I am currently reading Erwin Kreyszig-Introductory functional analysis with applications. I don't understand the proof below in it. Why is it true that $d(x_n,y_n) \leq d(x_n,y) + d(x,y) + d(y,y_n)$ ? I also don't understand similiar inequality is given by interchanging $x_n$ and $x$ as well as $y_n$ and $y$ ? why is that true ?
[Math] Suppose $x_n \rightarrow x$ and $y_n \rightarrow y$, then $d(x_n,y_n) \rightarrow d(x,y)$
functional-analysismetric-spacesproof-explanation
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I’ll expand on Kreyszig’s argument. I don’t have the book, so it’s possible that some of what I’m about to do duplicates earlier parts of his proof.
We start with two completions of $\langle X,d\rangle$, $\langle\hat X,\hat d\rangle$ and $\langle\tilde X,\tilde d\rangle$. This means that there are $W\subseteq\hat X$ and $\tilde W\subseteq\tilde X$ such that
- $W$ is dense in $\hat X$,
- $\tilde W$ is dense in $\tilde X$,
- there is an isometry $T:X\to\hat X$ such that $T[X]=W$, and
- there is an isometry $\tilde T:X\to\tilde X$ such that $\tilde T[X]=\tilde W$.
We want to show that there is an isometry $h:\tilde X\to\hat X$ such that $h\upharpoonright\tilde W=T\circ\tilde T^{-1}:\tilde W\to W$.
Let $\tilde x\in\tilde X$. $\tilde W$ is dense in $\tilde X$, so for each $n\in\Bbb Z^+$ there is a point $\tilde x_n\in B_{\tilde d}\left(\tilde x,\frac1n\right)\cap\tilde W$. Clearly the sequence $\langle\tilde x_n:n\in\Bbb Z^+\rangle$ converges to $\tilde x$ in $\tilde X$. For each $n\in\Bbb Z^+$ let $x_n=\tilde T^{-1}(\tilde x_n)$ and $\hat x_n=T(x_n)\in\hat X$. $\tilde T$ and $T$ are isometries, so for all $m,n\in\Bbb Z^+$ we have
$$\hat d(\hat x_m,\hat x_n)=\tilde d(\tilde x_m,\tilde x_n)\;.\tag{1}$$
The sequence $\langle\tilde x_n:n\in\Bbb Z^+\rangle$ is convergent in $\tilde X$, so it’s Cauchy, and it follows immediately from $(1)$ that $\langle\hat x_n:n\in\Bbb Z^+\rangle$ is Cauchy in $\hat X$. And $\hat X$ is complete, so $\langle\hat x_n:n\in\Bbb Z^+\rangle$ must converge to some $\hat x\in\hat X$.
Suppose that $\langle\tilde x_n':n\in\Bbb Z^+\rangle$ is another sequence in $\tilde W$ converging to $\tilde x$. For each $\epsilon>0$ there is an $m_\epsilon\in\Bbb Z^+$ such that $\tilde d(\tilde x_n,\tilde x),\tilde d(\tilde x_n',\tilde x)<\frac{\epsilon}2$ whenever $n\ge m_\epsilon$, and hence
$$\tilde d(\tilde x_n,\tilde x_n')\le\tilde d(\tilde x_n,\tilde x)+\tilde d(\tilde x,\tilde x_n')<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$$
whenever $n\ge m_\epsilon$. For each $n\in\Bbb Z^+$ let $x_n'=\tilde T^{-1}(\tilde x_n')\in X$ and $\hat x_n'=T(x_n')\in\hat X$. Again using the fact that $\tilde T$ and $T$ are isometries, we have $\hat d(\hat x_n,\hat x_n')=\tilde d(\tilde x_n,\tilde x_n')$ for each $n\in\Bbb Z^+$, and hence $\hat d(\hat x_n,\hat x_n')<\epsilon$ whenever $n\ge m_\epsilon$.
Let $\epsilon>0$. Since $\langle\hat x_n:n\in\Bbb Z^+\rangle$ converges to $\hat x$ in $\hat X$, there is a $k_\epsilon$ such that $\hat d(\hat x_n,\hat x)<\frac{\epsilon}2$ whenever $n\ge k_\epsilon$. Thus,
$$\hat d(\hat x_n',\hat x)\le\hat d(\hat x_n',\hat x_n)+\hat d(\hat x_n,\hat x)<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$$
whenever $n\ge\max\{k_\epsilon,m_{\epsilon/2}\}$, and $\langle\hat x_n':n\in\Bbb Z^+\rangle$ also converges to $\hat x$.
This shows that the map $h:\tilde X\to\hat X$ defined by $h(\tilde x)=\hat x$ is well-defined: no matter which sequence in $\tilde W$ converging to $\tilde x$ I use to define $\hat x$, I get the same point of $hat X$. Moreover, if $\tilde x\in\tilde W$, then I can let $\tilde x_n=\tilde x$ for each $n\in\Bbb Z^+$ and see that $h(\tilde x)=T\left(\tilde T^{-1}(\tilde x)\right)$, so that $h\upharpoonright\tilde W=T\circ\tilde T^{-1}$, as desired.
