Lately I've been studying Banach spaces and isometries, and encountered many explicit isometrys involving $c$, $c_0$, $c^*$, $c_o^*$, $\ell^1$, $\ell^\infty$, etc. (Here $c\subset\ell^\infty$ is the subspace of converging sequences, endowed with the $\|\cdot\|_\infty$-norm and $c_0$ is the space of those sequences converging to $0$.)

It is always pretty easy to show bijection/linearity, and it's allegedly easy to show the isometry as well. Yet I couldn't see any book explicitly proving it.

I'll give here an example that it would mean the world to me if you could show me how to prove that it preservers the norm. It's pretty generic and would help me understand how to do it myself.

Consider $T:\ell^1 \rightarrow c^*$ given by $Tx(y)=x_1 \lim y_n + \sum_{n=2}^{\infty} x_ny_{n-1}$.

It was pretty easy using the triangle inequallity to get $||Tx||\le ||x||_1$, but showing the equallity is the part I'm always struggling with.

Any assistance would be appreciated. Thanks in advance.


The proof that $T$ is an isometry typically consists of two parts:

  1. Show that $\|Tx\|\le \|x\|$ for all $x$
  2. Show that $\|Tx\|\ge \|x\|$ for all $x$

For the classical function and sequence spaces, 1) may involve Hölder's inequality, sometimes in its trivial $(1,\infty)$-form $\left|\sum a_nb_n\right|\le \sup_n|b_n| \sum_n |a_n|$.

Step 2) then involves the equality case of that inequality.

In your situation, $\|Tx\|\le \|x\|$ follows from $\|Tx(y)\|\le \|x\|_1\|y\|_\infty$, that is $$ \left|x_1 \lim y_n + \sum_{n=2}^{\infty} x_ny_{n-1}\right|\le |x_1| \lim |y_n| + \sum_{n=2}^{\infty} |x_ny_{n-1}| \le \|x\|_1\|y\|_\infty$$ To achieve equality above, we would like $y$ that makes everything positive and equal; more precisely:

$$\lim y_n = (\operatorname{sign}x_1)\|y\|_\infty\tag{3}$$ $$x_ny_{n-1} = |x_n| \|y\|_\infty\tag{4}$$

We can shoot for $\|y\|_\infty=1$ to simplify things in (3), (4). Problem is, (4) may force $y_n$ to alternate between $\pm 1$, and then $y$ will not be in $c$. So, in general we cannot attain equality everywhere in (3)-(4).

But we can come arbitrarily close using a truncated sequence: $y_n=\operatorname{sign}x_{n+1}$ for $n\le N$, and $y_n=\operatorname{sign}x_1$ for $n>N$. Then $$\begin{split} x_1 \lim y_n + \sum_{n=2}^{\infty} x_ny_{n-1} &= |x_1| + \sum_{n=2}^{N+1} |x_n| + \sum_{n=N+2}^{\infty} x_n \operatorname{sign}x_1 \\ &\ge |x_1| + \sum_{n=2}^{N+1} |x_n| - \sum_{n=N+2}^{\infty} |x_n| \\& > \|x\|_1-\epsilon \end{split}$$ if $N$ is large enough so that $\sum_{n=N+2}^{\infty} |x_n|<\epsilon/2$.

  • $\begingroup$ Wow, thanks. That was very helpful! Do you have a similar tip for a complex case? (in which sign is problematic) $\endgroup$ – user188400 Dec 12 '14 at 22:24
  • 1
    $\begingroup$ Sign function is defined for complex numbers too: $\operatorname{sign}z = z/|z|$ when $z\ne 0$, and $0$ otherwise. $\endgroup$ – user147263 Dec 12 '14 at 22:28

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