In Exercises 1 through 28, determine whether each of the given sets is a real linear space, if addition and multiplication by real scalars are defined in the usual way. For those that are not, tell which axioms fail to hold. The functions in Exercises 1 through 17 are real-valued. In Exercises 3, 4, and 5, each function has domain containing 0 and 1. In Exercises 7 through 12, each domain contains all real numbers.
10. All bounded functions.
Recall from Section 1.16 (Volume 1) that a function bounded on an interval $[a, b]$ is a function for which there exists a number $M > 0$ such that $-M \leq f(x) \leq M$ for every $x$ in $[a, b].$
To verify closure under addition, let $f$ and $g$ be bounded by $M$ and $N$ respectively, where $M>0$ and $N>0.$ Then, we have $-M \leq f(x) \leq M$ and $-N \leq g(x) \leq N,$ which means
\begin{align*}
-M - N \leq f(x) + g(x) \leq M + N
\end{align*}for all $x.$ But by definition, $f(x) + g(x) = (f + g)(x),$ which means that $f + g$ is bounded by $M + N.$ And since $f$ and $g$ are arbitrary, this shows that the set of all bounded functions satisfies closure under addition.
To show closure under multiplication by real numbers, let $a$ be some real constant, and let $f(x)$ be bounded by some $M > 0.$ Then, we can multiply the inequality $-M \leq f(x) \leq M$ by $a$ to get
\begin{align*}
-aM \leq af(x) \leq aM
\end{align*}for all $x,$ which means $af$ is bounded. This can also be extended to verify the associative law, distributive laws, and the existence of identity.
The zero function $f(x) = 0$ is bounded for any $M >0,$ and multiplying any bounded $f$ by $-1$ leaves the inequality unchanged, thus verifying the existence of negatives. The commutative and associative laws of addition can be verified by noting that all bounded $f$ are real-valued and any $M > 0$ bounding a $f$ is also a real number.
Thus, we have shown that the set of all bounded functions is a linear space. $\,\blacksquare$
