The space $πΏ^π(π) \cap πΏ^\infty(π)$, $p<\infty$, with the norm $||π||_{πΏ^π \cap πΏ^\infty}=||π||_π+||π||_\infty$ is a Banach space. I imagine that if we remove the norm $||π||_\infty$ and leave only the $||π||_π$, the space $πΏ^π(π) \cap πΏ^\infty(π)$ with norm $||π||_{πΏ^π \cap πΏ^\infty}=||π||_π$ is not Banach, but how can we show this, what is the counterexample? Or is he Banach? Here I am not assuming that $X$ has finite measure.
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2$\begingroup$ Clearly $L^p(X) \cap L^\infty(X)$ is a subspace of $L^p(X)$ with the norm $\|-\|_p$. Do you know the criterion for when a subspace of a Banach space $X$ is itself a Banach space in the (restriction of the) norm of $X$? $\endgroup$β While I AmCommented Jul 5 at 20:46
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$\begingroup$ @WhileIAm no, the criterion I know is that a subspace of a space is Banach if and only if it is closed. $\endgroup$β IlovemathCommented Jul 5 at 22:06
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1$\begingroup$ That's the criterion I was thinking of. You can see by @Steen82 's argument that the subspace is not closed in the $p$-norm. $\endgroup$β While I AmCommented Jul 6 at 0:36
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$\begingroup$ In general a subspace $Y$ of a Banach space $X$ is complete iif $Y$ is closed. In particular if $Y$ is complete and dense then $Y=X.$ $\endgroup$β Ryszard SzwarcCommented Jul 6 at 16:14
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Let $f\geq0$ be an unbounded $L^p$ function. Let $f_n(x)=\min\{f(x),n\}.$ Then $f_n$ is in $L^{\infty}$ and $f_n\to f$ in with respect to the $p-$ norm.
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1$\begingroup$ Since $f$ is in $L^p$ and $f_n \leq f$ it follows that $f_n$in $L^p.$ $\endgroup$β Steen82Commented Jul 5 at 23:09
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1$\begingroup$ @Ilovemath This is a valid answer. $(f_n)$ is Cauchy but not convergent in your space. $\endgroup$β geetha290krmCommented Jul 5 at 23:11
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1$\begingroup$ Got it, I agree. Interestingly, if you limit the function norm this becomes Banach math-stackexchange-com.translate.goog/questions/1672875/β¦ $\endgroup$β IlovemathCommented Jul 5 at 23:13