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Let $a,b\in \mathbb{C}$. If $|a+b|=|a|+|b|$ it can be proven that there must be a positive real number $\lambda $ such that $b=\lambda a$ (their size increases optimally when they are angle aligned!). In my case, I have the following:

$$\begin{cases}1-\varepsilon \leq |a|\leq 1 \\ 1-\varepsilon \leq |b|\leq 1 \\ 2-\varepsilon \leq |a+b|\leq 2 \end{cases} \text{and I want to conclude $|a-b|<f(\varepsilon)$ where $f(\varepsilon)\rightarrow 0$ when $\varepsilon\rightarrow 0$}$$

I am not sure if this is true but if it were (and I think it is because $a$ and $b$ must still be angle aligned) it would allow we to finish a proof.

This is mostly useless for the inequality stated above, but here is why I am actually interested in this problem. I am trying to solve this in the case the isometry $J$ is not onto:

I want to prove there is no isometry $J: c\rightarrow c_o$, where $c$ are sequence with limit and $c_o$ sequences whose limit is zero endowed with the sup norm.

Suppose by way of contradiction there is an isometry. Denote by $e$ the sequence of $1$s. In this case: $2=\lVert J(e+e_n)\rVert$ so there is a certain index such that $|J(e+e_n)(t_n)|\geq 2-\varepsilon$. Furthermore, because of $\lVert J(e_n)\rVert=1$ and of the triangle inequality:

$$2-\varepsilon\leq |J(e+e_n)(t_n)|\leq |J(e)(t_n)| +1 \Rightarrow 1-\varepsilon \leq |J(e)(t_n)| $$

If we prove there are infinitely different $t_n$ we are done, because this would contradict $J(e)$ being in $c_o$. The terms should eventually go to zero, and not stay close to $1$...

Suppose by way of contradiction there is $t_n=t_m$. Because we have: $2\geq |J(e+e_n)(t_n)|\geq 2-\varepsilon$ and $1\geq |J(e_n)(t_n)|$ , $1\geq |J(e)(t_n)|$, the only way for this to occur is if: $$1\geq |J(e_n)(t_n)|\geq 1-\varepsilon$$ $$1\geq |J(e)(t_n)|\geq 1-\varepsilon$$ Otherwise the norm gets too small from the usual triangle inequality and we cannot attain $|J(e+e_n)(t_n)|\geq 2-\varepsilon$. With this:

$$1=\lVert J(e_m+e_n)\rVert\geq |J(e_n)(t_n)+J(e_m)(t_m)|=$$ $$|2J(e)(t_n)+(J(e_n-e)(t_n)+J(e_m-e)(t_m))|\geq$$ $$ 2|J(e)(t_n)|-|J(e_n-e)(t_n)|-|J(e_m-e)(t_n)|$$

And for this to be a contradiction I need $|J(e_n-e)(t_n)|$ and $|J(e_m-e)(t_n)|$ to be small. To do this we use the paralelogram law in $\mathbb{C}$ which Aeschepler cleverly thought of, and which would have never occured to me. :)

$$1\geq2|J(e)(t_n)|-\sqrt{(4|J(e_n)(t_n)|^2+4|T(e_m)(t_m)|^2-|J(e_n+e)(t_n)|^2-|J(e_m+e)(t_n)|^2)}\geq 2(1-\varepsilon)-\sqrt{8-2(2-\varepsilon)^2}=2-2\varepsilon-\sqrt{4\varepsilon-2\varepsilon^2}$$

Which is a contradiction for small $\varepsilon$. Therefore, $t_n\not=t_m$ and we get ourselves the sequence $J(e)$ not in $c_o$, which is absurd.

Kadmos
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    You can't have $a=b+f(\varepsilon)$ always true if $a$ and $b$ can vary per the given inequalities for a certain $\varepsilon$. Maybe you mean $|a-b| < f(\varepsilon)$? – aschepler Jan 21 '24 at 00:07
  • Opps. You are absolutely right, @aschepler. Thanks. Noticed it now. – Kadmos Jan 21 '24 at 00:23

1 Answers1

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Using the complex identity $|z|^2 = z \bar z$, we have

$$ |a+b|^2 = a \bar a + \bar a b + a \bar b + b \bar b $$ $$ |a-b|^2 = a \bar a - \bar a b - a \bar b + b \bar b $$

Manipulating these a little,

$$ |a+b|^2 + |a-b|^2 = 2 |a|^2 + 2 |b|^2 $$ $$ |a-b|^2 = 2|a|^2 + 2|b|^2 - |a+b|^2 $$

Therefore (assuming $\varepsilon<1$),

$$ |a-b|^2 \leq 4 - (2-\varepsilon)^2 $$ $$ |a-b| \leq \sqrt{4\varepsilon - \varepsilon^2} $$

aschepler
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  • Paralelogram to the recue! I should have been more up to date with Hilbert spaces to think of this. Absolutely love your approach. :) – Kadmos Jan 21 '24 at 09:10