{x\in[0,1]} |f(x)|
\end{align*}

Yes, on $\mathbb R$ you are using absolute value as the norm. (This is the standard norm when you consider $\mathbb R$ as a linear normed space - so you can always assume this one, unless the text specifies explicitly that in some example a different norm on $\mathbb R$ is considered.

BAYMAX

Oh,I see.Also they would violate the properties o the norm (non-negativity) if I don't use

Absolute signs

Martin Sleziak

Exactly.

You also have equivalent characterization now: A linear function $T\colon X\to\mathbb R$ is continuous if and only if there is an $\alpha$ such that $\|x\|\le1$ implies $|T(x)|\le\alpha$.

BAYMAX

yes for that to use I must be having $||x|| \leq 1$

?

or here $||f||_{1} \leq 1$

Martin Sleziak

BTW where does the problem come from? Is this really what the problem says? Since I suspect that $T$ is continuous w.r.t $\infty$-norm but not w.r.t 1-norm.

BAYMAX

08:31

Its homework, actually I don't have the answer.

How you quickly see the continuity?

Martin Sleziak

Oh, I see, the assignment says that it is continuous in neither of the cases.

Let's have a look at $\infty$ norm first, ok?

BAYMAX

yes

Martin Sleziak

Since this is the one where I think what the assignments says is not correct - so I'm curious to see whether I made a mistake.

BAYMAX

oh I guessed it! the assignment has no answers! sorry for any trouble.

Martin Sleziak

So we have $T(f)=f(1)$. And we are wondering whether we can make some estimate on this using $\|f\|_\infty=\sup\{|f(x)|; x\in[0,1]\}$.

It should be quite clear that $|f(1)|\le\sup\{|f(x)|; x\in[0,1]\}$.

Do you see why this is true? Do you see how this relates to our problem?

BAYMAX

08:36

Yes

any element of the set will be less than or equal to the supremum of that set?

Martin Sleziak

ok

BAYMAX

so it is continuous wrt $||.||_{\infty}$ norm right?

Martin Sleziak

And I guess you can also see that the above is exactly $$|T(f)| \le \|f\|_\infty.$$ Which means that $T$ is continuous.

Yes, it is continuous.

BAYMAX

yes

Martin Sleziak

Ok, so I guess this was rather easy.

We would also like to show that $T$ is not continuous w.r.t. 1-norm.

To this end it suffices to find a sequence $f_n\in C[0,1]$ such that $\|f_n\|=1$ but $|T(f_n)|\to\infty$.

Can you think of an example of a sequence of continuous functions such that $\int_0^1 |f_n(x)|\,dx=1$ but $\lim\limits_{n\to\infty} |f_n(1)|=\infty$. (You can try positive functions so that you do not have to worry about absolute values.)

And $\|f_n\|_1 \le 1$ would also work quite fine.

BAYMAX

08:43

let me think about it

Martin Sleziak

So if you prefer, replace it with $\int_0^1 |f_n(x)|\,dx \le 1$.

I guess that for examples like this, it's much easier to draw an example than to come up with a function given by an actual formula.

I think that a picture can be enough of an argument for you to see that it indeed is not continuous. But for an assignment you'll probably need to write down a more formal argument, so you might need to try write down an explicit formula for $f_n$ for that.

BAYMAX

I was thinking of triangles growing taller and taller but the area or integration value will blow up

Martin Sleziak

Yes. That would work.

The tall end will be on the right, since you want $|f_n(1)|\to\infty$.

BAYMAX

tail end is on the top (directed upwards parallel to the positive y axis)

and as $n$ increases the top end of the traingle grows in the direction parallel to the positive $y$ axis and thus tending towards infinity as $n$ increases

but in this case how $\int_{0}^{1}|f_{n}| dx \leq 1 $ as $n \rightarrow \infty$

perhaps we could squeeze the base length as $n \rightarrow \infty$ tending the area towards 0

Martin Sleziak

Yes.

BAYMAX

08:56

so at last we will be having a straight perpendicular line standing at perhaps $x = \frac{1}{2}$ and the $y$ value is $\infty$

Martin Sleziak

I think you mean $x=1$?

Your functional is defined by $T(f)=1$.

You want $|T(f_n)|\to\infty$, i.e., $|f_n(1)|\to\infty$.

BAYMAX

ohk

a perpendicular line standing at $x = 1$

how $\int_{0}^{1} f_{n} dx \leq 1$ here?

Better here I found an online drawing tool

flockdraw.com/40wizg

Martin Sleziak

It requires flash player. I'm not sure if I really want that.

Of course, you can take a screenshot and post it here.

