Linear Continuous AE(u): Unique Solution for Functional Analysis Problem

[Jakob1]

Hi. I've just solved a problem from functional analysis and I would be very glad if you checked if everything is all right:

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

The fact that $||m_{xy}|| = d(x,y)$ follows straight from the definition of $|| \cdot||_d$, right?

We denote the completion of \(\displaystyle AE_0 (X) \ \text{by} \ AE(X)\).

Now, the main problem:

We are given a Banach space \(\displaystyle (E, || \cdot ||)\) and a Lipschitz function \(\displaystyle u: X \rightarrow E\).

I need to show that there exists a unique linear continuous \(\displaystyle AE(u) : AE(X) \rightarrow E\) s.t. \(\displaystyle AE(u) (m_{xy}) = u(x)-u(y), \ x,y \in X\).

I think it follows from this: \(\displaystyle AE(u)(f) = AE(u) (\sum_{k=1}^n a_km_{x_k, y_k}) = \sum_{k=1}^n a_k AE(u)( m_{x_k, y_k}) = \sum_{k=1}^n a_k(u(x_k) - u(y_k))\) - this determines the value uniquely.

As for existence, let's define the function \(\displaystyle AE(u)\) as above. We need to check that its values are in \(\displaystyle E\). But we know that it is si, because \(\displaystyle u\) is Lipschitz, and \(\displaystyle \exists Lip(u) \ge 0 : \forall x, y \in X : ||u(x) - u(y)|| \le Lip(u) d(x,y)\) (or \(\displaystyle AE(u)\) is linear and continuous, so it has to be bounded (?)).

Next, I need to prove that \(\displaystyle ||AE(u)|| = Lip(u)\).

\(\displaystyle ||AE(u)|| = \sup \{ ||AE(u)(f)|| \ : \ ||f|| = 1 \}\).

\(\displaystyle ||AE(u)(f)|| = || \sum_{k=1}^n a_k (u(x_k) - u(y_k)) || \le || \sum_{k=1}^n a_k Lip(u) d(x_k, y_k) || = Lip(u) ||f|| = Lip(u)\).

And for \(\displaystyle f= a m_{xy}, a \in \mathbb{R}\) we have \(\displaystyle \||AE(u)(f)|| = Lip(u) \).

Is that correct?

I'll be very grateful for all your insight.

Thanks.

 

[Euge]

Hi Jakob,

It looks like you have the general idea, but there are a couple of issues with the argument. Since $||\cdot ||_d$ has been defined on $AE_0(X)$ and $AE(X)$ is the $||\cdot ||_d$-completion of $AE_0(X)$, to define $AE(u)$ properly, you must first define $AE(u)$ as a mapping from $AE_0(X)$ to $E$ and prove that it is linear and continuous. Then you can claim that $AE(u)$ uniquely extends to a continuous linear mapping from $AE(X)$ to the completion of $E$ (i.e., $E$ itself, since $E$ is Banach).

The definition you have set up for $AE(u)$ makes sense as a mapping from $AE_0(X)$ to $E$. I leave it to you to verify linearity of $AE(u)$. I'll verify continuity of $AE(u)$; given $f = \sum_{k = 1}^n a_k m_{x_ky_k}\in AE_0(X)$,

\(\displaystyle \|AE(u)(f)\|_E = \left\|\sum_{k = 1}^n a_k [u(x_k) - u(y_k)]\right\|_E\)

\(\displaystyle \le \sum_{k = 1}^n |a_k|||u(x_k) - u(y_k)||_E\quad (\text{by the triangle inequality})\)

\(\displaystyle \le \mathcal{Lip}(u)\sum_{k = 1}^n |a_k|d(x_k,y_k)\quad (\text{since $u$ is Lipschitz})\)

\(\displaystyle = \mathcal{Lip}(u)\|f\|_d.\)

 

[Jakob1]

Thank you for all your help, Euge.
I wonder if you could also take a look at my proof of the fact that \(\displaystyle ||m_{xy}|| = d(x,y)\).

\(\displaystyle AE_0(X)\) is a linear space and for a fixed \(\displaystyle x_0 \in X\), \(\displaystyle m_{x x_0}, \ x \neq x_0\), form its basis.

Any real valued function \(\displaystyle f\) on \(\displaystyle X\) such that \(\displaystyle f(x_0)=0\) defines a linear map on \(\displaystyle X\): \(\displaystyle F: \ AE_0(X) \ni u= \sum_{k=1}^n a_k m_{x_0x_k} \rightarrow \sum_{k=1}^n a_k f(x_k)\).

Then if we take \(\displaystyle f(z)=d(z,y)\) (it satisfies the condition: \(\displaystyle f(y)=0\)), we let \(\displaystyle F \) be the corresponding linear functional, with \(\displaystyle x_0\) replaced by \(\displaystyle y\).

Given the representation of \(\displaystyle m_{xy} = \sum_{k=1}^n a_k m_{x_ky_k} \), we have \(\displaystyle d(x,y) = f(x) = F(m_{xy}) = \sum_{k=1}^n a_k F(m_{x_ky_k})\).

So \(\displaystyle d(x,y) \le \sum_1^n |a_k| |d(x_k, y) - d(y_k, y)| \le \sum_1^n |a_k| d(x_k, y_k)\)

Taking \(\displaystyle inf\) over all such representations, we get \(\displaystyle ||m_{xy}|| \ge d(x,y)\). We get the reverse inequality by plugging $n=1, a_1 = 1, x_1 = x, y_1 = y$ into the definition of the norm.

However, I have trouble proving that we can indeed write any \(\displaystyle u \in AE_0(X)\) as \(\displaystyle u= \sum_{k=1}^n a_k m_{x_0x_k}\), that is with fixed \(\displaystyle x_0\).

Moreover, could you help me prove that \(\displaystyle ||u|| = 0 \iff u=0\)?