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Let $\Omega \subset \mathbb{R}^3$ be a bounded region and consdier the Sobolev space $H^1(\Omega):=W^{1,2}(\Omega)$. For brevity, let us restrict to "real" Sobolev spaces.

By Sobolev embedding, we have $H^1(\Omega) \subset L^6(\Omega)$ so that $L^{6/5}(\Omega) \subset H^{-1}(\Omega)$.

Then, for any $f \in L^{6/5}(\Omega)$ and $g \in H^{1}(\Omega)$, is the following equality \begin{equation} \langle f,g\rangle_{H^{-1} \times H^1} =\int_{\Omega} fg \end{equation} just a definition? If not, how can one justify this equality?

Could anyone please clarify for me?

George Giapitzakis
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Keith
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  • see https://math.stackexchange.com/questions/740355/dual-of-h1-0-h-1-or-h-01/740695#740695 – daw Dec 21 '23 at 15:18
  • or https://math.stackexchange.com/questions/2205362/representations-of-the-dual-space-of-a-hilbert-space/2205386#2205386 – daw Dec 21 '23 at 15:18
  • @daw So, the "canonical" way to understand any $f \in L^{6/5}$ as an element of $H^{-1}$ is the $L^2$ inner product as in my post? – Keith Dec 21 '23 at 18:58
  • Yes. But one has to be carefull, as there are two incompatible ways of understanding $f\in H^1$ as an element in $H^{-1}$. – daw Dec 22 '23 at 07:12
  • @daw for $H^1$, yes I agree. But for $L^{6/5}$, I think the $L^2$ inner product seems like the only way. – Keith Dec 22 '23 at 09:46

1 Answers1

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The expression

$$ \langle f, g \rangle_{H^{-1} \times H^1} = \int_{\Omega} fg $$

is not just a definition; it represents the duality pairing between the spaces $ H^{-1}(\Omega) $ and $ H^1(\Omega) $. Here's a brief clarification:

  1. Sobolev Spaces: The space $ H^1(\Omega) $ is a Sobolev space that consists of functions that are square-integrable over a domain $ \Omega $ along with their first derivatives. The space $ H^{-1}(\Omega) $ is the dual space to $ H^1(\Omega) $, which consists of the bounded linear functionals on $ H^1(\Omega) $.

  2. Duality Pairing: The duality pairing $ \langle \cdot , \cdot \rangle_{H^{-1} \times H^1} $ is a way to relate elements of a space with elements of its dual. For Sobolev spaces, it generalizes the concept of the inner product to include functionals.

  3. Sobolev Embedding: The Sobolev Embedding Theorem states that under certain conditions, functions in the Sobolev space $ H^1(\Omega) $ can be considered as elements of a space of functions with more integrability, like $ L^p(\Omega) $ spaces for certain $ p $. In $ \mathbb{R}^3 $, the Sobolev Embedding Theorem tells us that $ H^1(\Omega) $ is continuously embedded in $ L^6(\Omega) $, and hence its dual space $ H^{-1}(\Omega) $ is embedded in $ L^{6/5}(\Omega) $.

  4. Equality Justification: If $ f \in L^{6/5}(\Omega) $ and $ g \in H^1(\Omega) $, then the integral $ \int_{\Omega} fg $ makes sense because $ f $ and $ g $ are integrable over $ \Omega $ due to Hölder's inequality. The integral $ \int_{\Omega} fg $ essentially represents the action of the functional $ f $ on the function $ g $.

  5. Hölder's Inequality: The integral $ \int_{\Omega} fg $ is finite and well-defined due to Hölder's inequality, which in this setting allows us to integrate the product of a function in $ L^{6/5} $ and another in $ L^6 $.

To summarize, the expression given is a way to represent the action of a functional in $ H^{-1}(\Omega) $ on a function in $ H^1(\Omega) $, and it is justified by the properties of Sobolev spaces and the Sobolev Embedding Theorem. It's not just a definition, but rather a consequence of the theory of Sobolev spaces and how these function spaces interact with one another.

Rowing0914
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  • Thank you. But I don't feel that your justification is fully rigorous. $\int_\Omega fg$ is a well-defined integral, of course. But why should it necessarily coincide with the dual pairing $\langle f, g \rangle_{H^{-1} \times H^1}$? – Keith Dec 21 '23 at 11:46