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

In summary: I'm not sure what the triangle inequality would be for a function that isn't continuous at a point, sorry.In summary, the absolute value of an integral of a function over an interval cannot be bigger than the max absolute value of the integrand on the interval/path. If the function and interval is continuous, the max absolute value is a constant and can be pulled out of the integral.
  • #1
PhDeezNutz
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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##
 
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  • #2
Sorry for the bad LaTeX disaster I think it's fixed now
 
  • #3
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.
 
  • #4
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.
 
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  • #5
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..
 

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

1. What is the Cauchy-Schwarz inequality and how is it used in evaluating integrals of vector fields?

The Cauchy-Schwarz inequality is a mathematical inequality that states the absolute value of the dot product of two vectors is less than or equal to the product of their magnitudes. In evaluating integrals of vector fields, this inequality is used to find an upper bound for the integral, making it easier to calculate.

2. What is the process for using the Cauchy-Schwarz inequality to evaluate integrals of vector fields?

The process involves first finding the dot product of the vector field and a chosen vector, then using the Cauchy-Schwarz inequality to find an upper bound for the integral. This upper bound can then be used to approximate the actual value of the integral.

3. Can the Cauchy-Schwarz inequality be used for any type of vector field?

Yes, the Cauchy-Schwarz inequality can be used for any type of vector field, as long as the dot product can be calculated.

4. Are there any limitations to using the Cauchy-Schwarz inequality in evaluating integrals of vector fields?

One limitation is that the Cauchy-Schwarz inequality can only provide an upper bound for the integral, not the exact value. It also requires the calculation of the dot product, which may not always be feasible for complex vector fields.

5. How does using the Cauchy-Schwarz inequality compare to other methods of evaluating integrals of vector fields?

The Cauchy-Schwarz inequality is a relatively simple and straightforward method for finding an upper bound for integrals of vector fields. However, it may not always provide the most accurate approximation compared to other methods such as numerical integration or using Green's theorem.

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