The proof shows that the set $\mathcal{F} := \{u_m\}$ is precompact in $L^p(0, T; B)$ by verifying the two conditions of the Arzelà–Ascoli-type criterion for Bochner spaces: (i) precompactness of the "spatial slices" $\{u_m(t)\}$ in $B$ for a.e. $t$, and (ii) equicontinuity in time in the $B$-norm. The compact embedding $X_0 \subset\subset X$ provides (i), and the bound on the time derivatives in $L^q(0,T; X_1)$ provides (ii).

**Step 1: Reduction to uniform bounds.** By hypothesis, there exists $C > 0$ such that for all $m$: \begin{align*} \|u_m\|_{L^p(0,T; X_0)} \le C, \qquad \|(u_m)_t\|_{L^q(0,T; X_1)} \le C. \end{align*} We must show that $\{u_m\}$ has a subsequence converging in $L^p(0, T; X)$.

**Step 2: Compactness of spatial translates.**

[claim:Equicontinuity In Time]

For every $\varepsilon > 0$ there exists $\delta > 0$ such that for all $m$ and all $|h| < \delta$: \begin{align*} \int_0^{T-h} \|u_m(t+h) - u_m(t)\|_X^p \, d\mathcal{L}^1(t) < \varepsilon. \end{align*}

[/claim]

[proof] By the fundamental theorem of calculus in Bochner spaces, for $u_m \in L^p(0,T; X_0)$ with $(u_m)_t \in L^q(0,T; X_1)$: \begin{align*} u_m(t+h) - u_m(t) = \int_t^{t+h} (u_m)_t(s) \, ds \quad \text{in } X_1. \end{align*} Therefore $\|u_m(t+h) - u_m(t)\|_{X_1} \le \int_t^{t+h} \|(u_m)_t(s)\|_{X_1} \, ds$. By Hölder's inequality in time (with exponents $q$ and $q' = q/(q-1)$): \begin{align*} \|u_m(t+h) - u_m(t)\|_{X_1} \le h^{1/q'} \left(\int_t^{t+h} \|(u_m)_t(s)\|_{X_1}^q \, ds\right)^{1/q} \le h^{1/q'} C. \end{align*}

Now we interpolate between the $X_0$ and $X_1$ norms using the compact embedding. For any $\eta > 0$, the interpolation inequality for the triple $X_0 \subset\subset X \subset X_1$ gives: \begin{align*} \|v\|_X \le \eta \|v\|_{X_0} + C(\eta) \|v\|_{X_1} \quad \text{for all } v \in X_0. \end{align*} (This follows from the compactness of $X_0 \hookrightarrow X$: if it failed, there would exist $v_k$ with $\|v_k\|_{X_0} = 1$, $\|v_k\|_X \ge \eta + C_k \|v_k\|_{X_1}$ with $C_k \to \infty$, so $\|v_k\|_{X_1} \to 0$; by compactness, $v_k \to v$ in $X$ with $\|v\|_X \ge \eta$ and $\|v\|_{X_1} = 0$, hence $v = 0$, contradicting $\|v\|_X \ge \eta$.)

Applying this to $v = u_m(t+h) - u_m(t)$: \begin{align*} \|u_m(t+h) - u_m(t)\|_X &\le \eta \|u_m(t+h) - u_m(t)\|_{X_0} + C(\eta) \|u_m(t+h) - u_m(t)\|_{X_1} \\ &\le \eta \cdot 2\sup_m \|u_m\|_{L^\infty(0,T; X_0)} + C(\eta) \cdot h^{1/q'} C. \end{align*} Raising to the $p$-th power, integrating over $t \in [0, T-h]$, and choosing first $\eta$ small (to make the first term $< \varepsilon/2$) and then $\delta$ small (to make the second term $< \varepsilon/2$) gives the claim.

Note: the bound $\|u_m\|_{L^\infty(0,T;X_0)}$ follows from the $L^p(0,T;X_0)$ bound and the time derivative bound by a standard embedding argument (the space $\{u \in L^p(0,T;X_0) : u_t \in L^q(0,T;X_1)\}$ embeds into $C([0,T]; [X_0, X_1]_\theta)$ for appropriate $\theta$, and in particular into $L^\infty(0,T; X)$). [/proof]

**Step 3: Precompactness of spatial slices.** For a.e. $t \in [0, T]$, the set $\{u_m(t)\}_{m=1}^\infty$ is bounded in $X_0$ (by the $L^p(0,T;X_0)$ bound and Chebyshev's inequality — the set of $t$ where $\|u_m(t)\|_{X_0}$ is large has small measure). Since $X_0 \subset\subset X$, any bounded subset of $X_0$ is precompact in $X$. Therefore $\{u_m(t)\}$ is precompact in $X$ for a.e. $t$.

**Step 4: Application of the Arzelà–Ascoli-type criterion.** By a generalisation of the Arzelà–Ascoli theorem to Bochner spaces (the Kolmogorov–Riesz–Simon compactness criterion), a subset $\mathcal{F} \subset L^p(0,T; X)$ is precompact if and only if:

(a) $\mathcal{F}$ is bounded in $L^p(0,T; X)$,

(b) the set $\{u(t) : u \in \mathcal{F}\}$ is precompact in $X$ for a.e. $t \in [0, T]$,

(c) $\sup_{u \in \mathcal{F}} \int_0^{T-h} \|u(t+h) - u(t)\|_X^p \, dt \to 0$ as $h \to 0$.

Condition (a) follows from the $L^p(0,T;X_0)$ bound and the continuous embedding $X_0 \hookrightarrow X$. Condition (b) is Step 3. Condition (c) is Step 2. Therefore $\{u_m\}$ is precompact in $L^p(0,T; X)$, and a subsequence converges strongly.