Suppose that I have an image with n pixels and the following equation, ∑ r ∈ P ( U ( r ) − ∑ s ∈ N ( r ) w r s U ( s ) ) 2 = 0 U = ( U ( 1 ) , U ( 2 ) , . . . , U ( n ) ) are the variables; P is the set of all the n pixels; N ( r ) is the set of all neighboring pixels of r. For example, for an image with 3 × 3 pixels, N ( 1 ) = {2, 4, 5} w r s is a weighting function where ∑ s ∈ N ( r ) w r s = 1Note: I just mentioned that it is an image to better explain the role of the set N and the function w r s .Well, it's easy to show that "one" solution for that equation are the scalar multiples of the all-ones vector. I mean, U = k ⋅ ( 1 , 1 , . . . , 1 ) is clearly a solution.My question is: there's a way to proof that this is the "only" solution for the above equation?

Question

Suppose that I have an image with   n pixels and the following equation,      ∑          r      ∈      P        (  U  (  r  )  −      ∑          s      ∈      N      (      r      )            w          r      s        U  (  s  )      )    2    =  0  U  =  (  U  (  1  )  ,  U  (  2  )  ,  .  .  .  ,  U  (  n  )  ) are the variables;  P is the set of all the   n pixels;  N  (  r  ) is the set of all neighboring pixels of   r. For example, for an image with   3  ×  3 pixels,   N  (  1  )  =  {2, 4, 5}      w          r      s       is a weighting function where       ∑          s      ∈      N      (      r      )            w          r      s        =  1Note: I just mentioned that it is an image to better explain the role of the set   N and the function       w          r      s      .Well, it&#039;s easy to show that &quot;one&quot; solution for that equation are the scalar multiples of the all-ones vector. I mean,   U  =  k  ⋅  (  1  ,  1  ,  .  .  .  ,  1  ) is clearly a solution.My question is: there&#039;s a way to proof that this is the &quot;only&quot; solution for the above equation?

Charlize Manning · Accepted Answer

Hint. Suppose you have a solution that is not constant. Then the maximum of the solution occurs at (at least one) location s, with a strictly smaller value in (at least one) neighbor. Can you get a contradiction? You can if the weights are all positive, but not in general.

Suppose that I have an image with n pixels and the following equation, <munder> ∑

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New Questions in Algebra I