- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
1.5 Exercises
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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.
6. All step functions defined on [0, 1].
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Recall from Volume 1, Section 1.8, the definition of a step function:
$\text{Definition of a Step Function.}$ A function $s,$ whose domain is a closed interval $[a, b],$ is called a step function if there is a partition $P = \{x_0, x_1, ..., x_n\}$ of $[a,b]$ such that $s$ is a constant on each open subinterval of $P.$ That is to say, for each $k = 1,\ 2,\ ...,\ n,$ there is a real number $s_k$ such that
\begin{align*}
\\
s(x) &= s_k
\quad
\text{if}
\quad
x_{k - 1} < x < x_k
\end{align*}
Note: At each of the endpoints $x_{k - 1}$ and $x_k$ the function must have some well-defined value, but this need not be the same as $s_k.$
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To verify closure under addition, we note that for any two step functions defined on $[0, 1],$ if $x$ is in the respective partitions of the domains of $f$ and $g$ such that $f(x) = s_k$ and $g(x) = t_k$ then $(f + g)(x) = s_k + t_k = f(x) + g(x).$ If we define the zero function as the step function $f(x) = 0$ for all $x$ in $[0, 1]$ then we can see that the set is a linear space since the remaining axioms are satisfied by all step functions defined on $[0,1]$ being real-valued. $\,\blacksquare$