The family $\left\{h_k(\cdot)\right\}_{k=0}^{\infty}$ of Haar functions are defined for $0 \leq t \leq 1$ as follows: $$ \begin{gathered} h_0(t):=1 \quad \text { for } 0 \leq t \leq 1 \\ h_1(t):=\left\{\begin{array}{lr} 1 & \text { for } 0 \leq t \leq \frac{1}{2} \\ -1 & \text { for } \frac{1}{2}<t \leq 1 \end{array}\right. \end{gathered} $$ If $2^n \leq k<2^{n+1}, n=1,2, \ldots$, we set $$ h_k(t):=\left\{\begin{array}{l} 2^{n / 2} \text { for } \frac{k-2^n}{2^n} \leq t \leq \frac{k-2^n+1 / 2}{2^n} \\ -2^{n / 2} \text { for } \frac{k-2^n+1 / 2}{2^n}<t \leq \frac{k-2^n+1}{2^n} \\ 0 \text { otherwise. } \end{array}\right. $$ For $k=0,1,2, \ldots$, $$ s_k(t):=\int_0^t h_k(s) d s \quad(0 \leq t \leq 1) $$ is the $k$-th Schauder function.
Let $\left\{a_k\right\}_{k=0}^{\infty}$ be a sequence of real numbers such that $$ \left|a_k\right|=O\left(k^\delta\right) \quad \text { as } k \rightarrow \infty $$ for some $0 \leq \delta<1 / 2$. Then the series $$ \sum_{k=0}^{\infty} a_k s_k(t) $$ converges uniformly for $0 \leq t \leq 1$.
Proof. Fix $\varepsilon>0$. Notice that for $2^n \leq k<2^{n+1}$, the functions $s_k(\cdot)$ have disjoint supports. Set $$ b_n:=\max _{2^n \leq k<2^{n+1}}\left|a_k\right| \leq C\left(2^{n+1}\right)^\delta $$ Then for $0 \leq t \leq 1$, $$ \begin{aligned} & \leq C \sum_{n=m}^{\infty}\left(2^{n+1}\right)^\delta 2^{-n / 2-1}<\varepsilon \\ & \end{aligned} $$ for $m$ large enough, since $0 \leq \delta<1 / 2$. This concludes the proof.
I would like to know if the series $$ \sum_{k=0}^{\infty} a_k s_k(t) $$ is unconditionally convergent. Remember that if $\left\{f_k\right\}_{k=1}^{\infty}$ is a sequence and $\sum_{k=1}^{\infty} f_{\sigma(k)}$ is convergent for all permutations $\sigma$, we say that $\sum_{k=1}^{\infty} f_k$ is unconditionally convergent. In that case, the limit is the same regardless of the order of summation.
Since $ \left|a_k\right|=O\left(k^\delta\right)$, there is a constant $D>0$ such that $$ \left|a_k\right|< Dk^\delta\quad\text{for all } k\ge1. $$
For $n\ge1$, let $$u_n(t)=\sum_{2^n \le k<2^{n+1}}|a_ks_k(t)|\quad\text{for }t\in [0,1].$$ Since for $2^n\le k<2^{n+1}$, the functions $s_k(\cdot)$ have disjoint supports,
$$ u_n(t)<D\left(2^{n+1}\right)^\delta2^{-n/2-1} $$
So$$\begin{aligned} &\quad\sum_{k=0}^{\infty} |a_k s_k(t)|\\ &\le |a_0| +|a_1|/2 + \sum_{n=1}^{\infty}u_n(t)\\ &<|a_0| +|a_1|/2 + \sum_{n=1}^{\infty} D\left(2^{n+1}\right)^\delta2^{-n/2-1}\\ &=|a_0| +|a_1|/2+ \frac{D2^{2\delta-3/2}}{1-2^{-1/2+\delta}}. \end{aligned}$$
The last expression is finite. Hence $\sum_{k=0}^{\infty} a_k s_k(t)$ converges unconditionally.