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Does someone know if the function resulting of the sum of two locally bounded functions is also locally bounded?

Thanks in advance!

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The answer is yes, and it is straightforward enough for you to try to work out for yourself.

Suppose that $X$ is a metric space and $f,g: X \rightarrow \mathbb{R}$ are two locally bounded functions. This means that for each $x \in X$, there exists an open set $U_f$ containing $x$ on which $f$ is bounded and an open set $U_g$ containing $x$ on which $g$ is bounded. We want to use $U_f$ and $U_g$ to construct an open set on which both $f$ and $g$ are bounded, for then $f+g$ will be bounded as well.

Can you think of how to do this?

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    Thank you Clark. The procedure must be analogous to that one showed by Magidin, i.e., suppose that $U_f+U_g\neq\emptyset$.2011-01-09
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A function $f$ is locally bounded if for every $x$ there exists an open neighborhood $U_x$ of $x$, and a constant $M_x$ (which may depend on $x$) such that $|f(y)|\leq M_x$ for all $y\in U_x$.

Suppose $f$ and $g$ are both locally bounded. Let $x$ be an element of the domain. Then there are open neighborhoods $U_x$ and $V_x$ of $x$, and constants $M_x$ and $N_x$ such that $|f(y)|\leq M_x$ for all $y\in U_x$, and $|g(z)|\leq N_x$ for all $z\in V_x$.

What can you say about $f(y)+g(y)$ if $y\in U_x\cap V_x$?

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    @Humberto: There is no "assumption" needed; it's a *fact*: both $U_x$ and $V_x$ are open neighborhoods of $x$, and therefore, they both contain $x$ and their intersection is open, hence the intersection is an open set that contains $x$.2011-01-09