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While doing one homework problem i encounter a difficulty.

Is it true, if X is Banach space equipped with $||.||_1$ and $||.||_2$. Suppose $||x_n-x||_1\to0$ and $||x_n-y||_2\to0$ then x=y? Here X is subspace of $L^p$ and $L^q$ but p and q are not conjugate. $||.||_1$ is $L^p$ norm and $||.||_2$ is $L_q$ is norm

I can show $x=y$ a.e but i don't know how to show x=y.

Any help please....

Toeplitz
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  • Notion a.e. is not defined for an abstract Banach space. You certainly know something more than you are telling us.. – A.Γ. Oct 12 '15 at 01:34
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    Also, I assume you mean $|x_n - x|_1 \to 0$, and not $=0$? – Prahlad Vaidyanathan Oct 12 '15 at 02:07
  • yes you are correct, than u – Toeplitz Oct 12 '15 at 02:19
  • @A.G. I edited my qurestion. yep I agree with you – Toeplitz Oct 12 '15 at 02:26
  • I'm confused. First, is there a difference between $L_q$ and $L^q$ ? Because you are using both notations. Second, usually $L^p$ stands for a space of functions, with the norm being in terms of an integral. Your notation suggests you are working in space of sequences, which is often denoted by $\ell ^p$. Finally, what you mean by $| . |_1$ is $L^p$ norm ? If there is any norm I would say that is the $L^p$ norm, it is the norm $| . |_p$. – Integral Oct 12 '15 at 03:29
  • $x=y$ almost everywhere is the best we can expect. – saz Oct 12 '15 at 05:51

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In general, this is not true. In fact, it is possible to construct an example with $x \ne y$, see the related post https://math.stackexchange.com/a/426499/58577.

However, in your situation, life is a little bit easier as pointed out by @saz. Every sequence converging in $L^*$ with $*=p$ or $*=q$ has a subsequence, which converges pointwise a.e. This can be used to show $x = y$ a.e. and this is $x = y$.

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gerw
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