# V Space With Norm $||*||$ - Fourier Series

• physics1000
In summary, the conversation is about a question regarding the convergence of a sequence in a space with a norm. The question asks to prove that if the sequence converges to two different vectors, then those vectors must be equal. The problem is not clear on which norm and definition to use, as the norms mentioned are just specific examples and the question asks to show it for an arbitrary norm. The conversation also includes a discussion on how to write mathematical expressions on the platform.f

#### physics1000

Homework Statement
I wrote below what is my question exactly, no need for answer, the question is not explainable well.
Relevant Equations
no equations here. only defintion, which we have not learned and I dont know what norm they want...
Hi, a question regarding something I could not really understand
The question is:
Let V be a space with Norm $||*||$
Prove if $v_n$ converges to vector $v$.
and if $v_n$ converges to vector $w$
so $v=w$
and show it by defintion.
The question is simple, the thing I dont understand, what definition? what norm exactly? we learnt only norm $L^1$ and $L^2$ and $L^\infty$ which is uniform convergence.
Please tell me what I wrote above.

EDIT: I dont know how to write latex here, at MSE its with $$, here it wont work. The definition of a norm. The norms you mentioned are just specific examples of norms. You need to show it for an arbitrary norm. vanhees71 The definition of a norm. The norms you mentioned are just specific examples of norms. You need to show it for an arbitrary norm. Well, I dont really know what is arbitary norm, we have not learned it :\I dont know what is the definition also, we learnt only what I said above. The definition of a norm. The norms you mentioned are just specific examples of norms. You need to show it for an arbitrary norm. If I knew the defintion of arbitrary norm, the question is easy, my problem is understanding what is the thing they want from me, I dont know how they expect us to solve something we have not learned... physics1000 HI thanks, I saw this already. Definitions should be with epsilon. I dont see it there. that why i skipped it. HI thanks, I saw this already. Definitions should be with epsilon. I dont see it there. that why i skipped it. When I mean definition. I mean definition of how to prove. Like for limit, there is for all \epsilon > 0 , exist n_0 such that... and such Homework Statement:: I wrote below what is my question exactly, no need for answer, the question is not explainable well. Relevant Equations:: no equations here. only defintion, which we have not learned and I dont know what norm they want... Hi, a question regarding something I could not really understand The question is: Let V be a space with Norm ||*|| Prove if v_n converges to vector v. and if v_n converges to vector w so v=w Sounds as if you should prove the uniqueness of limits according to an arbitrary norm topology. and show it by defintion. ... which is a bit strange to say. If something is true by definition, then there is nothing to show. If something has to be shown, then it can be used as a definition. I would phrase the problem as: Show that ##\;\text{ IF }\;\lim_{n \to \infty}v_n=v\;\text{ AND }\;\lim_{n \to \infty}v_n=w\;\text{ THEN }\;v=w.## Step 1: What does ##\lim_{n \to \infty}v_n=v## mean? Step 2: Use the triangle inequality to show ##\lim_{n \to \infty}\|v -w\|=0.## The question is simple, the thing I dont understand, what definition? what norm exactly? we learnt only norm L^1 and L^2 and L^\infty which is uniform convergence. Please tell me what I wrote above. EDIT: I dont know how to write latex here, at MSE its with$$, here it wont work.
Just replace $by ## for inline codes, and use $$for extra lines. Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/ vanhees71 and physics1000 Sounds as if you should prove the uniqueness of limits according to an arbitrary norm topology. ... which is a bit strange to say. If something is true by definition, then there is nothing to show. If something has to be shown, then it can be used as a definition. I would phrase the problem as: Show that ##\lim_{n \to \infty}v_n=v\text{ AND }\lim_{n \to \infty}v_n=w\text{ THEN }v=w.## Step 1: What does ##\lim_{n \to \infty}v_n=v## mean? Step 2: Use the triangle inequality to show ##\lim_{n \to \infty}\|v -w\|=0.## Just replace by ## for inline codes, and use$$ for extra lines. Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/ Hi, Thanks alot! The problem is, you told me how to answer it, but isnt there any definition? like the limit with epsilon? or definition for uniform covnergence with epsilon also? it sounds really weird for me there aint any formal definition for it. By saying definiton, I mean ( I will take for example the limit ): for all ##\epsilon > 0##, there exist a natural number ##N## such that for all ## n>N## we have ##|an-L|<\epsilon ## HI thanks, I saw this already. Definitions should be with epsilon. I dont see it there. that why i skipped it. If you saw that already why do you claim not to know what an arbitrary norm is? An arbitrary norm is any function that satisfies that definition. Your task is to show that — given the norm topology — a sequence converges to a single unique point (if it converges at all). If you saw that already why do you claim not to know what an arbitrary norm is? An arbitrary norm is any function that satisfies that definition. Your task is to show that — given the norm topology — a sequence converges to a single unique point (if it converges at all). ahh, my bad english comes here. when you say, a sequence converges to a single unique point, you mean the ##||u-v||=0##? What fresh said? Hi, Thanks alot! The problem is, you told me how to answer it, but isnt there any definition? like the limit with epsilon? or definition for uniform covnergence with epsilon also? it sounds really weird for me there aint any formal definition for it. The point is: In step 1 you have to make a definition in order to explain what convergence even means. In step 2, you show that limits are unique, which again means, that the definition in step one is well-defined. In other words: In order to make sense of step 1, you need to show step 2. The point is: In step 1 you have to make a definition in order to explain what convergence even means. In step 2, you show that limits are unique, which again means, that the definition in step one is well-defined. In other words: In order to make sense of step 1, you need to show step 2. ahh my bad english... I will try to understand it, thanks! You gave me enough information for me to try to solve it now, I will update :) Managed to solve it, problem is what is the defintion they want? I showed like that: ##||v-w|| = ||v-v_n +v_n - w|| <= ||v_n - v|| + ||v_n - w|| ##, now as limit of n goes to ##\infty## we get zero. problem is, again, the definition, what definition they want? Managed to solve it, problem is what is the defintion they want? I showed like that: ##||v-w|| = ||v-v_n +v_n - w|| <= ||v_n - v|| + ||v_n - w|| ##, now as limit of n goes to ##\infty## we get zero. problem is, again, the definition, what definition they want? The definition is $$\lim_{n \to \infty}v_n = v \quad :\Longleftrightarrow \quad \lim_{n \to \infty}\|v_n-v\|=0$$ In order to make sense of ##v_n \longrightarrow v## we need a distance: ##v_n## gets closer and closer to ##v.## What is closer? That's the part where the norm steps in. ##v_n## is closer to ##v## than ##v_m## if ##\|v_n-v\|<\|v_m-v\|.## This transports the uniqueness of the distance limit in ##\mathbb{R}## to the limit in ##(V\, , \,\|\;.\;\|).## Your explanation above is the core of the solution, but you should add the technical details, the epsilontic. physics1000 The definition is $$\lim_{n \to \infty}v_n = v \quad :\Longleftrightarrow \quad \lim_{n \to \infty}\|v_n-v\|=0$$ In order to make sense of ##v_n \longrightarrow v## we need a distance: ##v_n## gets closer and closer to ##v.## What is closer? That's the part where the norm steps in. ##v_n## is closer to ##v## than ##v_m## if ##\|v_n-v\|<\|v_m-v\|.## This transports the uniqueness of the distance limit in ##\mathbb{R}## to the limit in ##(V\, , \,\|\;.\;\|).## Your explanation above is the core of the solution, but you should add the technical details, the epsilontic. Regarding the limit, its pretty obvious no? I will mention in anyway. regarding the distance, to be honest, I havent saw this formula since linear algebra 2, I dont think we need to know it. Regarding the epsilontic, thats my problem, the question, which epsilon? because each thing has its own definition. there aint any definition for the limit of ##||v-w|| = 0## sorry if I am making you crazy, I really dont understand what you mean by the definition of epsilon that I am trying to find out. We know ##\displaystyle{\lim_{n \to \infty}v_n=v},## i.e. for any given real, positive number ##\varepsilon >0,## there is a natural number ##N(\varepsilon)\in \mathbb{N}## depending on ##\varepsilon ## such that ##\|v-v_n\|<\frac{\varepsilon }{2}## for all ##n>N(\varepsilon ).## We know ##\displaystyle{\lim_{n \to \infty}v_n=w},## i.e. for any given real, positive number ##\varepsilon >0,## there is a natural number ##M(\varepsilon)\in \mathbb{N}## depending on ##\varepsilon ## such that ##\|v_n-w\|<\frac{\varepsilon }{2}## for all ##n>M(\varepsilon ).## Hence for all ##n>\max\{N(\varepsilon ),M(\varepsilon )\}## we get (by the triangle inequality of the norm) $$\|v-w\|=\|v-v_n+v_n-w\|\leq \underbrace{\|v-v_n\|}_{<\frac{\varepsilon }{2}}+\underbrace{\|v_n-w\|}_{<\frac{\varepsilon }{2}} <\varepsilon .$$ We can choose ##\varepsilon ## arbitrarily small, so (by the non-negativity norm): $$0\leq \lim_{\varepsilon \to 0}\|v-w\| =\|v-w\| \leq \lim_{\varepsilon \to 0}\varepsilon = 0$$ Last but not least, we apply another axiom of the norm and conclude: $$\|v-w\|=0 \Longrightarrow v-w=0 \Longrightarrow v=w.$$ This altogether means that the axioms (= definition) of the norm guarantee, that the limit definition ##\displaystyle{\lim_{n \to \infty}v_n}=v## via the norm is uniquely defined, i.e. the limit is well-defined. Proofs are always structured that way: - list what we already know - transform this list into formulas by the definitions of the known - prove the statement with those formulas We know ##\displaystyle{\lim_{n \to \infty}v_n=v},## i.e. for any given real, positive number ##\varepsilon >0,## there is a natural number ##N(\varepsilon)\in \mathbb{N}## depending on ##\varepsilon ## such that ##\|v-v_n\|<\frac{\varepsilon }{2}## for all ##n>N(\varepsilon ).