Let $(B, \|\cdot\|_B)$ be a Banach space. Then, for any $n\in\mathbb{N},$ $(\boldsymbol{B}=B^n, \|\cdot\|_{\boldsymbol{B}})$ with $\|\boldsymbol{x}\|^2_{\boldsymbol{B}}\,\colon= \sum^n_{i=1}\|x_i\|^2_B$ for $\boldsymbol{x}\colon = (x_1, ..., x_n)^\top\in \boldsymbol{B}$ is also a Banach space. Now, let's consider $x'\in \boldsymbol{B}',$ where $\boldsymbol{B}'$ is the dual space of $\boldsymbol{B},$ i.e. the Banach space consisting of all linear, continuous mappings (functionals) from $\boldsymbol{B}$ to $\mathbb{R}.$
Are there for every $x'\in\boldsymbol{B}'$ functionals $x'_1, \dots, x'_n\in B'$ so that for any $\boldsymbol{b}=(b_1, \dots, b_n)^\top\in\boldsymbol{B}$ holds
$$ x'(\boldsymbol{b}) = x'_1(b_1) + \cdots + x'_n(b_n) \quad ? $$
We have $$\boldsymbol{b}=(b_1,b_2,\ldots, b_n)^{T}=(b_1,0,\ldots, 0)^T+(0,b_2,0,\ldots, 0)^T+\ldots +(0,\ldots, 0,b_n)^T$$ By the additivity we get $$x'(\boldsymbol{b})=x'((b_1,0,\ldots, 0)^T)+x'((0,b_2,0,\ldots, 0)^T)+\ldots+ x'((0,\ldots, 0,b_n)^T)$$ Hence for $$x'_j(b)=x'((0,\ldots,0,b,0,\ldots, 0)^T)$$ where $b$ is located on the $j$th position, we get $$x'(\boldsymbol{b})=x_1'(b_1)+x_2'(b_2)+\ldots +x_n'(b_n)$$