Now we have to check that $h$ is an isometry.
Let $\tilde x$ and $\tilde y$ be any two points in $\tilde X$, and let $\langle\tilde x_n:n\in\Bbb Z^+\rangle$ and $\langle\tilde y_n:n\in\Bbb Z^+\rangle$ be sequences in $\tilde W$ converging to $\tilde x$ and $\tilde y$, respectively. Then the sequence $\big\langle\langle\tilde x_n,\tilde y_n\rangle:n\in\Bbb Z^+\big\rangle$ converges to $\langle\tilde x,\tilde y\rangle$ in the space $\tilde X\times\tilde X$. The metric $\tilde d$ is a continuous function from $\tilde X\times\tilde X$ to $\Bbb R$, so the sequence $\langle\tilde d(\tilde x_n,\tilde y_n):n\in\Bbb Z^+\rangle$ converges to $\tilde d(\tilde x,\tilde y)$ in $\Bbb R$.
As before, for each $n\in\Bbb Z^+$ let $x_n=\tilde T^{-1}(\tilde x_n)\in X$, and let $\hat x_n=T(x_n)\in W$, so that $\langle\hat x_n:n\in\Bbb Z^+\rangle$ converges to $h(\tilde x)$ in $\hat X$. Similarly, for each $n\in\Bbb Z^+$ let $y_n=\tilde T^{-1}(\tilde y_n)\in X$ and $\hat y_n=T(y_n)\in W$, so that $\langle\hat y_n:n\in\Bbb Z^+\rangle$ converges to $h(\tilde y)$.
The same argument that I used to show that $\langle\tilde d(\tilde x_n,\tilde y_n):n\in\Bbb Z^+\rangle$ converges to $\tilde d(\tilde x,\tilde y)$ in $\Bbb R$ can be used to show that $\langle\hat d(\hat x_n,\hat y_n):n\in\Bbb Z^+\rangle$ converges to $\hat d(\hat x,\hat y)$ in $\Bbb R$. But $\tilde d(\tilde x_n,\tilde y_n)=d(x_n,y_n)=\hat d(\hat x_n,\hat y_n)$ for each $n\in\Bbb Z^+$, so
$$\hat d\big(h(x),h(y)\big)=\lim_{n\to\infty}\hat d(\hat x_n,\hat y_n)=\lim_{n\to\infty}\tilde d(\tilde x_n,\tilde y_n)=\tilde d(\tilde x,\tilde y)\;,$$
and $h$ is indeed an isometry.
The last step is to show that $h$ maps $\tilde X$ onto $\hat X$. Let $\hat x\in\hat X$ be arbitrary. $W$ is dense in $\hat X$, so there is a sequence $\langle\hat x_n:n\in\Bbb Z^+\rangle$ in $W$ converging to $\hat x$. Use the ideas above to show that the sequence
$$\left\langle\tilde T\left(T^{-1}(\hat x_n)\right):n\in\Bbb Z^+\right\rangle$$
in $\tilde W$ is Cauchy and therefore converges to some $\tilde x\in\tilde X$; then show that $\hat x=h(\tilde x)$.
Just calculate it directly; $$f(x)=\sum_k \xi_k\gamma_k = \sum_k \gamma_k\gamma_k = \sum_k \gamma_k^2=\left(\sum_k \gamma_k^2\right)^{\frac12}\cdot\left(\sum_k \gamma_k^2\right)^{\frac12}=\|x\|\cdot\left(\sum_k \gamma_k^2\right)^{\frac12}$$ since $\|x\|=\left(\sum_k\xi_k^2\right)^{\frac12} = \left(\sum_k\gamma_k^2\right)^{\frac12}$ when the $\xi$ and $\gamma$ are equal. It's not norm $1$, but we can just multiply by a constant for that.
Best Answer
Recall that any metric space $M$ has a metric $d$ defined on it where: $$d:M\times M\to \mathbb R_{\geq 0}$$ and $d$ fulfills the following axioms:
Symmetry: $$d(x,y) = d(y,x)$$
Non-negativity $$d(x,y)\geq 0\qquad d(x,y)=0\iff x = y$$
Triangle inequality: $$d(x,y)\leq d(x,z)+d(z,y)$$ We can apply the triangle inequality twice to $d(x_n,y_n)$ as follows:
$$d(x_n,y_n) \leq d(x_n,y)+d(y,y_n)$$
We also have that $$d(x_n,y)\leq d(x_n,x)+d(x,y)$$
We can combine these to get that:
$$d(x_n,y_n)\leq d(x,x_n)+d(x,y)+d(y,y_n)$$ Here I implicitly used the symmetry condition to say that $d(x,x_n)=d(x_n,x)$. It wasn't required, but it's good to recognize that these are equal.
We can interchange $x_n$ and $x$ and $y_n$ and $y$ because throughout this argument we've been treating them as points, and "forgetting" that they are sequences. We could have used points called $a,b,c,d$ and still obtained a valid relation between them. So, interchange the points is just saying "this argument is true for any $4$ points, and as we don't need to repeat it, we'll just take the end result and modify it to how we want it").