Obtaining this estimate How do I obtain this following estimate: 

$$\max_{0\le t \le T} \| \mathbf{u}(t) \|_{L^2(U)} \le C(\|\mathbf{u}\|_{L^2(0,T;H_0^1(U))}+\|\mathbf{u'}\|_{L^2(0,T;H^{-1}(U))}), \tag{10}$$ the constant $C$ depending only on $T$.

This is from PDE Evans, 2nd edition, page 304.
I am following the procedure of the textbook:

To obtain $\text{(10)}$, we integrate $\text{(12)}$ with respect to $s$, recall the inequality $|\langle \mathbf{u'},\mathbf{u}\rangle| \le \|\mathbf{u'}\|_{H^{-1}(U)}\|\mathbf{u}\|_{H_0^1(U)}$, and make some simple estimates.

By the way, this is $\text{(12)}$ from the textbook:

$$\| \mathbf{u}(t) \|^2_{L^2(U)} = \| \mathbf{u}(s) \|^2_{L^2(U)} + 2 \int_s^t \langle \mathbf{u'}(\tau),\mathbf{u}(\tau)\rangle d\tau \tag{12}$$ for all $0 \le s,t\le T$.

I couldn't show much work on this, but here is what I have done so far:
\begin{align}
(T-0)  \| \mathbf{u}(t) \|^2_{L^2(U)} &= \int_0^T \| \mathbf{u}(t) \|^2_{L^2(U)} \, ds \\
 &= \int_0^T \| \mathbf{u}(s) \|^2_{L^2(U)} \, ds + 2 \int_0^T \int_s^t \langle \mathbf{u'}(\tau),\mathbf{u}(\tau)\rangle d\tau \, ds \\
&\le \int_0^T \| \mathbf{u}(s) \|^2_{L^2(U)} \, ds + 2 \int_0^T \int_s^t \|\mathbf{u'}(\tau)\|_{H^{-1}(U)}\|\mathbf{u}(\tau)\|_{H_0^1(U)} d\tau \, ds
\end{align}
But how can I continue on from here?
 A: Your inequality (12) is missing some squares, it should be

$$\| \mathbf{u}(t) \|_{L^2(U)}^2 = \| \mathbf{u}(s) \|_{L^2(U)}^2 + 2 \int_s^t \langle \mathbf{u'}(\tau),\mathbf{u}(\tau)\rangle d\tau \tag{12}$$

With this small correction you get
$$ \begin{align}
T \| \mathbf{u}(t) \|_{L^2(U)}^2 &\le \int_0^T \| \mathbf{u}(s) \|_{L^2(U)}^2 \, ds + 2 \int_0^T \int_s^t \|\mathbf{u'}(\tau)\|_{H^{-1}(U)}\|\mathbf{u}(\tau)\|_{H_0^1(U)} d\tau \, ds
\end{align} $$
The nested integrals are nasty, so we estimate them by integrating over $[0,T]$ instead of $[s,t]$ in the inner integral:
$$ \begin{align}
T \| \mathbf{u}(t) \|_{L^2(U)}^2 &\le \int_0^T \| \mathbf{u}(s) \|_{L^2(U)}^2 \, ds + 2 \int_0^T \int_0^T \|\mathbf{u'}(\tau)\|_{H^{-1}(U)}\|\mathbf{u}(\tau)\|_{H_0^1(U)} d\tau \, ds\\
&\le \int_0^T \| \mathbf{u}(s) \|_{L^2(U)}^2 \, ds + 2 T \int_0^T \|\mathbf{u'}(\tau)\|_{H^{-1}(U)}\|\mathbf{u}(\tau)\|_{H_0^1(U)} d\tau\\
\end{align} $$
Now apply Young's inequality to the product and you get
$$ \begin{align}
T \| \mathbf{u}(t) \|_{L^2(U)}^2 
&\le \int_0^T \| \mathbf{u}(s) \|_{L^2(U)}^2 \, ds + T \int_0^T \|\mathbf{u'}(\tau)\|_{H^{-1}(U)}^2 + \|\mathbf{u}(\tau)\|_{H_0^1(U)}^2 d\tau\\
&\le (1+T) \| \mathbf{u} \|_{L^2(0, T; L^2(U))}^2 \, ds + T \|\mathbf{u'}\|_{L^2(0, T; H^{-1}(U))}^2
\end{align} $$
Now put the $T$ on the right hand side, take the square root and use $\sqrt{a+b} \leq \sqrt a + \sqrt b$. As the right hand side is independent of $t$ you can take the maximum and arrive at your inequality.
