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(1) ##f## is continuous
(2) ##f## is defined on a sequentially compact interval ##K##
Prove that ##f## is uniformly continuous.
Suppose ##f## is not uniformlycontinuous. Then there is ##\epsilon>0## such that for any ##\delta>0##, there is ##x,y\in K## such that if ##xy<\delta##, ##f(x)f(y)\geq \epsilon##.
Choose ##\delta=1##. Then there is a pair of real numbers which we will denote as ##x_1,y_1## such that if ##x_1y_1<\delta##, uniformcontinuity fails to hold.
Now choose ##\delta=\frac{1}{2}## and find a pair of real numbers ##x_2,y_2## that also fail this condition. Choose a sequence of real numbers in this fashion and denote them as: ##\{x_n\},\{y_n\}##
These are sequences contained in ##K##. Therefore, for ##\{x_n\}##, there exists a subsequence ##x_{n_k}\rightarrow z\in K##.
Moreover, if ##\epsilon>0##, choose ##N_\epsilon\in\mathbb{N}## that is at least ##\frac{1}{\epsilon}##. Then if ##k\geq N_\epsilon##, ##n_k\geq n_{N_\epsilon}##. Note that ##n_{N_\epsilon}\geq \frac{1}{\epsilon}##. Hence, by construction:
##x_{n_k}y_{n_k}<\frac{1}{n_k}\leq \epsilon##
By the Triangle Inequality, ##y_{n_k}\rightarrow z##.
By the Triangle Inequality, ##f(y_{n_k})f(x_{n_k})<\epsilon##, which contradicts the initial claim that every pair of images of the points ##x_i,y_i## constructed thus are not Cauchy. Hence, ##f## must be uniformlycontinuous.
Choose ##\delta=1##. Then there is a pair of real numbers which we will denote as ##x_1,y_1## such that if ##x_1y_1<\delta##, uniformcontinuity fails to hold.
Now choose ##\delta=\frac{1}{2}## and find a pair of real numbers ##x_2,y_2## that also fail this condition. Choose a sequence of real numbers in this fashion and denote them as: ##\{x_n\},\{y_n\}##
These are sequences contained in ##K##. Therefore, for ##\{x_n\}##, there exists a subsequence ##x_{n_k}\rightarrow z\in K##.
Moreover, if ##\epsilon>0##, choose ##N_\epsilon\in\mathbb{N}## that is at least ##\frac{1}{\epsilon}##. Then if ##k\geq N_\epsilon##, ##n_k\geq n_{N_\epsilon}##. Note that ##n_{N_\epsilon}\geq \frac{1}{\epsilon}##. Hence, by construction:
##x_{n_k}y_{n_k}<\frac{1}{n_k}\leq \epsilon##
By the Triangle Inequality, ##y_{n_k}\rightarrow z##.
By the Triangle Inequality, ##f(y_{n_k})f(x_{n_k})<\epsilon##, which contradicts the initial claim that every pair of images of the points ##x_i,y_i## constructed thus are not Cauchy. Hence, ##f## must be uniformlycontinuous.
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