# How is the inner product in $H^{-1/2}$ defined?

Since $H^{1/2}$ is a Hilbert space, $H^{-1/2}$ must also be a Hilbert space by the isomorphism of Riesz representation theorem. How is the inner product defined there?

We know there is a nice Cauchy-Schwarz $\| f\cdot g\|_{L^2(\Gamma )}\leq \| f\|_{L^2(\Gamma )}\| g\|_{L^2(\Gamma )}$. Is there a counterpart of this in $H^{-1/2}$? If $f,g\in L^2(\Gamma )$, is it true that $\| f\cdot g\|_{H^{-1/2}(\Gamma )}\leq \| f\|_{H^{-1/2}(\Gamma )}\| g\|_{H^{-1/2}(\Gamma )}$?

Or is $\| f\cdot g\|_{H^{-1/2}(\Gamma )}\leq \| f^2\|_{H^{-1/2}(\Gamma )}\| g^2\|_{H^{-1/2}(\Gamma )}$ the best we can do?

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The title is about inner product, but in the text you write strange things like $\| f\cdot g\|_{L^2(\Gamma )}\leq \| f\|_{L^2(\Gamma )}\| g\|_{L^2(\Gamma )}$. What does $f\cdot g$ mean here? The pointwise product of two $L^2$ functions is not in $L^2$ in general. – 40 votes Jul 28 '13 at 20:38
The inner product on the dual of a general Hilbert space $H$ is described on Wikipedia. One more way to write it is $\langle f,g\rangle_{H^*} =\sum_j f(e_j)g(e_j)$ where $e_j$ is any orthonormal basis of $H$. – 40 votes Jul 28 '13 at 21:38
So I'm asking two questions. Isn't that inequality Cauchy-Schwarz? – user33869 Jul 28 '13 at 22:20
The Cauchy-Schwarz inequality is $\| f\cdot g\|_{L^1(\Gamma )}\leq \| f\|_{L^2(\Gamma )}\| g\|_{L^2(\Gamma )}$. The inequality you stated for $L^2$ is false. – 40 votes Jul 28 '13 at 22:39
I believe the answer to your question about concretely defining the inner product for fractional exponents is "use the Fourier transform." mathoverflow.net/questions/25293/dual-spaces-of-sobolev-spaces – dls Jul 30 '13 at 5:01

## 1 Answer

If I understood you first question well, there is some ways to define inner product in $H^{-1/2}$. For example by using Riesz theorem, if we indetify $F,G\in H^{-1/2}$ with $f,g\in H^{1/2}$, then we can define the inner product $(\cdot,\cdot)$ in $H^{-1/2}$ by $$(F,G)=((f,g))$$

where $((\cdot,\cdot))$ is the inner product in $H^{1/2}$.

Another way is: define in $H^{-1/2}$ the norm $$\|F\|_{H^{-1/2}}=\sup_{f\in H^{1/2},\ \|f\|_{H^{1/2}}=1}\langle F,f\rangle$$

You can verify that $\|\cdot\|_{H^{-1/2}}$ satisfies the paralelogram law and hence it is possible to define a inner product with it (do you know how to do it?). Note that, if we assume Riesz theorem here, we have the first definition.

For your second question, first note that your statement is wrong, as pointed out by @40votes. The right statement is: if $(\cdot,\cdot)$ denotes the inner product in $H^{-1/2}$ then for all $f,g\in L^2$ we have that $$|(f,g)|\leq \|f\|_{H^{-1/2}}\|g\|_{H^{-1/2}}$$

This is true because every inner product satisfies Cauchy-Schwarz inequality and also $L^2\subset H^{-1/2}$.

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An example of the first definition is the minus norm used in the least-square finite element in the computation community, when converting a second order elliptic problem to a first order system, in case anyone is interested in "real life" application. +1. – Shuhao Cao Aug 21 '13 at 2:35