1

i'm looking for a non-continuous linear mapping f:$V \to \mathbb{R}$ with $V$ a normed linear space. I found that I needed to use a infinite-dimensional V because if V is finite-dimensional then f would automaticaly be continuous.

I maybe thought to this one: 1 if $x \le 0$ and 0 if $x>0$ but i'm not sure if i used that V is normed linear space.

As followed i wanted to prove that if f is continuous in zero that f is continuous overall. The implication of "f is continuous" from "f is continuous in zero" is trivial. But I'm not sure if i can prove the reverse implication because i don't understand the problem with the linear mapping, so i need some help there.

I also thought that if f is uniformly continuous that f is normal continuous and so on but the other implication needs a counter example. I'm just a little stuck that i need to work with linear mappings because i know the implication for a normal function. So maybe someone can help me understanding the connection between continuous and linear mappings?

Thanks a lot.

Jean Marie
  • 81,803
questmath
  • 875
  • 5
  • 11
  • Every linear functional on a finite dimensional normed linear space is continuous and every infinite dimensional normed linear space supports a discontinuous linear functional. – Kavi Rama Murthy Jul 28 '20 at 09:28
  • yeah that's what i said that I need to find an example where V is infinite dimensional – questmath Jul 28 '20 at 09:30
  • You can construct one using a Hamel basis. This is a standard construction which you can find on the net and also on this site. – Kavi Rama Murthy Jul 28 '20 at 09:36
  • 2
    Does this answer your question? Discontinuous linear functional – Kavi Rama Murthy Jul 28 '20 at 09:38
  • No i already looked into that and i just don't understand that Hamel Basis. I saw that post and all the answers and i'm still stuck with my question after a day so i thougt i posted one myself – questmath Jul 28 '20 at 09:39
  • 1
    @mathmath As the second answer on that page states, any construction of a discontinuous linear functional on a Banach space (a complete normed linear space) will require some version of the axiom of choice – Ben Grossmann Jul 28 '20 at 09:42
  • 1
    The counterexample (1 if $x≤0$ and 0 if $x>0$) you have been thinking about cannot work for 2 reasons : $x>0$ has a meaning only on the vector space $\mathbb{R}$ 2) even on this space, it is not linear. – Jean Marie Jul 28 '20 at 09:44
  • 1
    @mathmath The example constructed in the answer below can be similarly made if we take $V$ to be the set of all polynomials – Ben Grossmann Jul 28 '20 at 09:49
  • The only thing i don't get is why it would be normalized? is it because we just say that ||($x_{n}$)||=...? – questmath Jul 28 '20 at 09:51
  • @mathmath As far as your proof goes: instead of thinking about discontinuous linear functionals, try to prove directly that if $f$ is continuous at $0$, then $f$ is continuous at all points. – Ben Grossmann Jul 28 '20 at 09:51
  • @mathmath I don't understand what you mean by "normalized" – Ben Grossmann Jul 28 '20 at 09:52
  • maybe Normed inestead of normalized? i'm sorry i don't talk english normally – questmath Jul 28 '20 at 09:53
  • is that true? I really thought that if f is continuous at zero it can be discontinuous somewhere else? – questmath Jul 28 '20 at 09:54
  • @mathmath That's ok, thanks for clarifying. It needs to be normed (or at least have some kind of topology) in order for the notion of "continuity" to make sense. For instance, the $\epsilon$-$\delta$ definition of continuity requires that we have a notion of the distance between two inputs. In a normed space, $d(x,y) = |x - y|$. – Ben Grossmann Jul 28 '20 at 09:56
  • 1
    @mathmath Yes, it is true. If a linear map over a normed space is continuous at $0$, then it is necessarily continuous at all points. – Ben Grossmann Jul 28 '20 at 09:57
  • okey then i'm going to try to prove that because i was only looking for a counter example – questmath Jul 28 '20 at 09:59
  • does this mean that also this also implicate uniformly continuous? – questmath Jul 28 '20 at 10:00
  • 1
    @mathmath Yes. So in fact, what you should prove instead is that if $f$ is continuous at $0$, then it is uniformly continuous over the entire space. – Ben Grossmann Jul 28 '20 at 10:00
  • Yes so to conclude; if i have (1) f is continuous in 0 (2) f is continuous (3) f is uniformly continuous then (1),(2) and (3) are equivalent with each other because we're talking about a normed vectorspace and a lineaire map? – questmath Jul 28 '20 at 10:02
  • 1
    @mathmath That is completely correct. – Ben Grossmann Jul 28 '20 at 10:04
  • wow thanks a lot, i'm going to try answering my own question with a proof then :) thank you – questmath Jul 28 '20 at 10:04

1 Answers1

3

Example: Let $V=\bigoplus^\infty\mathbb{R}=c_{00}(\mathbb{R})$ be the space of sequences of reals with only finitely many nonzero terms (i.e., eventually constant 0 sequences), equipped with, e.g. the norm $$ \lVert (x_n)\rVert = \sqrt{\sum_{n=1}^\infty x_n^2} $$ and let $f\colon V\to\mathbb{R}$ be $$ f((x_n))=\sum_{n=1}^\infty nx_n. $$ Then $f$ is linear, but discontinuous since $f((0,0,\dots,0,\epsilon,0,0,\dots))=n\epsilon\to\infty$ as $n\to\infty$ but $\lVert(0,0,\dots,0,\epsilon,0,0,\dots)\rVert=\epsilon$.

Note: A linear map $f$ is uniformly continuous iff continuous iff continuous at $0$. That is just $f(x)-f(y)=f(x-y)=f(x-y)-f(0)$ so if you can control $f$ on the ball of radius $r$ near $0$, then you have the same control on how much $f$ can vary on the ball of radius $r$ near any other points. This proves the only nontrivial implication "continuous at $0$ $\Rightarrow$ uniformly continuous" (the other two: "uniformly continuous $\Rightarrow$ continuous" and "continuous $\Rightarrow$ continuous at $0$" are trivial).

user10354138
  • 33,239