We wish to use the equation to show the estimate
$$ \lVert u'_m(0) \rVert_{L^2(U)}^2 \leq C\left(\lVert u_m(0) \rVert_{H^2(U)}^2 + \lVert f_m \rVert_{H^1(0,T;L^2(U))}^2\right). $$
The strategy is to consider the equation at $t=0,$ and show the other terms are bounded. This requires some care however, as a-priori the equation only holds almost everywhere in $t.$
To do this, first observe by Section 5.9, Theorem 2 (Calculus on spaces involving time) that we have a continuous embedding
$$ H^1(0,T;L^2(U)) \hookrightarrow C([0,T],L^2(U)). $$
Hence for each $k$ the mapping $t \mapsto f_m^k(t) = \langle f_m(t), w \rangle$ is continuous on $[0,T]$ and for each $m$ we have $d_m^k(t)$ satisfies the ODE system
$$ (d_m^k)'(t) + \sum_{j=1}^m d_m^j(t) B[w_j,w_k;t] = f_m^k(t). $$
As each $B[w_j,w_k;t]$ is smooth in $t$ (differentiating under the integral sign), by standard ODE theory we deduce that the unique solution $d_m^k(t)$ must be continuously differentiable on $[0,T].$ Therefore the equation holds pointwise on $[0,T],$ and evaluating at $t=0$ we obtain the identiy
$$ u_m'(0) = - \sum_{k=1}^m B[u_m(0),w_k;0]w_k + f_m(0). $$
To conclude observe we can control both terms on the right hand side as
\begin{align*}
\left| B[u_m(0),w_k;t]\right| &\leq C \lVert u_m(0) \rVert_{H^2(U)} \\
\lVert f_m(0) \rVert_{L^2(U)} &\leq C\lVert f_m \rVert_{H^1(0,T;L^2(U))},
\end{align*}
where we used the continuous embedding above to estimate the $f_m$ term. Hence putting everything together we get
\begin{align*}
\lVert u_m'(0) \rVert_{L^2(U)} &\leq \sum_{k=1}^m \left| B[u_m(0),w_k;t]\right| \lVert w_k\rVert_{L^2(U)} + \lVert f_m(0) \rVert_{L^2(U)} \\
&\leq \left(\lVert u_m(0) \rVert_{H^2(U)}^2 + \lVert f_m \rVert_{H^1(0,T;L^2(U))}^2\right),
\end{align*}
as required.
First I feel like we are missing some information here. With the regularity of $u_m$ and $u$ you stated, namely $u, u_m \in \mathcal{W}(0,T, H^1_0, H^{-1})$ the term $\sup\limits_{0 \leq t \leq T} \| u_m \|_{H^1_0}^2$ (for which I assume you mean $\sup\limits_{0 \leq t \leq T} \| u_m(t) \|_{H^1_0}^2$) doesn't make sense since you only have the embedding
\begin{align}
\mathcal{W}(0,T, H^1_0, H^{-1}) \hookrightarrow \mathcal{C}([0,T], L^2).
\end{align}
and not an embedding
\begin{align}
\mathcal{W}(0,T, H^1_0, H^{-1}) \hookrightarrow \mathcal{C}([0,T], H^1_0).
\end{align}
So for that reason let us switch the supremum for the essential supremum. Now let us fix a measurable set $\Xi \subseteq (0,T)$. From the convergence of $u_m$ we then especially have
\begin{align}
u_m \rightharpoonup u \,\,\, \text{ in } \,\,\, L^2(\Xi, H^1_0).
\end{align}
From the weak sequential lower semicontinuity of the norm in $L^2(\Xi, H^1_0)$ we find
\begin{align}\tag{1}
\| u \|_{L^2(\Xi, H^1_0)}^2 \leq \liminf_{m} \| u_m \|_{L^2(\Xi, H^1_0)}^2 \leq |\Xi| C \left( \| g \|_{H^1_0}^2 + \| f \|_{L^2(0,T;L^2)}^2 \right).
\end{align}
Assuming there exists some measurable set $M \subseteq (0,T)$ with $|M| >0$ and such that
\begin{align}
\| u(t) \|_{H^1_0}^2 > C \left( \| g \|_{H^1_0}^2 + \| f \|_{L^2(0,T;L^2)}^2 \right) \quad \forall t \in M
\end{align}
we can take $\Xi = M$ in (1) and get an immediate contradiction. (Actually we could have straight away started with the set $M$ but whatever)
We can now at least conclude
\begin{align}
\underset{t \in (0,T)}{\mathrm{esssup}} \| u(t) \|_{H^1_0}^2 \leq C \left( \| g \|_{H^1_0}^2 + \| f \|_{L^2(0,T;L^2)}^2 \right).
\end{align}
What do you think? Another approach would be to look for more information on $u_m, u$ which would grant $u_m(t) \to u(t)$ in $H^1_0$ for almost any $t \in (0,T)$. One could then use the weak sequential lower semicontinuity of the norm in $H^1_0$ instead of $L^2(0,T;H^1_0)$. With the given information I only see how one could maybe prove $u_m(t) \to u(t)$ in $L^2$ for almost any $t \in (0,T)$. I might be missing something though.
Best Answer
On the one hand, $$ u_t + K u \leq 0 $$ on $\{ u \geq 0\}$. On the other hand, $$ v_t + K v = 0, $$ with $v = u^+$ on $\Delta_T$.
If we subtract second equation from the first one, we get $$ (u-v)_t + K(u-v) \leq 0 \quad \implies \quad w_t + K w \leq 0, $$ where $w := u-v$. (We can do that, since $\frac{d}{dt}$ and $K$ are linear operators). Moreover, $w = 0$ on $\Delta_T$.
Hence, applying Weak Maximum Principle for $w$ (Theorem 8), we obtain $$ \max_{W_T} w = \max_{\Delta_T} w = 0, $$ and therefore, $w \leq 0$. Thus, $u \leq v$.