Let f:[a,b]→R be continuous. Does max{|f(x)|:a≤x≤b}
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 |x−y|<δ, it implies |f(x)−f(y)|<ϵ.
Then, for a≤x≤b,
f(x)=f(b)+[f(x)−f(b)]
|f(x)|≤|f(b)|+|f(x)−f(b)|
maxa≤x≤b|f(x)|≤|f(b)|+maxa≤x≤b|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(limk→∞xnk)|=limk→∞|f(xnk)|=limn→∞|f(xn)|=sup[a,b]|f|.
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