BAYMAX

for me too it showed at first instance

perhaps if you refresh that page it can show the page

again

NV-US

hello

BAYMAX

09:03

Hi

NV-US

i hv a question, but its not about functional analysis

Martin Sleziak

@BAYMAX Sorry, I can't get it working.

BAYMAX

Its ok

Martin Sleziak

As I said, you can post a screenshot here if you draw something elsewhere.

BAYMAX

ok

Martin Sleziak

09:05

@NV-US Well, if the question is not about functional analysis, then probably this is not the right place... You have main chat room and also several discipline-specific chat rooms.

NV-US

but they are inactive

Martin Sleziak

Main chat room is far from inactive. (8k messages posted last week.)

NV-US

the last msg was 10 min ago

Martin Sleziak

You have rather strange definition of inactive :-) I see that I've been inactive for the most of my life :-)

NV-US

sorry but i hv exam tommorow, so i'm worried and all

i am having problem in a definition

can i ask? if its long, dont answer

:)

Martin Sleziak

09:08

Why not asking in a suitable room?

NV-US

no users present there (currently)

BAYMAX

sorry but I while drawing i found that the area also tends to infinity

Martin Sleziak

Well, you have to make it small enough.

We agreed that $f_n$ is going to be a triangle.

What about the triangle determined by the points $(0,0)$, $(1/2,0)$, $(0,2)$.

And then $(0,0)$, $(2/3,0)$, $(0,3)$.

In general, $(0,0)$, $(1-1/n,0)$, $(0,n)$.

BAYMAX

hmm..I have to think and draw a bit.. will take some time .sorry

Martin Sleziak

I.e., it is a triangle with height $n$ and the lenght of the base is $1/n$.

BAYMAX

09:20

if we see the area is

$\frac{1}{2} * (n-1)$

and as $n \rightarrow \infty$ this area tends to $\infty$

but we want $\int_{0}^{1}|f_{n}| dx \leq 1$

Martin Sleziak

Sorry my bad.

I mean $(0,0)$, $(1-1/n,0)$, $(1,n)$.

The area of such triangle is $1/2$.

What I wrote above was not even a function...

BAYMAX

hmm I see

any other approach we can use to show discontinuity?

here

Martin Sleziak

There are probably many possibilities, but this seemed to me to be the easist one.

BAYMAX

hmm will try this a bit more

Feeds

11:11

Martin Sleziak has made a change to the feeds posted into this room

Martin Sleziak

As an experiment I have tried to add the feed with featured (bounty) questions in functional-analysis tag to this room.

I hope I did it correctly - we'll see whether it works or not.

If we see that it is too disruptive, we can remove the feed.

BAYMAX

Ok,sure!

Feeds

0

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BAYMAX

how about the feeds not posting here but appearing at the top left side so that one can see the title of the question too!

@MartinSleziak

Martin Sleziak

Ok, so it seems that the feed at least post something in the room. We will see what happens if some of those questions is bumped or if a new question is featured.

@BAYMAX Well, I (and probably most users) will only visit the room from time to time - as opposed to being here all the time. So if they are posted in the form of ticker, they are quite likely to remain unnoticed.

BAYMAX

11:24

ok,got it!

Martin Sleziak

I don't think bounties are that frequent. (If we wanted to add another feed with many items, than the ticker is much better option.)

For example, if we want also another feed with all questions tagged functional-analysis, then adding this as chat messages would result to room drowned with post from feeds.

BAYMAX

i see

Martin Sleziak

But adding this feed as ticker would be ok I guess.

Feeds

Martin Sleziak has made a change to the feeds posted into this room

Martin Sleziak

We might as well try it - just to see how it looks.

I have added feed for all questions with (functional-analysis) tag; via ticker (as opposed to chat messages).

BAYMAX

11:28

let us see!

BAYMAX

12:59

How are you getting the area $\frac{1}{2}$ for the triangles?

@MartinSleziak

Martin Sleziak

@BAYMAX $\frac12\cdot\frac1n\cdot n$.

1/2*(base)*(height)

base is $1/n$ and height is $n$

The fuctions $f_n$ I had in mind are the triangles which are in your picture on the right.

BAYMAX

The points must be then $(1-\frac{1}{n}) , (1,0) ,(1,n)$

Martin Sleziak

$$f_n(x)=
\begin{cases}
0, & \text{for }x\le 1-\frac1n, \\
n^2x-n(n-1) & \text{for }x\ge 1-\frac1n.
\end{cases}
$$

@BAYMAX You have probably some typo there. (The first of your three "points" only has one coordinate.

When I wrote the three points I meant the points which determine the graph of the piecewise linear function. (Not the vertices fo the triangle.)

BAYMAX

Now I gotcha

nice!

thank you!

