Is this inequality in $L^2$ true?

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Suppose we have functions $f \in L^2([0,T],H^{-1}),g \in L^2([0,T],H^{1})$, i would like to see how can i get the following inequality

\begin{equation} \left|(f,g)_{L^2([0,T] \times \mathbb{T})}\right| \leq ||f||_{ L^2([0,T],H^{-1})}||g||_{L^2([0,T],H^{1})} \end{equation}

where $\mathbb{T} = \mathbb{R}/\mathbb{Z}$. I'm trying to solve this for a specific problem in which i have not realized how to do that, so i would like to know if in general i can get that inequality, if u think that some info is missing or needed, i could show you the specific problem that i have.

Thank you in advance.

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Let $f \in L^2([0,T] \times \mathbb T)$ be given. Then, we typically identify $f$ with the functional $\hat f \in L^2(0,T; H^{-1})$, which is defined via $$ \langle \hat f, g\rangle_{L^2(H^{-1}), L^2(H^1)} := \int_0^T \int_\Omega f(t,x) g(t,x)\,\mathrm dx \, \mathrm{d}t = (f, g)_{L^2([0,T]\times \mathbb T)} $$ for all $g \in L^2(0,T,H^1)$. Similarly, one identifies $f(t) \in L^2$ with $\hat f(t) \in H^{-1}$ defined via $$ \langle \hat f(t), g\rangle_{H^{-1},H^1} = \int_\Omega f(t,x) \, g(t,x) \, \mathrm{d}x $$ for all $g \in H^1$. Thus, $$ (f, g)_{L^2([0,T]\times \mathbb T)} = \int_0^T \langle \hat f(t), g(t)\rangle_{H^{-1},H^1} \, \mathrm{d}t. $$ This last identity (together with the identification of $f$ and $\hat f$) yields your inequality: $$ |(f, g)_{L^2([0,T]\times \mathbb T)}| \le \int_0^T \|\hat f(t)\|_{H^{-1}} \, \|g(t)\|_{H^1} \, \mathrm{d}t \le \| \hat f\|_{L^2(H^{-1})} \, \|g\|_{L^2(H^1)}. $$