◀ ▲ ▶Branches / Analysis / Proof

Proof

(related to Proposition: Step Functions as a Subspace of all Functions on a Closed Real Interval)

According to the definition of a subspace, we have to verify the following properties:

1 $0\in T[a,b].$

• Obviously, the constant function defined by $f(x)=0$ for all $x\in[a,b]$ is step function.
• Thus, it is an element of $T[a,b].$

2 If $\phi,\psi\in T[a,b]$ then $\phi+\psi\in T[a,b].$

• Let $\phi$ be a step function defined with respect to the partition $$a=x_0 < x_1 < \ldots < x_{n-1} < x_n=b$$ and let $\psi$ be a step function defined with respect to the partition $$a=y_0 < y_1 < \ldots < y_{m-1} < y_m=b.$$
• We define the set union of both partions: $$\{t_0,t_1\ldots t_k\}:=\{x_0, x_1, \ldots, x_n\}\cup\{y_0, y_1, \ldots, y_m\},$$ and observe that $\phi$ and $\psi$ are both constant functions on every real interval $]t_{i-1},t_{i}[$, $i=1,\ldots,k.$
• Thus, $\phi+\psi$ is a step function and therefore $\phi+\psi\in T[a,b].$

3 If $\phi\in T[a,b]$ and $\lambda\in\mathbb R$ then $\lambda\phi\in T[a,b].$

• Obviously, if $\phi$ is a step function, so is $\lambda\phi.$
  ∎

Thank you to the contributors under CC BY-SA 4.0!

Github:

References

Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983