# Why duality pairing $\langle w,v \rangle=\int_{[0,T]}\langle w(t), v(t)\rangle_{V^*\times V}dt$ can be defined this way?

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 $v:[0,T]\to V$, $\int_{[0,T]}\|v(t)\|_V^2dt<\infty$),

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

$\mathcal{W}=\{w\in \mathcal{V}\,|\,w^{'}\in\mathcal{V}^* \}$ 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 $(1)$?

• 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 – Xander Henderson Aug 9 '17 at 15:40
• I am not sure if it is so obvious. – zorro47 Aug 9 '17 at 17:09
• Ugh... the link got mangled: goo.gl/4JUigw And no, I am not sure that it is obvious, but it has that flavour. – Xander Henderson Aug 9 '17 at 17:19
• I don't think Riesz - Markov - Kakutani theorem is the best idea in here. Instead we can use duality relation. – 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 $u\in X$.