myntfalskj4

Answered

2022-07-16

Suppose that I have an image with $n$ pixels and the following equation,

$\sum _{r\in P}(U(r)-\sum _{s\in N(r)}{w}_{rs}U(s){)}^{2}=0$

$U=(U(1),U(2),...,U(n))$ 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 $3\times 3$ pixels, $N(1)=\text{{2, 4, 5}}$

${w}_{rs}$ is a weighting function where $\sum _{s\in N(r)}{w}_{rs}=1$

Note: I just mentioned that it is an image to better explain the role of the set $N$ and the function ${w}_{rs}$.

Well, it's easy to show that "one" solution for that equation are the scalar multiples of the all-ones vector. I mean, $U=k\cdot (1,1,...,1)$ is clearly a solution.

My question is: there's a way to proof that this is the "only" solution for the above equation?

$\sum _{r\in P}(U(r)-\sum _{s\in N(r)}{w}_{rs}U(s){)}^{2}=0$

$U=(U(1),U(2),...,U(n))$ 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 $3\times 3$ pixels, $N(1)=\text{{2, 4, 5}}$

${w}_{rs}$ is a weighting function where $\sum _{s\in N(r)}{w}_{rs}=1$

Note: I just mentioned that it is an image to better explain the role of the set $N$ and the function ${w}_{rs}$.

Well, it's easy to show that "one" solution for that equation are the scalar multiples of the all-ones vector. I mean, $U=k\cdot (1,1,...,1)$ is clearly a solution.

My question is: there's a way to proof that this is the "only" solution for the above equation?

Answer & Explanation

Charlize Manning

Expert

2022-07-17Added 12 answers

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.

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