Navier Stokes Energy Estimate I'm currently taking a course in Fluid Mechanics which I'm struggling quite a bit with. I don't have the best background in maths. In particular, I can't seem to deduce the following energy estimate for the Navier Stokes equation. If anyone could help me to understand where I'm going wrong, I would appreciate it. 
Consider the incompressible Navier–Stokes equations on a bounded domain $\mathbb{T}_d$ with periodic boundary conditions. We know the Navier Stokes equations can be written as 
$$\frac{\partial\mathbf{u}}{\partial t}+\mathbf{\omega}\times\mathbf{u}=\nu\Delta\mathbf{u}-\nabla(p+\frac{1}{2}|\mathbf{u}|^2)+\mathbf{f}$$
We can then find $\frac{d}{dt}\|\mathbf{u}\|^2_{L^2}$ and use Holder's inequality along with Young's and Poincare's inequality to deduce that
$$\frac{d}{dt}\|\mathbf{u}\|^2_{L^2}+2\nu\|\nabla\mathbf{u}\|_{L^2}^2\leq\delta\|\mathbf{u}\|_{L^2}^2+\frac{1}{\delta}\|\mathbf{f}\|_{L^2}^2$$
From here we can integrate the differential equation to find. 
$$\|\mathbf{u}(\cdot,t)\|^2_{L^2}\leq \|\mathbf{u}(\cdot,0)\|^2_{L^2}\exp(-\nu t/c^2)+\bigg(\frac{c^2}{\nu}\bigg)^2\|\mathbf{f}(\cdot,t)\|_{L^2}^2(1-\exp(-\nu t/c^2))$$
But what I don't understand is that from here we can establish that for $T>0$
$$\mathbf{u}\in L^{\infty}([0,T];L^2(\mathbb{T}^d,\mathbb{R}^d))$$
Also, If we integrate the differential equation before applying Poincare's inequality we can find 
$$\|\mathbf{u}(\cdot,t)\|^2_{L^2}+\frac{\nu}{c^2}\int_0^T\|\nabla\mathbf{u}(\cdot,\tau)\|_{L^2}^2d\tau\leq \nu \int_0^T\|\mathbf{u}(\cdot,\tau)\|_{L^2}^2d\tau+\frac{1}{\nu}\int_0^T\|\mathbf{f}(\cdot,\tau)\|_{L^2}^2d\tau$$
Again I don't follow why this means that for any $T>0$
$$\mathbf{u} \in L^2([0,T];H^1(\mathbb{T}^d,\mathbb{R}^d))$$ and thus 
$$\mathbf{u} \in L^{\infty}([0,T];L^2(\mathbb{T}^d,\mathbb{R}^d))\cap L^2([0,T];H^1(\mathbb{T}^d,\mathbb{R}^d))$$
Many thanks. 
 A: First of all the equation that you writed is not the well-known Navier-Stokes equations since you included the term $\omega$, the term $\omega$ represent the vorticity and you get it when you take the curl of the velocity $u$. But, no problem let is return to the famous Navier Stokes equation which can be written as 
\begin{align}
\frac{\partial u}{\partial t}+(u\cdot\nabla)u&= \Delta u -\nabla p\\
      \nabla\cdot u &=0.
\end{align}
Before one pass to the energy inequality, one can get rid of the term $p$ which represent the pressure by applying the Helmholtz-Leray projection or by any other mathematical tool. Also, you have to know that there is a whole analysis one should make before talking about the energy estimates and here's a sketsh of this analysis:
A global Lipschitz condition leads to global existence and uniqueness,
whereas (since one have not a global Lipschitz condition), one needs to establish a local Lipschitz condition generally leads only to local existence
and uniqueness. However, a local Lipschitz condition supplemented with some additional a
priori bounds, if appropriate bounds exist and can be found, leads
to global existence and uniqueness. And finally, without a local Lipschitz condition our construction of a continuous
solution via Picard iteration is suspect and moreover, even
if a continuous solution can be found by some other means, it is
not necessarily unique. Now, since Navier-Stokes equation is not an ordinary differential equation but rather a PDE you have to approximate the equation by using as for example the
Galerkin approximations sheme. Now, formally we return to the energy estimate:
To obtain the energy estimate you have to take the inner product in $L^2(\mathbb{T}^d)$ of the aformentioned Navier-Stokes equation against $u$, one infers that 
\begin{align}
\langle\frac{\partial u}{\partial t}, u\rangle_{L^2}+\langle(u\cdot\nabla)u,u\rangle_{L^2}&= \langle\Delta u,u\rangle_{L^2}.
\end{align}
Now by using the fact that $\langle\frac{\partial u}{\partial t}, u\rangle_{L^2}=\frac{1}{2}\frac{d}{dt}\|u(t)\|_{L^2}^2$ and by integrating by parts $\langle\Delta u,u\rangle_{L^2}$, one obtains:
\begin{align}
\frac{d}{dt}\|u(t)\|_{L^2}^2+2\|\nabla u(t)\|_{L^2}^2\leq |\langle(u\cdot\nabla)u,u\rangle_{L^2}|.
\end{align}
As you mentioned in your question you can apply the Holder's and Young's product inequalities several times and use the cancellation law to get rid of the term $|\langle(u\cdot\nabla)u,u\rangle_{L^2}|$ to get the energy inequality
\begin{align}
\|u(s)\|_{L^2}^2+\int_0^s\|\nabla u(t)\|_{L^2}^2dt\leq \|u(0)\|_{L^2}^2.
\end{align} 
The answer of your question consists in informing you that for any function $f\in \mathbb{R}_+\times\mathbb{T}^d$, one has:
\begin{align}
\|f\|_{L^\infty(\mathbb{R_+};L^2)}^2&=\sup\limits_{t\in \mathbb{R}_+}\|f(t)\|_{L^2}^2\\
\|f\|_{L^2(\mathbb{R_+};\dot{H}^1)}^2&=\int_{\mathbb{R}_+}\|\nabla f(t)\|_{L^2}^2dt.
\end{align}
Adding the information that $\|u(0)\|_{L^2}$ is constant. I hope that my answer was clear.
