Lower semicontinuity of $ \Delta(u):=\sup_{r\in[0,T]}\int_0^r\frac{|u(r)-u(s)|^{\delta}}{(r-s)^{\alpha+1}}\,ds\;\;. $ Let $0<\alpha<1/2$ and consider the space $W_0^{\alpha,\infty}(0,T,\Bbb R^d)$ of all measurable functions $f:[0,T]\to\Bbb R^d$ such that
$$
||f||_{\alpha,\infty}:=\sup_{t\in[0,T]}\left(|f(t)|+\int_0^t\frac{|f(t)-f(s)|}{(t-s)^{\alpha+1}}\,ds\right)<+\infty\;\;.
$$
Now $||f||_{\alpha,\infty}$ is a norm and $\left(W_0^{\alpha,\infty}(0,T,\Bbb R^d),||f||_{\alpha,\infty}\right)$ is a Banach normed space.
Consider then the operator
$
\Delta:W_0^{\alpha,\infty}(0,T,\Bbb R^d)\to[0,+\infty]
$
defined as
$$
\Delta(u):=\sup_{r\in[0,T]}\int_0^r\frac{|u(r)-u(s)|^{\delta}}{(r-s)^{\alpha+1}}\,ds\;\;
$$
where $\delta$ is fixed in $]0,1]$.
How can I show that $\Delta$ is lower semicontinous? I.e. how can I prove that
$$
\Delta(u)\le\liminf_{||u-v||_{\alpha,\infty}\to0}\Delta(v)
$$
holds for every $u\in W_0^{\alpha,\infty}(0,T,\Bbb R^d)$?
The paper I'm reading, says that it follows from the fact that the convergence in $W_0^{\alpha,\infty}(0,T;\Bbb R^d)$ implies the uniform convergence, but I don't know how to exploit this.
 A: Let $u\in W_0^{\alpha,\infty}(0,T;\Bbb R^d)$ fixed; then if $v\in W_0^{\alpha,\infty}(0,T;\Bbb R^d)$ we have that
\begin{align*}
\Delta(v)-\Delta(u)
&=\sup_{t\in[0,T]}\int_0^t\frac{|v(t)-v(s)|^{\delta}}{(t-s)^{\alpha+1}}\,ds-\sup_{t\in[0,T]}\int_0^t\frac{|u(t)-u(s)|^{\delta}}{(t-s)^{\alpha+1}}\,ds\\
&\ge\sup_{t\in[0,T]}\left(\int_0^t\frac{|v(t)-v(s)|^{\delta}}{(t-s)^{\alpha+1}}\,ds-\int_0^t\frac{|u(t)-u(s)|^{\delta}}{(t-s)^{\alpha+1}}\,ds\right)\\
=&\sup_{t\in[0,T]}\left(\int_0^t\frac{|v(t)-v(s)|^{\delta}-|u(t)-u(s)|^{\delta}}{(t-s)^{\alpha+1}}\,ds\right)
\end{align*}
from which we have that
\begin{align*}
&\liminf_{||v-u||_{\alpha,\infty}\to0}\left(\Delta(v)\right)-\Delta(u)\ge\\
&\ge\liminf_{||v-u||_{\alpha,\infty}\to0}\left[
\sup_{t\in[0,T]}\left(\int_0^t\frac{|v(t)-v(s)|^{\delta}-|u(t)-u(s)|^{\delta}}{(t-s)^{\alpha+1}}\,ds\right)\right]\\
&=\sup_{t\in[0,T]}\left[
\liminf_{||v-u||_{\alpha,\infty}\to0}\left(\int_0^t\frac{|v(t)-v(s)|^{\delta}-|u(t)-u(s)|^{\delta}}{(t-s)^{\alpha+1}}\,ds\right)\right]\\
\mbox{(Fatou)}&\ge
\sup_{t\in[0,T]}
\left(\int_0^t(t-s)^{-\alpha-1}
\underbrace
{\liminf_{||v-u||_{\alpha,\infty}\to0}\left(|v(t)-v(s)|^{\delta}-|u(t)-u(s)|^{\delta}\right)}_{=0}\,ds\right)=0
\end{align*}
recalling that the convergence in $W_0^{\alpha,\infty}(0,T;\Bbb R^d)$ implies the uniform convergence; thus
$$
\liminf_{||v-u||_{\alpha,\infty}\to0}\Delta(v)\ge\Delta(u)
$$
as wanted.
