Fundamental complex inequality

In summary, to prove the inequality for complex valued functions, we can use the expression Re[e^{-i \theta} \int_a^b f(t) dt] and set \theta = arg(\int_a^b f(t) dt) to reduce it to the absolute value. This is achieved by rotating the complex number \int_a^b f(t) dt to align with the real axis. This approach is a clever and unique way of proving the inequality.
  • #1
Zorba
77
0
To proove the inequality:

[tex]\left | \int_a^b f(t) dt \right | \le \int_a^b | f(t) | dt[/tex]

for complex valued [tex]f[/tex], use the following:

[tex]\textrm{Re}\left [ e^{-\iota\theta}\int_a^b f(t) dt\right ] = \int_a^b \textrm{Re}[e^{-\iota\theta}f(t)] dt \le \int_a^b | f(t) | dt[/tex]

and then if we set:

[tex]\theta = \textrm{arg}\int_a^b f(t) dt[/tex]

the expression on the left reduces to the absolute value [tex]\square[/tex]So the last part, where he says that it reduces to the modulus by using that substitution of theta, I don't get it... how does that reduce to the modulus? :smile:
 
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  • #2
Let [tex]z=\int_a^b f(t)dt[/tex]

Let [tex]\theta=arg(z)[/tex]

Then we can write [tex]z=Re^{i \theta}[/tex] where [tex]R=|z|[/tex]

Now look at [tex]e^{-i \theta} z = e^{-i \theta} R e^{i \theta} = R = |z|[/tex]
 
  • #3
Excellent, thanks for the quick reply. :smile:
 
  • #4
To give a bit more of an exposition, you can think of multiplying a complex number by [tex]e^{i \theta}[/tex] as being the same thing as rotating it in the complex plane by an angle of [tex] \theta[/tex]. In this case what they did was rotate the number so that it lined up with the real axis

It's pretty clever; I've never seen this proof done like this before
 

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