Variational methods - conjugate of function

  • Thread starter Thread starter braindead101
  • Start date Start date
  • Tags Tags
    Conjugate Function
braindead101
Messages
158
Reaction score
0
Let F:H->R bar be a function and F*:H->R bar its conjugate. Fix aEH and show that the conjugate of the new function G(u)=F(u-a) is G*(u*)=F*(u*)+<a,u>
Verify the case where F:R^2->R, F(x)=1/2(x)^2 and a = (2,-1)

I don't really know how to show this. please help
 
Physics news on Phys.org
First, define your terms. What kind of objects are H and R? Euclidean spaces? General vector spaces? Hilbert spaces? Is "R", at least, the set of real numbers? The fact that you then use R2 as a specific case implies that it is. Is <a, u> the inner product in H? Finally, what, precisely, is your definition of "conjugate"?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top