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 exist?




MY WORK



I believe it does and I want to prove it.



Since f:[a,b]\to\Bbb{R} is continuous, then f is uniformly continuous. Let \epsilon> 0 be given, then \exists\, \delta> such that \forall x,y\in [a,b] with |x-y|<\delta, it implies |f(x)-f(y)|<\epsilon.



Then, for a\leq x\leq b,



\begin{align} f(x)=f(b)+[f(x)-f(b)]\end{align}
\begin{align} |f(x)|\leq |f(b)|+|f(x)-f(b)|\end{align}

\begin{align} \max\limits_{a\leq x\leq b}|f(x)|\leq |f(b)|+\max\limits_{a\leq x\leq b}|f(x)-f(b)|\end{align}



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 x_n \in [a,b] such that \lim_{n \rightarrow \infty} |f(x_n)| = \sup_{[a,b]}|f|. Now get a convergent subsequence x_{n_k} that converges to some x \in [a,b].



By continuity of |f|,



|f(x)| = |f( \lim_{k\rightarrow \infty}x_{n_k})| = \lim_{k\rightarrow \infty} |f(x_{n_k})| = \lim_{n \rightarrow \infty} |f(x_n)| = \sup_{[a,b]}|f|.


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