## We know ##\displaystyle{\lim_{n \to \infty}v_n=w},## i.e. for any given real, positive number ##\varepsilon >0,## there is a natural number ##M(\varepsilon)\in \mathbb{N}## depending on ##\varepsilon ## such that ##\|v_n-w\|<\frac{\varepsilon }{2}## for all ##n>M(\varepsilon ).## Hence for all ##n>\max\{N(\varepsilon ),M(\varepsilon )\}## we get (by the triangle inequality of the norm) $$\|v-w\|=\|v-v_n+v_n-w\|\leq \underbrace{\|v-v_n\|}_{<\frac{\varepsilon }{2}}+\underbrace{\|v_n-w\|}_{<\frac{\varepsilon }{2}} <\varepsilon .$$ We can choose ##\varepsilon ## arbitrarily small, so (by the non-negativity norm): $$0\leq \lim_{\varepsilon \to 0}\|v-w\| =\|v-w\| \leq \lim_{\varepsilon \to 0}\varepsilon = 0$$ Last but not least, we apply another axiom of the norm and conclude: $$\|v-w\|=0 \Longrightarrow v-w=0 \Longrightarrow v=w.$$ This altogether means that the axioms (= definition) of the norm guarantee, that the limit definition ##\displaystyle{\lim_{n \to \infty}v_n}=v## via the norm is uniquely defined, i.e. the limit is well-defined. Proofs are always structured that way: - list what we already know - transform this list into formulas by the definitions of the known - prove the statement with those formulas ohh, with epsilon like that. to be honest, I did something similliar, way shorter ( just wrote definition ) and said that each of the limits ( like you did ) is smaller than ##\frac {\epsilon} 2## and thus the sum of two is smaller than epsilon. The question is the max needed? I did without max, just took another ##n## But ahh I see, i guess I needed to elaborate more in my answer, I did really just 3 lines of definition proof. Thank you very much, now I understand! Regarding the limit, its pretty obvious no? I assume that the above exercise is part of Legendre's proof of Dirichlet's theorem, which by the way seems to be rather complicated: The problem was first formulated in full by Adrien-Marie Legendre in 1798. This was linked to a first attempt of a proof. In the second edition of his book "Essai sur la théorie des nombres" (published in 1808) he gave an erroneous proof. In the third edition of 1830 he repeated the same mistake. Legendre's error lay behind the words "As you can easily see..." appearing at the end of Section 409 of the third edition. ... So much to "pretty obvious". I remember that an "obvious" took me an entire day and three variable transformations until it was obvious to me, too. Yes, it was pretty obvious. But only to either incredibly gifted persons or to those who did the handcraft dozens of times before. physics1000 So much to "pretty obvious". I remember that an "obvious" took me an entire day and three variable transformations until it was obvious to me, too. Yes, it was pretty obvious. But only to either incredibly gifted persons or to those who did the handcraft dozens of times before. oh by the limit is obvious, I mean regarding the limit of norm is equal to 0 if and if only the sub itself is zero. not the other thing, or you talked about it too, hehe. anyway, thank you very much for staying in patience with me :) fresh_42 A nice result to have : A sequence in a Hausdorff space can converge to only one limit. A norm || .|| gives rise to a metric d(x,y):= || x-y|| , so that this satisfies the Hausdorff condition Also, If your metric d(x,y) arises from a norm then d is translation invariant, so that d(x+a,y+a)=d(x,y) for all a, and homogeneous, i.e., d(cx,cy)=cd(x,y). Then you recover the norm || .|| by ||x||:=d(x,0) physics1000 and fresh_42 A nice result to have : A sequence in a Hausdorff space can converge to only one limit. A norm || .|| gives rise to a metric d(x,y):= || x-y|| , so that this satisfies the Hausdorff condition Also, If your metric d(x,y) arises from a norm then d is translation invariant, so that d(x+a,y+a)=d(x,y) for all a, and homogeneous, i.e., d(cx,cy)=cd(x,y). Then you recover the norm || .|| by ||x||:=d(x,0) Wow, its way above my study :\ Its a course in fourier series. but thanks :) WWGD Wow, its way above my study :\ Its a course in fourier series. but thanks :) Don't let it intimidate you. I may have written it somewhat unclearly. physics1000 Homework Statement: I wrote below what is my question exactly, no need for answer, the question is not explainable well. Relevant Equations: no equations here. only defintion, which we have not learned and I dont know what norm they want... Hi, a question regarding something I could not really understand The question is: Let V be a space with Norm$||*||$Prove if$v_n$converges to vector$v$. and if$v_n$converges to vector$w$so$v=w$and show it by defintion. The question is simple, the thing I dont understand, what definition? what norm exactly? we learnt only norm$L^1$and$L^2$and$L^\infty$which is uniform convergence. Please tell me what I wrote above. EDIT: I dont know how to write latex here, at MSE its with $$, here it wont work.$$ _always_ works. you just need enough of it ;).$\$ _always_ works. you just need enough of it ;).
So it is like duck tape.

WWGD