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Let $1<p<\infty$, and let $x$ and $y$ be vectors in $l_p$ with $\left \|x \right \|=\left \|y \right \|=1$ and $\left \|x +y \right \|=2$, how to prove $x=y$?

I know how to prove for $p=2$ using paralleogram law, but for $p\neq 2$, $l_p$ has no inner product, so I don't know how to prove those cases.

I know this is not true for $l^1$ and $l^\infty$, I would be grateful if you could give me a hint on the difference between general $l^p$ and these two spaces which could help me come up with a solution myself.

JonathanZ
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HIH
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    https://math.stackexchange.com/questions/80139/why-is-the-l-p-norm-strictly-convex-for-1p-infty – Evangelopoulos Foivos Jun 23 '24 at 15:19
  • @EvangelopoulosFoivos Like we can use Minkowski equality case for $L^p$ as if $l^p$ is $L^p$ space with some weird measure which turns integral into sum? – HIH Jun 23 '24 at 15:24
  • @HIH Yes, one can. Namely, we consider $L^p(\mathbb{N}, dm)$ where $dm$ is the counting measure (i.e. $dm(S)$ is the cardinality of $S$). Then we get for $f\in L^p(\mathbb{N}, dm)$ $$ \int_{\mathbb{N}} \vert f(n) \vert^p dm(n)= \sum_{n\in \mathbb{N}} \vert f(n) \vert^p. $$ Thus, we can identify $L^p(\mathbb{N}, dm)$ and $\ell^p(\mathbb{N})$. – Severin Schraven Jun 30 '24 at 12:18
  • It is actually an equality. Simply because a sequence is a function from $\mathbb{N}$ to the space of where your coefficients live. – Severin Schraven Jun 30 '24 at 12:22
  • If you are looking for a more general class of spaces with the property you describe, you can google uniformly convex spaces. The Milman-Pettis theorem is telling us that every uniformly convex Banach space is reflexive (i.e. the map $X\rightarrow X^{**}, x \mapsto (\varphi\mapsto \varphi(x))$ is an isomorphism of Banach spaces). However, $\ell^1$ and $\ell^\infty$ are not reflexive. One could say that the underlying geometry of the space is very different for $p\in {1,\infty}$. – Severin Schraven Jun 30 '24 at 12:29

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