61
$\begingroup$

I'm trying to find a discontinuous linear functional into $\mathbb{R}$ as a prep question for a test. I know that I need an infinite-dimensional Vector Space. Since $\ell_2$ is infinite-dimensional, there must exist a linear functional from $\ell_2$ into $\mathbb{R}$. However, I'm having trouble actually coming up with it.

I believe I'm supposed to find an unbounded function (although I'm not sure why an unbounded function is necessarily not continuous; some light in that regard would be appreciated too), so I thought of using the vectors $e^i$, which have all entries equal to zero, except for the $i$-th one. Then, you can define $f(e^i)=i$. That'd be unbounded, but I'm not sure if it'd be linear, and even if it is, I'm not sure how to define it for all the other vectors in $\ell_2$.

A friend mentioned that at some point the question of whether the set $E=\{e^i:i\in\mathbb{Z}^+\}$ is a basis would come up, but I'm not sure what a basis has to do with continuity of $f$.

I'm just learning this topic for the first time, so bear with me please.

The space of sequences that are eventually zero (suggested by a few people) turned out to be exactly what I needed. It also helped to cement the notions of Hamel basis, not continuous, etc.

$\endgroup$
6
  • 4
    $\begingroup$ +1. Did you try other infinite-dimensional spaces? E.g., your idea of $f(e^i) = i$ will work automatically if you take the space of sequences with finitely many nonzero elements. $\endgroup$
    – Srivatsan
    Jan 15 '12 at 7:55
  • 2
    $\begingroup$ A linear map between two normed vector spaces is continuous if and only if it's bounded (in the operator norm). This is a well know result and the proof can be found in every book about functional analysis. $\endgroup$
    – math
    Jan 15 '12 at 8:00
  • $\begingroup$ @Srivatsan thanks a lot. I really like this idea of space of sequences with finitely many nonzero elements. I'll look into it and post back here what I find. $\endgroup$
    – FPP
    Jan 15 '12 at 8:40
  • 3
    $\begingroup$ You can try to find an unbounded linear functional on a subspace and extend it using Hahn-Banach theorem. \\ If you are familiar with the notion of Hamel basis, it can also be used to prove the existence of unbounded functional on infinite dimensional normed spaces. See e.g. Example 4.2 in Heil: A basis theory primer. In fact, Srivatsan's comment above is a special case of this, since $\{e^i; i\in\mathbb N\}$ is a Hamel basis of the space of sequences that are eventually zero. $\endgroup$ Jan 15 '12 at 8:42
  • $\begingroup$ A similar question was also posted at MO: Unbounded linear operator defined on $l^2$ $\endgroup$ Oct 21 '12 at 10:35
44
$\begingroup$

$\newcommand{\Zobr}[3]{#1:#2\to#3}\newcommand{\R}{\mathbb R}$ A different approach to show existence of unbounded functionals is using the notion of Hamel basis.

Definition: Let $V$ be a vector space over a field $K$. We say that $B$ is a Hamel basis in $V$ if $B$ is linearly independent and every vector $v\in V$ can be obtained as a linear combination of vectors from $B$. (By linearly independent we mean that if a finite linear combinations of elements of $B$ is zero, then all coefficients must be zero.)

This is equivalent to the condition that every $x\in V$ can be written in precisely one way as $$\sum_{i\in F} c_i x_i$$ where $F$ si finite, $c_i\in K$ and $x_i\in B$ for each $i\in F$.

This is probably better known in the finite-dimensional case, but many properties of bases remain true in the infinite-dimensional case as well:

  • Every vector space has a Hamel basis. In fact, every linearly independent set is contained in a Hamel basis.
  • Any two Hamel bases of the same space have the same cardinality.
  • Choosing images of basis vector uniquely determines a linear function, i.e., if $B$ is a basis of $V$ then for any vector space $W$ and any map $\Zobr gBW$ there exists exactly one linear map $\Zobr fVW$ such that $f|_B=g$.

Claim: If $X$ is an infinite-dimensional linear normed space, then there exist non-continuous linear function $\Zobr fX{\R}$.

See also Example 4.2 in Heil: A basis theory primer.

Proof. Choose an infinite linearly independent set $\{x_n; n\in\mathbb N\}$ such that $\|x_n\|=1$. (An infinite linearly independent set exists, since $X$ is infinite-dimensional. Normalizing the vectors does not influence the linear independence.) There is a Hamel basis $B$ containing this set.

