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Let us consider the sequence space $c_0$ with the equivalent norm $$\Vert x \Vert^2 = \max_{i\ge1} \vert x^i \vert^2 + \sum_{i=2}^{\infty} 2^{-i+1} \vert x^i \vert^2 $$ for $x=(x^1,x^2,\ldots)\in c_0$.

Let us take two sequences $(x_{2n})$ and $(x_{2n+1})$ in $c_0$ such that $x_{2n} \xrightarrow{w} x$ and $x_{2n+1} \xrightarrow{w} y$ and $I_n:=2\Vert x_{2n} \Vert^2 + 2 \Vert x_{2n+1} \Vert^2 - \Vert x_{2n} + x_{2n+1} \Vert^2 =0$ for all $n$. Does it then follow that $x=y$?

This question can be restated simply as follows: Take any $x$ and $y$ in $c_0$ such that $I(x,y):=2\Vert x \Vert^2 + 2 \Vert y \Vert^2 - \Vert x+y \Vert^2 =0$. Does it then follow that $x=y$?

Or even more simply: Take any $x$ and $y$ in $c_0$ such that $\Vert x+y \Vert=\|x\|+\|y\|$. Does it then follow that $x$ and $y$ are linearly dependent?

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  • $\begingroup$ I found your question incomprehensible. So, I have tried to make it comprehensible. Does this editing correspond to your original intent? $\endgroup$
    – Iosif Pinelis
    Commented Jul 19, 2023 at 18:06
  • $\begingroup$ Thank you, sir. But, my query was: for the above-mentioned norm on $c_0$, if we take two subsequences with different weak limits, and then if we make $I_n=0$, we get the weak limit points to be equal or not. And if not, how the sequence will look like? $\endgroup$
    – PPB
    Commented Jul 20, 2023 at 2:56
  • $\begingroup$ (i) What do you mean by "if we make $I_n=0$"? Do you mean here "if we assume that $I_n=0$ for all $n$"? (ii) You say "with different weak limits", and then you say "we get the weak limit points to be equal or not". What could that possibly mean? Overall, do you agree with my edits -- at least with the first paragraph of the edited post? $\endgroup$
    – Iosif Pinelis
    Commented Jul 20, 2023 at 15:24

1 Answer 1

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The answer is no. E.g., let $x_{2n}=x:=(\frac12,\frac12,\frac13,\frac14,\ldots)$ and $x_{2n+1}=y:=(0,\frac12,\frac13,\frac14,\ldots)$ (for all $n$). Then (i) $x_{2n}=x\to x$ and $x_{2n+1}=y\to y$ in any sense and (ii) $\|x_{2n}+x_{2n+1}\|=2\|x_{2n}\|=2\|x_{2n+1}\|$ and hence $I_n=0$ for all $n$. However, $x\ne y$. $\quad\Box$

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  • $\begingroup$ I appreciate your answer. However, I was looking for two sub-sequences and their weak limits to verify. I would be obliged if you could give me a hint on how to proceed to get the answer to be YES for $\sum_{i=1}^\infty$. $\endgroup$
    – PPB
    Commented Jul 20, 2023 at 2:44
  • $\begingroup$ Also, I think the norm will be rotund if we take $\sum_{i=1}^\infty$. And the norm mentioned above is not rotund. $\endgroup$
    – PPB
    Commented Jul 20, 2023 at 2:50
  • $\begingroup$ @PriyankaPriyadarshiniBehera : (i) You wrote "I was looking for two sub-sequences and their weak limits to verify. I would be obliged if you could give me a hint on how to proceed to get the answer to be YES for $\sum_{i=1}^\infty$. I can supply these details if this is all that you need. (ii) What is your definition of rotundity and what is the purpose of your comment about the rotundity? . $\endgroup$
    – Iosif Pinelis
    Commented Jul 20, 2023 at 15:35
  • $\begingroup$ @PriyankaPriyadarshiniBehera : I have added the details on the "No", which completes the answer to your question. $\endgroup$
    – Iosif Pinelis
    Commented Jul 20, 2023 at 16:01
  • $\begingroup$ We can also take $x=(1, 1,0,...)$ and $y=(-1, 1,0,...)$ to check that $I_n=0$ but $x \neq y$. $\endgroup$
    – PPB
    Commented Jul 21, 2023 at 10:27

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