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MATH 421 DISCUSSION WEEK 5 LESSON PLAN CHRIS JANJIGIAN We are now going to move on to continuous functions now. Since we have an exam, I want to focus on doing problems this week. Let's start o by recalling the denition of continuity: Denition. A function f (x) is said to be continuous at a if ∀ > 0, ∃δ > 0 such that for |x − a| < δ |f (x) − f (a)| < Notice that this is basically the same as the denition of a limit, only now we do care what happens at a. Since the proofs here are very similar to the proofs involving limits, I'm just going to dive in and start doing some harder examples. The rst one is a result that is helpful on the review. Theorem. If f (x) and g(x) are continuous functions, then max(f (x), g(x)) is a continuous function. This result has an easy, but slick proof if you know the algebraic identity max(a, b) = a+b+|a−b| , but that's a bit unreasonable to expect from this class. Let's 2 do the nitty gritty proof instead. The basic idea of this proof is that if f (a) > g(a) this is easy, since we know that there is a δ > 0 so that max(f (x), g(x)) = f (x) on |x − a| < δ (why is this true?). It takes only a little more work if f (a) = g(a). Proof. Let > 0 and x a. If f (a) > g(a) then since f (x) − g(x) is continuous, there is δ1 > 0 so that on |x − a| < δ1 we have that Remark. |f (a) − g(a)| 2 In particular then, for |x − a| < δ1 we have that max(f (x), g(x)) = f (x). Then there is δ2 > 0 so that for |x−a| < δ2 we have |f (x)−f (a)| < . Now, for |x−a| < δ = min(δ1 , δ2 ) we have | max(f (x), g(x)) − max(f (a), g(a))| = |f (x) − f (a)| < . The case of g(a) > f (a) is the same, so it only remains to be seen what happens if f (a) = g(a). If f (a) = g(a) then there is δ1 > 0 so that for |x − a| < δ1 |f (x) − f (a)| < and there is δ2 > 0 so that for |x − a| < δ2 we have |g(x) − g(a)| < . Then for |x − a| < δ = min(δ1 , δ2 ) we have |f (x) − g(x)| > | max(f (x), g(x)) − max(f (a), g(a))| < since the above term is either |f (x) − f (a)| or |g(x) − g(a)|. Example. There is no continuous function on f (x) [0, 1] with f (x) = sin( x1 ) on (0, 1]. 1 MATH 421 DISCUSSION WEEK 5 LESSON PLAN Proof. 2 It suces to show that limx→0+ sin( x1 ) does not exist. So suppose that 1 lim+ sin( ) = L x→0 x exists and set = 12 . Then there is δ > 0 so that for 0 < x < δ we have 1 < δ and therefore | sin( x1 ) − L| < 12 . There is N ∈ N suciently large that π +2πN 2 1 < δ as well. Notice that 3π +2πN 2 ! 1 sin = sin 1 π π +2πN 2 ! 1 sin = sin 1 3π +2πN 2 so 3π + 2πN 2 ! 1 2 = | sin 2 | sin 1 ! − L| + | sin π +2πN 2 a contradiction. = −1 ! 1 − sin 1 π +2πN 2 1 + 2πN = 1 |≤ 1 3π +2πN 2 1 1 ! − L| < 3π +2πN 2 1 1 + =1 2 2 It is useful to spend some time thinking about what can go wrong with continuity. Example. Show that if f is continuous on [a, b] then there is a continuous function g on R so that g(x) = f (x) on [a, b]. Show that the previous result need not hold if f is only required to be continuous on (a, b). Solutions: f (a) x ≤ a g(x) = f (x) a ≤ x ≤ b f (b) b ≤ x f (x) = sin( x1 ) is continuous on (0, 1) but there is no limit as x → 0+ so it cannot be continuously extended to [0, 1] or R. Problem. Show that if f (x) and g(x) are continuous, then min(f (x), g(x)) is continuous. MATH 421 DISCUSSION WEEK 5 LESSON PLAN 3 You can do this in either of the ways discussed above, but there is a nice sneaky trick mathematicians like that you can use here: Proof. − max(−f (x), −g(x)) = min(f (x), g(x)) Problem. Show that every continuous function f (x) can be written f (x) = E(x) + O(x) where E(x) is even and continuous and O(x) is odd and contin- uous. Proof. E(x) = f (x)+f (−x) 2 O(x) = f (x)−f (−x) 2 Problem. Show that if f (x) is continuous at a then ∀ > 0 ∃δ > 0 so that for |x − a| < δ and |y − a| < δ we have |f (x) − f (y)| < . Fix > 0. There is δ > 0 so that for |x − a| < δ we have |f (x) − f (a)| < 2 . Then for |x − a| < δ and |y − a| < δ we have Proof. |f (x) − f (y)| ≤ |f (x) − f (a)| + |f (a) − f (y)| < + = 2 2 Problem. If |f (x)| ≤ |x| show that f is continuous at 0. Give an example of such a function which is continuous only at zero. Notice that 0 ≤ |f (0)| ≤ 0 so f (0) = 0. Take > 0 and let |x| < . Then |f (x) − 0| ≤ |x| < and therefore f is continuous at zero. Proof. An example of such a function is ( x if x is rational f (x) = −x if x is irrational