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Proof: (related to "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. $0\in T[a,b].$

  1. 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 $\phi+\psi\in T[a,b].$
  1. 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 $\lambda\phi.$
q.e.d

| | | | created: 2020-01-09 06:43:22 | modified: 2020-01-09 06:43:22 | by: bookofproofs | references: [581]

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Bibliography (further reading)

[581] Forster Otto: “Analysis 1, Differential- und Integralrechnung einer Veränderlichen”, Vieweg Studium, 1983