Tuesday, July 2, 2019

real analysis - Let f:[a,b]toBbbR be continuous. Does max|f(x)|:aleqxleqb exist?



Let f:[a,b]R be continuous. Does max{|f(x)|:axb}

exist?




MY WORK



I believe it does and I want to prove it.



Since f:[a,b]R is continuous, then f is uniformly continuous. Let ϵ>0 be given, then δ> such that x,y[a,b] with |xy|<δ, it implies |f(x)f(y)|<ϵ.



Then, for axb,



f(x)=f(b)+[f(x)f(b)]


|f(x)||f(b)|+|f(x)f(b)|


maxaxb|f(x)||f(b)|+maxaxb|f(x)f(b)|



I am stuck at this point. Please, can anyone show me how to continue from here?


Answer



Because [a,b] is compact, every sequence in [a,b] has a subsequence that limits to a point in [a,b]. Pick a sequence xn[a,b] such that limn|f(xn)|=sup[a,b]|f|. Now get a convergent subsequence xnk that converges to some x[a,b].



By continuity of |f|,



|f(x)|=|f(limkxnk)|=limk|f(xnk)|=limn|f(xn)|=sup[a,b]|f|.


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