I am referring to this paper, p. 7.
It is said im Lemma 1 that $$ \frac{1}{2}\frac{d}{dt}\Vert f^\varepsilon(t)\rVert^2_{L^2(dx d\gamma)}+\frac{1}{\varepsilon^2}\Vert f^\varepsilon(t)-p^\varepsilon (t)M\Vert^2_{L^2(dxd\gamma)}\leqslant 0\tag{2.2} $$ for each $t\geq 0$, in particular means that $$ \max\left(\Vert f^\varepsilon\Vert^2_{L_t^\infty L^2(dxd\gamma)},\frac{2}{\varepsilon^2}\Vert f^\varepsilon-p^\varepsilon M\Vert^2_{L_t^2L^2(dxd\gamma)}\right)\leqslant\Vert f_0\Vert^2_{L^2(dxd\gamma)}\tag{2.3}. $$
I do not completely understand how (2.2) implies (2.3).
First of all, we can of course multiply (2.2) by $2$ to get $$ \frac{d}{dt}\Vert f^\varepsilon(t)\rVert^2_{L^2(dx d\gamma)}+\frac{2}{\varepsilon^2}\Vert f^\varepsilon(t)-p^\varepsilon (t)M\Vert^2_{L^2(dxd\gamma)}\leqslant 0 $$
Now, I guess they integrate with respect to time. But I do not know what the notations $$ L_t^\infty L^2(dxd\gamma),\qquad L_t^2L^2(dxd\gamma) $$ are supposed to mean and why in the first expression we have $L_t^\infty$ while in the second one we have $L_t^2$.
The notation was already in the comments. For convenience, let $B=L^2(dx\,d\gamma)$. The inequality follows immediately once you integrate in time. You get for all $t\geq 0$, \begin{align} \|f^{\epsilon}(t)\|_B^2-\|f^{\epsilon}(0)\|_B^2+\frac{2}{\epsilon^2}\int_0^t\|f^{\epsilon}(s)-p^{\epsilon}(s)M\|_B^2\,ds&\leq 0, \end{align} or rearranging, \begin{align} \|f^{\epsilon}(t)\|_B^2+\frac{2}{\epsilon^2}\int_0^t\|f^{\epsilon}(s)-p^{\epsilon}(s)M\|_B^2\,ds&\leq \|f_0\|_B^2. \end{align} Since this is true for all $t\geq 0$, we find that \begin{align} \sup_{t\geq 0}\|f^{\epsilon}(t)\|_B^2+\frac{2}{\epsilon^2}\int_0^{\infty}\|f^{\epsilon}(s)-p^{\epsilon}(s)M\|_B^2\,ds&\leq \|f_0\|_B^2. \end{align} Hence, \begin{align} \|f^{\epsilon}\|_{L^{\infty}([0,\infty);B)}^2+\frac{2}{\epsilon^2}\|f^{\epsilon}-p^{\epsilon}M\|_{L^2([0,\infty);B)}^2&\leq \|f_0\|_B^2. \end{align} For the first term, I used the fact that sup of square of non-negative numbers is square of sup, and for the second term, it’s just definition.