Consider evolution triple $(V,H,V^*)$ ($H$ is a pivot space). I introduce new spaces

$\mathcal{V}=L^2([0,T], V)$ (a space of measurable functions $v:[0,T]\to V$, $\int_{[0,T]}\|v(t)\|_V^2dt<\infty$),


$\mathcal{W}=\{w\in \mathcal{V}\,|\,w^{'}\in\mathcal{V}^* \}$ is endowed with graph norm, i.e., $\|w\|_{\mathcal{W}}=\|w\|_{\mathcal{V}}+\|w^{'}\|_{\mathcal{V^{*}}}$, (separable and reflexive Banach space)

$\mathcal{H}=L^2([0,T], H)$.

Then we obtain continuous embeddings $$\mathcal{W}\subset \mathcal{V}\subset\mathcal{H}=\mathcal{H}^*\subset \mathcal{V}^*.$$ The duality pairing between $\mathcal{V}^*$ and $\mathcal{V}$ is denoted by $$\langle w,v \rangle_{\mathcal{V}^*\times\mathcal{V}}=\int_{[0,T]}\langle w(t), v(t)\rangle_{V^*\times V}dt\qquad (1)$$ for $w\in \mathcal{V}^*$ and $v\in \mathcal{V}$. My question is: Can we denoted this way? - I mean we define it in very particular way $(1)$ - why can we define it this way? Yet $\langle w,v \rangle_{\mathcal{V}^*\times\mathcal{V}}=w(v)$. So, what allows us to define $(1)$ and what stays behined uniqueness of $(1)$?

  • $\begingroup$ Are you familiar with the Riesz Representation Theorem? Your example seems to be an application thereof. en.wikipedia.org/wiki/Riesz–Markov–Kakutani_representation_theorem $\endgroup$ – Xander Henderson Aug 9 '17 at 15:40
  • $\begingroup$ I am not sure if it is so obvious. $\endgroup$ – zorro47 Aug 9 '17 at 17:09
  • $\begingroup$ Ugh... the link got mangled: goo.gl/4JUigw And no, I am not sure that it is obvious, but it has that flavour. $\endgroup$ – Xander Henderson Aug 9 '17 at 17:19
  • $\begingroup$ I don't think Riesz - Markov - Kakutani theorem is the best idea in here. Instead we can use duality relation. $\endgroup$ – zorro47 Aug 11 '17 at 21:11

Look at A question about duality relation $L_p([0,T];V)^*=L_q([0,T];V^*)$. Naturally, one can also prove the opposite, i.e.,

To each $f\in X^*$ there corresponds exactly one $v\in L_q([0,T];V^*)$ with $$\langle f,u \rangle=\int_{0}^{T}\langle v(t), u(t)\rangle_{V^*\times V}dt$$ for all $u\in X$.


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.