Answer to Question #249064 in Real Analysis for dasuuu

Question #249064

Show that π‘ˆ(βˆ’π‘“, 𝑝) = βˆ’πΏ(𝑓, 𝑝) and 𝐿(βˆ’π‘“, 𝑝) = βˆ’π‘ˆ(𝑓, 𝑝). 


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Expert's answer
2021-10-11T16:03:03-0400

π‘ˆ(𝑓, 𝑝) and 𝐿(𝑓, 𝑝) are upper and lower Riemann sums for the partition p.

π‘ˆ(𝑓,𝑝)=βˆ‘MiΞ”xi, L(𝑓,𝑝)=βˆ‘miΞ”xiπ‘ˆ(𝑓, 𝑝)=\sum M_i\Delta x_i,\ L(𝑓, 𝑝)=\sum m_i\Delta x_i

Mi=sup{f(x):xiβˆ’1≀x≀xi}, mi=inf{f(x):xiβˆ’1≀x≀xi}M_i=sup\{f(x):x_{i-1}\le x \le x_i\},\ m_i=inf\{f(x):x_{i-1}\le x \le x_i\}


sup{βˆ’f(x):xiβˆ’1≀x≀xi}=βˆ’inf{f(x):xiβˆ’1≀x≀xi}=βˆ’misup\{-f(x):x_{i-1}\le x \le x_i\}=-inf\{f(x):x_{i-1}\le x \le x_i\}=-m_i

U(βˆ’f,p)=βˆ’βˆ‘miΞ”xi=βˆ’L(f,p)U(-f,p)=-\sum m_i\Delta x_i=-L(f,p)


inf{βˆ’f(x):xiβˆ’1≀x≀xi}=βˆ’sup{f(x):xiβˆ’1≀x≀xi}=βˆ’Miinf\{-f(x):x_{i-1}\le x \le x_i\}=-sup\{f(x):x_{i-1}\le x \le x_i\}=-M_i

L(βˆ’f,p)=βˆ’βˆ‘MiΞ”xi=βˆ’U(f,p)L(-f,p)=-\sum M_i\Delta x_i=-U(f,p)


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