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: f∼g if {x:f(x)≠g(x)} has measure zero. By construction of the Lebesgue integral, for all 1≤p<∞ we have ‖f‖p=‖g‖p if f∼g; we would like ‖f‖∞ to have the same property. sup|f| won't work because we can have f∼g 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 f∼g 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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