Let $b:H \times H \to \mathbb{R}$ be a bounded bilinear form on Hilbert space $H$. Fix $u \in H$. Then $b(u, \cdot):H \to \mathbb{R}$ is bounded so $b(u,\cdot) \in H^*.$ Then $b(u,\cdot) = F_u(\cdot)$ where $F_u \in H^*$.

Right? Isn't this a kind of Riesz representation theorem without coercivity?


1 Answer 1


The Riesz Representation Theorem holds for any bounded linear functional on a Hilbert space. There is no mention of coercivity in the theorem.

Besides the fact that a functional can never be coercive, as it always has a nontrivial kernel.

Edit: attempt at clarification.

These are the three relevant results to the discussion:

Riesz Representation Theorem: given $f\in H^*$, there exists $v\in H$ such that $f(x)=\langle x,v\rangle$ for all $x\in H$.

Corollary of the RRT: If $b$ is bilinear and bounded on $H$ and $u\in H$, then there exists $v\in H$ such that $b(u,x)=\langle x,v\rangle$ for all $x\in H$.

Lax-Milgram: given $b$ is bilinear, bounded, and coercive, and $f\in H^*$, then there exists $v\in H$ such that $f(x)=b(x,v)$ for all $x\in H$.

So, the RRT is a particular case of LM, where $b$ is taken to be the scalar product.

On the other hand, the corollary---which seems to be the result quoted in the question---does not attempt to represent every functional using $b$, but rather represent $b$ when one coordinate is fixed.

  • $\begingroup$ Sorry, I don't understand your last assertion. The concept of coercive functional is well defined, as far I known. May you give me some further explaination? $\endgroup$
    – user91126
    Sep 11, 2013 at 21:13
  • $\begingroup$ My bad. Like I said, the functional cannot be coercive; the bilinear form can. In any case, the RRT does not require the bilinear form to be coercive. $\endgroup$ Sep 11, 2013 at 21:27
  • $\begingroup$ @MartinArgerami But the Riesz Representation Theorem uses the inner product as the bilinear form, which is obviously coercive. Right? And I've never seen a proof of any kind of Lax-Milgram type result which doesn't have a coercivity hypothesis. Do you have a reference for this? $\endgroup$
    – user38355
    Sep 12, 2013 at 14:44
  • $\begingroup$ @brom: I fail to follow your comment. The inner product in a Hilbert space is coercive in a vacuous way ($\langle x,x\rangle=\|x\|^2$ by definition). The Riesz Representation Theorem is a characterization of the bounded functionals on a Hilbert space. There is no "coercive" in the statement. A proof of the RRT can be found in any Functional Analysis book. There is even one in Wikipedia (which I don't recommend, because the notation used there makes it look more complicated than it is). $\endgroup$ Sep 12, 2013 at 15:23
  • $\begingroup$ @MartinArgerami I agree with you. I just also look at the RRT as a special case of the Lax-Milgram Lemma concerning representing the linear functionals on a Hilbert space by a more general bilinear form (and given the OP mentioning a bilinear form I think this is also their point of view). This does require the bilinear form to be coercive. But I'm sure you know all of this; I was just a little confused by your previous sentence "the RRT does not require the bilinear form to be coercive." $\endgroup$
    – user38355
    Sep 12, 2013 at 17:06

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