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:

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

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

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

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

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

Now, the main problem:

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

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

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

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

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

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

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

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

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 $f = \sum_{k = 1}^n a_k m_{x_ky_k}\in AE_0(X)$,

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

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

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

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

 

[Jakob1]

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

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

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

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

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

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

Taking $$inf$$ over all such representations, we get $$||m_{xy}|| \ge d(x,y)$$. 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 $$u \in AE_0(X)$$ as $$u= \sum_{k=1}^n a_k m_{x_0x_k}$$, that is with fixed $$x_0$$.

Moreover, could you help me prove that $$||u|| = 0 \iff u=0$$?