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Let $\{f_n\}_{n\in\mathbb{N}}$ be a family of functions from $\mathbb{R}$ to $\mathbb{R}$.

a.) $\sup\{\sum_{n}f_n(x):x\in\mathbb{R}\}\leq \sum_{n}\sup\{f_n(x):x\in\mathbb{R}\}$.

b.) $\sup\{\sum_{n}f_n(x_n):x_n\in\mathbb{R}\} = \sum_{n}\sup\{f(x_n):x_n\in\mathbb{R}\}$.

Attempted proof a.) - Since $\sum_{n}f_n(x)\leq \sum_{n}\sup\{f_n(x):x\in\mathbb{R}\}$ then $\sum_n f_n(x)$ is bounded above by $\sum_{n}\sup\{f_n(x):x\in\mathbb{R}\}$. Thus taking the supremum over all $x$ we get, $$\sup\{\sum_{n}f_n(x):x\in\mathbb{R}\}\leq \sum_{n}\sup\{f_n(x):x\in\mathbb{R}\}$$

I am not exactly sure how to proceed with b.) I am being thrown off a bit by the $f(x_n)$ it seems that we have $f_n\rightarrow f$ but I am not sure. Any suggestions is greatly appreciated.

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Wolfy
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1 Answers1

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For part a, let $x\in\mathbb R$. Then for each $n$, we can choose $x_n\in\mathbb R$ such that $f_n(x)\leqslant f_n(x_n)$, which yields $$\sum_n f_n(x)\leqslant \sum_n f_n(x_n) $$ and hence $$\sup_{x\in\mathbb R}\sum_n f_n(x)\leqslant \sum_n\sup_{x\in\mathbb R} f_n(x).$$

For part b, assume both quantities are finite to avoid trivialities. Define \begin{align} A_m &= \sup\left\{\sum_{n=1}^m f_n(x_n):x_n\in\mathbb R\right\}\\ B_m &= \sum_{n=1}^m \sup\{f_n(x_n):x_n\in\mathbb R\}.\end{align} From the equality $\sup S+\sup T=\sup(S+T)$ for bounded above sets $S,T\subset \mathbb R$ we see that $A_m=B_m$ for all $m$. Therefore the equality holds in the limit, so that $$\sup\left\{\sum_n f_n(x_n):x_n\in\mathbb R\right\}= \sum_n\sup\{f_n(x_n):x_n\in\mathbb R\}.$$

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