Proving ||u||_d = 0 and u=0 in Metric Spaces

[Linux]

Let \(\displaystyle (X, d)\) be a metric space, \(\displaystyle AE_0(X) = \{ u : X \rightarrow \mathbb{R} \ : \ u^{-1} (\mathbb{R} \setminus \{0 \} ) \ \ \text{is finite}, \ \sum_{x \in X} u(x)=0 \}\),

for \(\displaystyle x,y \in X, \ x \neq y, \ m_{xy} \in AE_0(X), \ \ m_{xy} (x)=1, \ m_{xy}(y)=-1, \ m_{xy}(z)=0\) for \(\displaystyle z \neq x, y\) and \(\displaystyle m_{xx} \equiv 0\)

for \(\displaystyle u \in AE_0(X), \ ||u||_d = \inf \{ \sum_{k=1}^n |a_k| d(x_k, y_k) \ : \ u= \sum_{k=1}^n a_km_{x_k, y_k}, a_k \in \mathbb{R}, x_k, y_k \in X, n \ge 1 \}\)

How to prove that \(\displaystyle ||u||_d = 0 \ \iff \ u=0\)?

I know it is not so obvious that then all terms in the sum \(\displaystyle \sum_{k=1}^n |a_k| d(x_k, y_k)\) must be zero.

Could you help me out a bit?

Thank you.

 

[Opalg]

Let \(\displaystyle (X, d)\) be a metric space, \(\displaystyle AE_0(X) = \{ u : X \rightarrow \mathbb{R} \ : \ u^{-1} (\mathbb{R} \setminus \{0 \} ) \ \ \text{is finite}, \ \sum_{x \in X} u(x)=0 \}\),

for \(\displaystyle x,y \in X, \ x \neq y, \ m_{xy} \in AE_0(X), \ \ m_{xy} (x)=1, \ m_{xy}(y)=-1, \ m_{xy}(z)=0\) for \(\displaystyle z \neq x, y\) and \(\displaystyle m_{xx} \equiv 0\)

for \(\displaystyle u \in AE_0(X), \ ||u||_d = \inf \{ \sum_{k=1}^n |a_k| d(x_k, y_k) \ : \ u= \sum_{k=1}^n a_km_{x_k, y_k}, a_k \in \mathbb{R}, x_k, y_k \in X, n \ge 1 \}\)

How to prove that \(\displaystyle ||u||_d = 0 \ \iff \ u=0\)?

I know it is not so obvious that then all terms in the sum \(\displaystyle \sum_{k=1}^n |a_k| d(x_k, y_k)\) must be zero.

Could you help me out a bit?

Thank you.

Hi Linux, and welcome to MHB!

The problem of showing that $\|u\|_d = 0$ implies $u=0$ seems difficult, and I have not been able to solve it. See http://mathhelpboards.com/analysis-50/could-you-check-my-solution-13108.html.

 

[Fallen Angel]

Hi,

I have not much time so I may be wrong but doing it fast I think that the set $\{(a_{k},x_{k},y_{k}) \ : \ u=\displaystyle\sum_{k=1}^{n}a_{k}m_{x_{k},y_{k}}\}$ is finite, so the infimum is always reched (i.e., it's a minimum) and your norm is given by series of positive terms, si the minimum being zero implies that all terms are zero.

I will think on it with caution later.

 

[Opalg]

I have not much time so I may be wrong but doing it fast I think that the set $\{(a_{k},x_{k},y_{k}) \ : \ u=\displaystyle\sum_{k=1}^{n}a_{k}m_{x_{k},y_{k}}\}$ is finite, so the infimum is always reached (i.e., it's a minimum) and your norm is given by series of positive terms, so the minimum being zero implies that all terms are zero.

The problem is that there may be different representations of $u$ using points other than those at which $u$ is nonzero.

Suppose for example that the underlying metric space is the real line, and that $u = m_{0,1}$ (so that $u(0) = 1$, $u(1) = -1$ and $u$ is zero at all other points). Then $u$ can also be expressed as $u = m_{0,1/2} + m_{1/2,1}$, or as \(\displaystyle u = \sum_{k=0}^{999}m_{k/1000,(k+1)/1000}.\) It is intuitively clear that $\|u\|_d$ cannot be decreased by such procedures, but a formal proof seems very elusive.

 

[blue_raver22]

To prove that ||u||_d = 0 \ \iff \ u=0, we can start by considering the definition of ||u||_d. From the given definition, we can see that ||u||_d = 0 if and only if the infimum of the set is equal to 0. This means that for any given \epsilon > 0, there exists a sum \sum_{k=1}^n |a_k| d(x_k, y_k) such that ||u||_d < \epsilon.

Now, if ||u||_d = 0, then this means that for any \epsilon > 0, there exists a sum \sum_{k=1}^n |a_k| d(x_k, y_k) such that ||u||_d < \epsilon. However, since ||u||_d = 0, this sum must be equal to 0. This implies that all terms in the sum must be equal to 0, since otherwise, the sum would be greater than 0. In other words, we must have a_k = 0 for all k, which means that u = 0.

Conversely, if u = 0, then all the terms in the sum \sum_{k=1}^n |a_k| d(x_k, y_k) must be equal to 0, since u = \sum_{k=1}^n a_km_{x_k, y_k} and m_{x_k, y_k} \neq 0 for any k. This means that ||u||_d = 0, since the infimum of the set is equal to 0.

Therefore, we have shown that ||u||_d = 0 \ \iff \ u=0, which proves the desired statement.