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| -rw-r--r-- | lecture_3.typ | 58 |
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diff --git a/lecture_3.typ b/lecture_3.typ index 72d0c8b..9318bc0 100644 --- a/lecture_3.typ +++ b/lecture_3.typ @@ -180,10 +180,10 @@ s.t. \ $ forall epsilon > 0 qquad B_epsilon (x) inter A != emptyset and B_epsilo We have shown both directions of containment, and so the sets are equivalent. *Definition* (bounded sets): A set $A in RR^d$ is _bounded_ if $exists R > 0 : A subset.eq B_R^d := {x in RR^d : -norm(x) < R}$. I.e., a set is bounded if there exists a ball that contains the entire set. + norm(x) < R}$. I.e., a set is bounded if there exists a ball that contains the entire set. #pagebreak() -*Definition* (compact sets): A set $A subset.eq RR^d$ is _compact_ if every sequence ${x_n}_(n = 1) ^ (infinity) +*Definition* (compact sets): A set $A subset.eq RR^d$ is _compact_ if every sequence ${x_n}_(n = 1)^(infinity) subset.eq A$ has a convergent subsequence whose limit is in $A$. *Examples* @@ -195,13 +195,13 @@ subset.eq A$ has a convergent subsequence whose limit is in $A$. - $(0, 1)$ (consider $1 / n -> 0 in.not (0, 1)$) *Theorem*: A set $K subset.eq RR^d$ is compact *if and only if* it is closed, and bounded. \ -_Proof_: -- $==>$: we have assumed that $K$ is compact, let us prove that it is closed. Let ${x_n}_(n = 1) ^ (infinity) - subset.eq K$ converging to $x in RR^d$. We need to prove that $x in K$. Since $K$ is compact, $exists {x_(n_k)}_(k=1) ^ - (infinity)$ subsequence of ${x_n}$ that converges to $y in k$. Since ${x_n}$ is convergent, so every subsequence - converges to $ lim_(n -> infinity) x_n & = x \ => y & = x \ => x & in K $ \ - If $K$ were _not_ bounded, then $forall n in NN qquad exists x_n in K : norm(x_n) >= n$. We claim that ${x_n}_(n = 1) ^ - (infinity)$ has no convergent subsequence. Indeed, if $lim_(k -> infinity) x_(n_k) = x$, then $lim_(k -> infinity) +_Proof_: +- $==>$: we have assumed that $K$ is compact, let us prove that it is closed. Let ${x_n}_(n = 1)^(infinity) + subset.eq K$ converging to $x in RR^d$. We need to prove that $x in K$. Since $K$ is compact, $exists {x_(n_k)}_(k=1)^(infinity)$ subsequence of ${x_n}$ that converges to $y in k$. Since ${x_n}$ is convergent, so every subsequence + converges to $ lim_(n -> infinity) x_n & = x \ + => y & = x \ + => x & in K $ \ + If $K$ were _not_ bounded, then $forall n in NN qquad exists x_n in K : norm(x_n) >= n$. We claim that ${x_n}_(n = 1)^(infinity)$ has no convergent subsequence. Indeed, if $lim_(k -> infinity) x_(n_k) = x$, then $lim_(k -> infinity) norm(x_n_k) = norm(x)$. However; $lim_(k -> infinity) norm(x_n_k) >= lim_(k -> infinity) n_k = infinity$. \ - $<==$: Suppose that $K$ is closed and bounded. Let ${x_n} subset.eq K$ be a sequence. Since $K$ is closed, it suffices to prove that ${x_n}$ has a convergent subsequence. We proceed by induction on $d$: The base case is @@ -209,4 +209,44 @@ _Proof_: bounded sequence in $RR$, and by Bolzano-Weierstrass, has a convergent subsequence ${x_d^n_k_l}$, but then ${x^(n_k_l)}$ is a convergent subsequence of ${x^n}$. +== Continuity + +*Definition* Let $A subset.eq RR^d$ and $f: A -> RR^k$. $f$ is continuous at $x in A$ if one of the following holds: ++ For every ${x_n} subset.eq A : x_n -> x in A, lim_(n -> infinity) f(x_n) = f(x)$ ++ $forall epsilon > 0 qquad exists delta > 0 : forall y in B_delta (x) inter A, abs(f(y) - f(x)) < epsilon$ +We say that $f$ is continuous if it is continuous at every $x in A$. + +*Notation*: $C(A) =$ class of continuous functions from $A$ to $RR$ + +*Lemma*: Fix $A subset.eq RR^d, f, g : A -> RR^k$ that are continuous at $x in A$. Then: ++ $f + g$ is continuous at $x$ ++ $f dot g$ is continuous at $x$ ++ If $forall y in A, g(y) != 0 => f / g$ is continuous at $x$. +Note, all these operations are pointwise. \ +_Proof_: ++ Let ${x_n} subset.eq A$ with $x^n -> x$. + $ + (f + g)(x^n) & = f(x^n) + g(x^n) \ + & --> f(x) + g(x) \ + & = (f + g)(x) + $ ++ Left as an exercise for the reader ++ Left as an exercise for the reader + +*Proposition*: $C(A)$ is closed to addition, and multiplication (i.e. it is an "algebra"). \ +_Proof_: Follows immediately from the above lemma. + +*Proposition*: If $f: A -> RR^k, g: RR^k -> RR^d$ are continuous, then $g compose f: A -> RR^d$ is also continuous. \ +_Proof_: Left as an exercise for the reader. + +*Theorem*: Let $K subset.eq RR^d$ be compact, and let $f in C(K)$. Then $f$ is bounded and attains both a maximum and a +minimum. \ +_Proof_: We first prove that $f$ is bounded from above. Let us assume towards contradiction that there is some ${x_n} +subset.eq K$ s.t. $ forall n qquad f(x_n) >= n$. $K$ is compact, so there is a convergent subsequence ${x_n_k}$ with +limit $x in K$. However, then $f(x_n_k) underbrace(-->, k -> infinity) f(x)$. On the other hand $f(x_n_k) >= n_k$, and +so does not converge, which is a contradiction. \ +Since $f$ is bounded, the set ${f(x) : x in k}$ has a supremum $M$. We must show that $exists x in K : f(x) = M$. For +every $n qquad exists x_n in K$ with $f(x_n) >= M - 1 / n$. Since $K$ is compact, there is a convergent subsequence +${x_n_k}$ with limit $x in K$. Since $f$ is continuous, $f(x) = lim_(k -> infinity) f(x_n_k) cases(<= M, >= lim_(n -> +infinity) M - 1 / n) = M$. |