Martin Sleziak

I.e., I wrote three points A,B,C and I meant that graph consists of two linear segments A--B (here the function is zero) and B--C (here it increases).

BAYMAX

13:08

Yes

BAYMAX

14:17

*) Let $X$ be the space of real sequences having finitely many non-zero terms with $||.||_{p}$ , $1 \leq p \leq \infty$

then $f$ is continuous for which value of $p$ ?

Martin Sleziak

What is $f$?

BAYMAX

perhaps a typo in the assignment or it must be $f : X \rightarrow X$

Why i didnot notice that

><

Martin Sleziak

Still, without any information on $f$ we cannot say whether or not it is continuous.

BTW is that assignment available online?

BAYMAX

No,they are usually written in our boards and we note them down!

Martin Sleziak

I wish this does not devolve into me solving your homework for you, but to do that, probably I have to learn how to ask right questions and give reasonable hints.

BAYMAX

14:21

No

its not like that

Martin Sleziak

Isn't something more given. Like $f(x_n)=(x_1,x_2/2,x_3/3,\ldots)$ or something similar.

BAYMAX

I ask those questions which I am unable to do after I think for a bit1

Martin Sleziak

Anyway, if the information you wrote here is all you've been given about this problem, then it's not answerable.

BAYMAX

its ok,will chk about it

Like thiso ne is from some PhD entrance exam

All norms on a normed vector space $X$ are equivalent provided

1)X is relexive

2)X is complete

3)X is finite dimensional

4)X is an inner product space

and answer given is 1

Martin Sleziak

3 is certainly true.

BAYMAX

14:27

Ok,sorry but let us discuss this after I read about the equivalence of norms

I am searching for some MCQ type questions which involve mind tickling as well as in a short period of time we can tackle over a large problems

due to which the doubts can come and can be defended!

Martin Sleziak

15:22

Looking at some posts on the main, 1) does not seem to be correct choice.

If any two norms on a vector space are equivalent then the space is finite-dimensional

Understanding of the theorem that all norms are equivalent in finite dimensional vector spaces

Inequivalent norms

Oh, I maybe misunderstood the question.

So the question is basically: Let X be a reflexive linear normed space. If we take any norm on X which generates the same topology, is this norm equivalent to the original one?

Anyway, I am not really sure what is correct answer to this problem. (And not even whether this is the intended formulation.)

What I wrote above is non-sense. Two norms are equivalent iff they generate the same topology.

So I'll have to admit that the question above is rather unclear to me.

If the same question was asked for "vector space X" (not for "norme vector space X"), then the answer would be finite dimensional.

BAYMAX

15:39

it's interesting to know that norms generate topology!

Martin Sleziak

Of course, every normed space is also a metric space, so it generates a topology.

When I trie to search for "All norms on a normed vector space X are equivalent provided", Google finds some pdf files. It seems that "inner product space" is given there as the correct answer.

BAYMAX

I still don't get the relation between a norm and a metric like which generates which?

Martin Sleziak

This is quite strange - I have to admit that it's rather unclear to me what the question actually asks.

If you have a norm $\|\cdot\|$ on $X$, then $d(x,y)=\|x-y\|$ is metric on $X$.

And metric gives us a topology.

In fact, whenever we talk about convergence, open set, closed set in a linear normed space, we mean convergence/open/closed w.r.t. this topology (=w.r.t. this metric).

BAYMAX

Nice!

So $|||| , ||||^{'}$ are equivalent

on $X$ if there exists positive scalars $\alpha , \beta$

such that $\alpha||x|| \leq ||x||^{'}\leq \beta||x||$

Martin Sleziak

15:54

Yes.

BAYMAX

Since $X$ is finite dimensional,let us say dim($X$) = $k$

and let $B = \{U_{1},U_{2},...,U_{k}\}$ be a basis of $X$

After this what should I do?

ok

Let any $x \in X$

then $x = a_{1}U_{1}+a_{2}U_{2}+...+a_{k}U_{k}$

Next $||x||_{B} = max\{ |a_{1}|,|a_{2},...,|a_{k}| \}$

is a norm

on $X$

so when we consider $||x|| = ||a_{1}U_{1} + a_{2}U_{2}+ ... +a_{k}U_{k}||$

$||x|| \leq ||x||_{B}(|| U_{1}||+||U_{2}||+...+||U_{k}||)$

let $\beta = ||U_{1}|| + ||U_{2}|| + ... +||U_{k}||$

so $||x|| \leq \beta ||x||_{B}$

BAYMAX

16:26

next how do i show $||x|| \geq \alpha||x||_{B}$

Transcript for

$Mathematics$ Modern Abstract Analysis