Let $B$ be a standard Brownian motion. I have showed that $$\sup_{t\in [0,1]}|B_t|\leq \sum_{n=1}^\infty \sup_{0\leq k\leq 2^n-1}\big|B_{\frac{k+1}{2^n}}-B_{\frac{k}{2^n}}\big|$$ now I want to deduce from this that for $p\geq 1$ $$\Bbb{E}\left[\sup_{t\in [0,1]}|B_t|^p\right]^{1/p}\leq \sum_{n=1}^\infty \left( \sum_{k=0}^{2^n-1} \Bbb{E}\left[|B_{\frac{k+1}{2^n}}-B_{\frac{k}{2^n}}\big|^p\right]\right)^{1/p}$$
I have found a proof in a book, but I don't get the first inequality. They start by $$\begin{align} \Bbb{E}\left[\sup_{t\in [0,1]}|B_t|^p\right]^{1/p}&\leq\sum_{n=1}^\infty \Bbb{E}\left[ \sup_{0\leq k\leq 2^n-1}\big|B_{\frac{k+1}{2^n}}-B_{\frac{k}{2^n}}\big|^p\right]^{1/p}\\&\leq...\end{align}$$ But I don't get the first inquality. I mean if I substitute $\sup_{t\in [0,1]}|B_t|^p$ with the first formuly from above I get $$\sup_{t\in [0,1]}|B_t|^p\leq\left( \sum_{n=1}^\infty \sup_{0\leq k\leq 2^n-1}\big|B_{\frac{k+1}{2^n}}-B_{\frac{k}{2^n}}\big|\right)^p$$ But then taking the conditional expectation why can I pull the sum out and take the power inside? I first thought about Minkowki inequality but I also don't think this helps. Can someone help me justifying this step?
Let $Y_{n,k}:=\big|B_{\frac{k+1}{2^n}}-B_{\frac{k}{2^n}}\big|$. It has been shown that $$ \sup_{t\in [0,1]}\lvert B_t\rvert\leqslant \sum_{n=1}^\infty \max_{0\leqslant k\leqslant 2^n-1}Y_{n,k} $$ hence an application of the triangle inequality for $\mathbb L^p$-norms ($\lVert X\rVert_p:=\left(\mathbb E\left[\lvert X\rvert^p\right]\right)^{1/p}$ gives $$ \left\lVert \sup_{t\in [0,1]}\lvert B_t\rvert\right\rVert_p \leqslant \sum_{n=1}^\infty \left\lVert \max_{0\leqslant k\leqslant 2^n-1}Y_{n,k}\right\rVert_p. $$ Since $$ \left\lvert \max_{0\leqslant k\leqslant 2^n-1}Y_{n,k}\right\rvert^p\leqslant \sum_{k=0}^{2^n-1}\left\lvert Y_{n,k}\right\rvert^p, $$ we derive the wanted result after having taken the expectation.