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math104-s22:notes:lecture_18

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math104-s22:notes:lecture_18 [2022/03/16 21:49]
pzhou created
math104-s22:notes:lecture_18 [2022/03/16 22:37] (current)
pzhou
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 ====== Lecture 18: Sequence of functions ====== ====== Lecture 18: Sequence of functions ======
 "How to measure the distance between two functions?" "How to measure the distance between two functions?"
 +
 +====== Sequence of functions ======
 +Just as you can have a sequence 
 +  * of number in $\R$
 +  * of vectors in $\R^n$
 +  * of points in a general metric space $X$. 
 +You can have a sequence of functions. $f_n(x)$ 
 +
  
 ===== The space of functions ===== ===== The space of functions =====
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 Q: can you find a metric on the space of functions $V$ so that metric convergence means pointwise convergence? (No, you but you can still define a topology on $V$ so that convergence in that topology means pointwise convergence) Q: can you find a metric on the space of functions $V$ so that metric convergence means pointwise convergence? (No, you but you can still define a topology on $V$ so that convergence in that topology means pointwise convergence)
 +
 +In general, if $M = \{f| f: X \to Y, f(X) \text{bounded}\}$ is the space of maps with bounded images, then for any $f, g \in M$, we can define $d_\infty(f,g) = \sup_{x} d_\infty(f(x), g(x))$. 
 +
 +==== Pointwise Convergence vs Uniform Convergence ====
 +  * The running bump $f_n = 1_{[n,n+1]}(x)$, which is 1 on $[n,n+1]$ and $0$ elsewhere. They converge to $0$ pointwise, but not uniformly. 
 +  * The shrinking and rising bump $f_n = n 1_{(0,1/n)}$ converge to $0$ pointwise, but not uniformly. 
 +
 +===== Uniform Convergence Preserves Continuity =====
 +Thm: If $f_n: \R \to \R$ are continuous and bounded,  and $f_n \to f$ uniformly, then $f$ is continuous. 
 +
 +Proof: we need to show that, for any $x \in \R$, for any $\epsilon>0$, there is $\delta > 0$, such that for any $x'$ with $|x' - x| < \delta$, we have $|f(x') - f(x)|<\epsilon$. 
 +
 +First, we choose $n$ large enough, such that $d_\infty(f_n, f) < \epsilon/3$. Then, for $f_n$ and $x$ and $\epsilon/3$, we find $\delta$, such that any $|x'-x|<\delta$ implies $|f_n(x) - f_n(x')|<\epsilon/3$. Finally, if any $|x'-x|<\delta$, then 
 +$$ |f(x) - f(x')| \leq |f(x) - f_n(x)| + |f_n(x) - f_n(x')| + |f_n(x') - f_n(x)| < \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon. $$
 +Done
 +
 +  
 +===== Examples =====
 +  * Devil's staircase
 +  * power series
 +  * Weierstrass M-test
 +
 +
 +
  
  
math104-s22/notes/lecture_18.1647492557.txt.gz · Last modified: 2022/03/16 21:49 by pzhou