Riesz representation theorem is very useful to discuss characteristic of functional on Hilbert Space $H$. It says that bounded linear operator $f : H \rightarrow \mathbb{F}$ can be represented as an inner product as $f(x) = \langle x , z\rangle$, where $z$ depends on $f$ can be determined uniquely such that $\|f\| = \|z\|$. Boundedness is only required to prove $\|f\| = \|z\|$.

There should be at least one unbounded linear functional which can not be represented by Riesz representation theorem. I am looking for a counterexample of such a functional.

Can we generalize this theorem for unbounded linear functional?

Thank you for your help.


What do you mean "Boundedness is only required to prove $||f|| = ||z||$?"

I claim that on the contrary, every linear functional of the form $f(x) = \langle x, z \rangle$ is bounded (equivalently, continuous), and thus any unbounded linear functional cannot be represented. Unbounded linear functionals are easy to construct if you know that every function on a linearly independent set extends to a linear functional.

Added: In more detail: if $S \subset H$ is a linearly independent subset, then (using Zorn's Lemma) there is a ("Hamel") basis $B$ containing $S$. Then if $\ell: S \rightarrow \mathbb{C}$ is any function whatsoever, we can extend it to a function $\ell: B \rightarrow \mathbb{C}$ in many ways -- e.g. by $\ell(x) = 0$ for all $x \in B \setminus S$ -- and then $\ell$ extends uniquely to a linear functional on $H$ by linearity. If $S$ is infinite and consists of elements of uniformly bounded norm (which we can always achieve just by renormalizing) and $\ell$ is unbounded on $S$, then every extension of $\ell$ to $H$ is an unbounded linear functional.

  • $\begingroup$ @Clark. First of all thank you for your answer. While I reading the proof of the theorem, I was thinking that boundedness is useful only to prove $\|f\| = \|z\|$. The remaining proof has no use of the boundedness of $f$. Are we using the property anywhere else? $\endgroup$ – Supriyo Oct 21 '13 at 11:38
  • $\begingroup$ @Samprity: As I said in my answer, that cannot be true, since every linear functional of the form $x \mapsto \langle x,z \rangle$ is bounded. I think you should look at the proof again. $\endgroup$ – Pete L. Clark Oct 21 '13 at 12:49

There are several ways to construct such a functional. Let $H$ be a separable Hilbert space, and $\{ e_n \}$ be an orthonormal basis for $H$. For suitable $f = \sum a_n e_n \in H$ we define the functional $L(f) = \sum n a_n$.

This will clearly only work for those $f$ in $H$ where the sum converges, but this is a common problem that arrises with unbounded operators.

$L$ is not bounded, since $$\|L\| = \sup_{\|f\| = 1} \|L(f)\| \ge \sup_{n} \|L(e_n)\| = \sup_{n} n = \infty$$

If this were to be represented by inner product against some vector in $H$, as the Riesz theorem states in the bounded case, then $L(f) = \langle f, g \rangle$ for some $g$. We can get the Fourier expansion for $g$ by evaluating $L(e_n) = n = \langle e_n,g\rangle$. Thus we have $g = \sum_{n} n e_n$. This would mean $$\|g\| = \sqrt{ \langle g,g \rangle } = \sqrt{ \sum n^2 } = \infty$$ and this is a contradiction.

There are generalizations for the Riesz theorem, but in a different direction than you mention. The theorem has been applied to more general Banach spaces than Hilbert spaces. For instance every linear functional, L, over the vector space $C(X)$ equipped with the supremum norm can be represented as integration against some measure $\mu_L$.

  • $\begingroup$ It seems to me that by defining a functional only on a proper subspace you are sidestepping the OP's question. $\endgroup$ – Pete L. Clark Oct 21 '13 at 6:07
  • $\begingroup$ Not really, this is a problem you would run into with any unbounded operator. $\endgroup$ – Joel Oct 21 '13 at 6:19
  • $\begingroup$ For instance a really common unbounded operator is the operation of differentiation, which is what this operator really is. $\endgroup$ – Joel Oct 21 '13 at 6:20
  • $\begingroup$ If you mean that in practice one often finds that naturally defined unbounded operators are not defined on the entire Hilbert space $H$, then yes, I agree. But the OP is asking about unbounded operators defined on all of $H$. There are only too many of these -- the cardinality of the set of them is $2^{2^{\aleph_0}}$ if $H = \ell^2(\mathbb{C})$ -- so when you say "[T]his is a problem you would run into with any unbounded operator." I'm not sure what you mean. $\endgroup$ – Pete L. Clark Oct 21 '13 at 6:58
  • $\begingroup$ @Joel Thank you for your answer. $\endgroup$ – Supriyo Oct 21 '13 at 11:41

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