Saturday, April 2, 2016

lp spaces - convergence of Lp norm





If I define |f|L=limn|f|Ln. How can I prove that this limit is esssup |f|?


Answer




The main reason to choose esssup|f| over sup|f| is that "functions" in Lp are in fact equivalence classes of functions: fg if {x:f(x)g(x)} has measure zero. By construction of the Lebesgue integral, for all 1p< we have fp=gp if fg; we would like f to have the same property. sup|f| won't work because we can have fg but sup|f|sup|g|, i.e. two functions in the same equivalence class will have different norm. Since esssup|f| "ignores" sets of measure zero, we will have esssup|f|=esssup|g| if fg and hence the norm will be well defined on our equivalence classes.


Edit: I guess the question changed as I was writing this. This is more the reason for esssup rather than the proof requested.


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