Then there is a linear function $\Zobr fX{\R}$ such that $f(x_n)=n$ and $f(b)=0$ for $b\in B\setminus\{x_n; n\in\mathbb N\}$. This function is obviously unbounded. $\square$

In fact, Srivatsan's comment above is a special case of this result, since $\{e^i; i\in\mathbb N\}$ is a Hamel basis of the space $c_{00}$ of sequences that are eventually zero.


This answer is very similar to this one.

The above was taken from these notes of mine. Several more results and references can be found there. I have also mentioned some basic facts about Hamel basis in another answer at this site.

You can also find much more information about Hamel bases at other posts at this site: "Hamel basis", hamel basis site:math.stackexchange.com.

$\endgroup$
4
  • 1
    $\begingroup$ Why is $f$ linear. I am making a mistake but I don't see it. $f(e_i+e_j)\neq f(e_i)+f(e_j)$, since $e_i+e_j$ is not in $B$, $f(e_i+e_j)$=0. $\endgroup$
    – simon
    Oct 7 '13 at 15:46
  • 1
    $\begingroup$ If we define values of a map on basic elements, then there is unique linear extension. (See the paragraph starting: Choosing images of basis vector uniquely determines...) It works similarly in finite-dimensional case. $\endgroup$ Oct 7 '13 at 17:02
  • $\begingroup$ Now, I see @Martin. By which theorem is this implied? $\endgroup$
    – simon
    Oct 7 '13 at 17:22
  • $\begingroup$ I have tried to give some explanation in chat, in order to avoid long discussion in comments here. $\endgroup$ Oct 7 '13 at 17:43
24
$\begingroup$

You won't find an explicit example of a discontinuous linear functional defined everywhere on a Banach space: these require the Axiom of Choice. However, you can find a discontinuous linear functional on a normed linear space. A typical scenario would be that you have Banach space $X$ (whose norm I'll denote $\|.\|_X$) which is a dense linear subspace of Banach space $Y$ (under a different norm $\|.\|_Y$, where $\|x\|_X \ge \|x\|_Y$ for all $x \in X$), and a linear functional $\phi$ on $X$ which is continuous for the norm $\|.\|_X$ but not for the norm $\|.\|_Y$. Thus if you take $X$ with the norm $\|.\|_Y$, you have a normed linear space with a discontinuous linear functional $\phi$. For example, take $X = \ell_2$, $Y = \ell_\infty$, and $\phi(x) = \sum_{i=1}^\infty x_i/i$.

$\endgroup$
4
  • $\begingroup$ Thanks a lot. Two things: 1) What if we do use AC? Is it true then that such a discontinuous linear functional always exists? 2) Is the last sum supposed to be indexed by $i$ instead of $j$? $\endgroup$
    – FPP
    Jan 15 '12 at 8:38
  • $\begingroup$ 1) Yes, in any infinite-dimensional vector space, if $e_1, e_2, \ldots$ is a linearly independent sequence you can define $\phi(e_j)$ arbitrarily and use Axiom of Choice to extend this to a linear functional on the whole space. 2) Yes, thanks for correcting this typo. $\endgroup$ Jan 15 '12 at 9:05
  • $\begingroup$ When you say X is a subspace of Y should you not specify with respect to which norm X is a subspace, especially given that you take the two spaces with different norms? $\endgroup$
    – gen
    Apr 29 '18 at 9:39
  • $\begingroup$ Being a subspace has nothing to do with norms: $X$ is a linear subspace of $Y$ if it is a subset that is closed under addition and under multiplication by scalars. $\endgroup$ Apr 30 '18 at 1:23
16
$\begingroup$

As Robert Israel already mentioned, you cannot write down an explicit (free of the axiom of choice) unbounded linear functional on a Banach space. But it's generally not hard for incomplete normed spaces. Nobody has mentioned my favorite example: the functional $\ell: C^1[-1,1] \to \mathbb{R}$ given by $\ell(f) = f'(0)$.

$\endgroup$
4
  • 10
    $\begingroup$ Where $C^1[-1,1]$ has the uniform norm. (If it has the $C^1$ norm, then this functional is continuous.) $\endgroup$ Jan 19 '12 at 22:54
  • $\begingroup$ Is the uniform norm the same as the supremum norm? $\endgroup$
    – gen
    Apr 29 '18 at 9:43
  • $\begingroup$ Also, what do you mean by $C^1 [-1,1]$? $\endgroup$
    – gen
    Apr 29 '18 at 9:49
  • 1
    $\begingroup$ @gen Yes, the uniform norm is the sup-norm and $C^1$ is the set of differentiable functions with continuous derivatives. $\endgroup$
    – Andrei Kh
    Dec 8 '19 at 8:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.