How to prove $|\int \limits_a^b f(x) dx|\leq\int \limits_a^b |f(x)|dx$ for f continuous? This is a step in the solution of a problem from Mendelson's introduction to topology. This book assumes the reader has only a background in first-year calculus, not measure theory or advanced calculus.
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2At least at an intuitive level, I believe the basic idea follows from the formulation of the definite integral as a Riemann sum, and the use of the triangle inequality ($|a+b| \leq |a|+|b|$ - but imagine that for any number of terms). I haven't checked the formality of it but it's a place to start. – PrincessEev Nov 02 '18 at 08:12
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You can write: $$f(x)=f_+(x)-f_-(x)$$ where $f_+(x)=\max(f(x),0)$ and $f_-(x)=-\min(f(x),0)$
Note that: $$|f(x)|=f_+(x)+f_-(x)$$
The functions $f_+$ and $f_-$ are both continuous and nonnegative so that: $$\left|\int_a^bf(x)dx\right|=\left|\int_a^bf_+(x)dx-\int_a^bf_-(x)dx\right|\leq \int_a^bf_+(x)dx+\int_a^bf_-(x)dx=\int_a^b|f(x)|dx$$
drhab
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If $f$ is complex values the argument is a little more involved. Use of Riemann sums will work in complex case also. – Kavi Rama Murthy Nov 02 '18 at 08:32