Theorem 1.21. A monotonic function f : [a b] → R on a compact interval is. Riemann integrable. Proof. Suppose that f is
A function f : A → R is continuous on A if and only if f. −1(V ) is open in prove that a continuous function maps compact intervals to compact intervals.
This completes the proof. Theorem 1.3. Let f be a bounded function from a bounded closed interval [a b] to IR. If the set of
To explain this point in more detail note that if a function f is continuous on Then f(I) is an interval. Proof. Suppose that f(I) is not a interval. Then ...
If f is a continuous function on the closed interval [a b]
show that f(x) is continuous on the interval (−∞∞)
Theorem 21. A continuous function on a compact metric space is bounded and uniformly continuous. Proof. If X is a compact metric space and f :
If a sequence (fn) of continuous functions fn : A → R converges uniformly on A ⊂ R to f : A → R then f is continuous on A. Proof. Suppose that c ∈ A and ϵ
We begin by proving a special case. Theorem 8.32 (Rolle). Suppose that f : [a b] → R is continuous on the closed
Theorem 3.4. If f and g are absolutely continuous on the interval I then f + g is absolutely continuous on I. Proof. Let ϵ
a. Show that the absolute value function. F(x) =
Prove that the function g so define on X is the desired extension of f. 14. Let I = [01] be the closed unit interval. Suppose f is continuous mapping of I into
7. Continuous Functions. Example 7.3. If f : (a b) ? R is defined on an open interval
If f : (a b) ? R is defined on an open interval
If f is continuous on the interval I then it is bounded and attains its maximum and minimum values on each subinterval
Surely this holds also for complex functions. 8. Suppose f (x) is continuous on [a
This completes the proof. Theorem 1.3. Let f be a bounded function from a bounded closed interval [a b] to IR. If the set of
x0 /? E. Show that there is an unbounded continuous function f : E ? R. (b) Show that if f1f2
Proof: It is clear that fn (x) = xn converges NOT uniformly on [01] since each term of {fn (x)} is continuous on [0
?1 at f(c). One can show that if f : I ? R is a continuous and one-to-one function on an interval I then f is strictly monotonic and f. ?1 is also.
It is obvious that a uniformly continuous function is continuous: if we can nd a which works for allx0 we can nd one (the same one) which worksfor any particularx0 We will see below that there are continuous functionswhich are not uniformly continuous Example 5 LetS=Randf(x) = 3x+ 7 Thenfis uniformly continuousonS Proof Choose" >0 Let ="=3
Here is another interesting and useful property of functions which are continuous on a closed interval [a;b]: the function must run through every y-value between f(a) and f(b) This makes sense since a continuous function can be drawn without lifting the pen from the paper
EXAMPLE Assume f(x) is continuous on the interval [ab] Then R b a f(x)sin?xdx ?0 PROOF: If f ? C1([ab]) this is easy to show by an integra-tion by parts using f?(x)?M for some constant M If f is only continuous use Theorem 2 to ?nd a smooth g(x) with kf ?gk?