Evaluating the Integral of a Vector Field Using Cauchy-Schwarz Inequality

[PhDeezNutz]

Homework Statement

Prove that $\left| \int_C f \left(z \right) \, dz \right| \leq \left|f \right|_{max} \cdot L$ where $\left|f \right|_{max}$ is the maximum value o $\left|f(z) \right|$ on the contour and $L$ is the arc length of the contour

Relevant Equations

Any complex valued function is of the form $f \left(z\right) = u \left(x,y\right) + i v\left(x,y\right)$

I think throughout this I need to use the triangle inequality repeatedly

$\left|\vec{a} + \vec{b} \right| \leq \left| \vec{a}\right| + \left|\vec{b} \right|$

Also the Cauchy Schwarz Inequality

$\left| \vec{a} \cdot \vec{b} \right| \leq \left| \vec{a} \right| \left| \vec{b} \right|$

Just for good measure

$\left| f \left(z\right) \right| = \sqrt{u^2 + v^2}$

Here is my attempt (Note:

$\left| \int_{C} f \left( z \right) \, dz \right| \leq \left| \int_C udx -vdy +ivdx +iudy \right|$

$= \left| \int_{C} \left( u+iv, -v +iu \right) \cdot \left(dx, dy \right) \right|$

Here I am going to surround the above expression with another set of absolute value bars

$\leq \left| \left| \int_{C} \left( u+iv, -v +iu \right) \cdot \left(dx, dy \right) \right| \right|$

Appealing to Cauchy-Schwarz
$\leq \left| \left| \int_{C} \left| \left( u+iv, -v +iu \right)\right| \left| \left(dx, dy \right)\right| \right| \right|$

taking the dot product of the newly defined vector field with itself and noting that cross terms vanish

$= \left| \left| \int_{C} \sqrt{u^2 + v^2}\left| \left(dx,dy\right)\right| \right| \right| \leq \left| \left(\sqrt{u^2 + v^2} \right)_{max} \int_{C} \left| \left(dx,dy\right) \right| \right|$

$\leq f_{max} L$

 

[PhDeezNutz]

Sorry for the bad LaTeX disaster I think it's fixed now

 

[PhDeezNutz]

Actually this problem is more trivial than I thought! An absolute value of an integral of a function over an interval (even if the function and interval is complex) cannot be bigger than the max absolute (a real constant) value of the integrand (on the interval/path) integrated over the contour. Since the max absolute value is a constant we can pull it out of the integral. The only thing left in the integrand is the arc length.

 

[Office_Shredder]

For the purposes of a rigorous proof, if you write the integral as the limit of a sum, this is just the triangle inequality, plus the absolute value can pass through the limit since it's a continuous function.

Your intuition in your second post sounds right to me.

 

[WWGD]

Just some technical points: I assume your f is Analytic, or at least continuous over the contour,. Contour is closed and bounded, therefore contact. That way U,V, and therefore f is guaranteed to have a maximum, as a continuous function defined in a compact set..