The Boundedness Theorem, Extreme Value Theorem and Intermediate Value Theorem

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These are very important theorems. They all have very close connection with the completeness property of the real numbers. They can be shown to be equivalent to the completeness property of the real numbers.

Theorem 1 (Boundedness Theorem).  If f : [a, b] ® R is continuous, then f is bounded.

Proof. If a = b, we have nothing to prove. We assume a < b, that is, [a, b] is a non-trivial interval. Let K = { x Î [a, b]: the restriction of f to [a, x], f | [a, x]:[a, x] ®R is bounded }. Then obviously K is bounded and K is non-empty for it obviously contains a. Let C = supremum of K , sup K. C exists by the completeness property of R because K is bounded above by b. By definition of C, C £ b because b is an upper bound for K. We claim that C > a. Why is this so? Since f is continuous at x = a, given any e > 0, there exists a d > 0 such that for any x in [a, b] with a £ x < a + d, we get f (a) - e < f (x) < f (a) + e. Now choose any d ' > 0 with d ' < min(d, b - a) then for any x with a £ x £ a + d ', we get then f (a) - e < f (x) < f (a) + e. That is to say f | [a, a + d ']:[a, a + d '] ®R is bounded. Hence a + d ' Î K and so a < a + d ' £ sup K = C. This proves the claim.

We next claim that C = b. Suppose on the contrary that C < b, then a < C < b. Then since f is continuous at C, following the same argument as above we can show that there exists a d > 0 such that f is bounded on the interval [C -d, C + d] Í [a, b], where C + d < b. That means there exists a M > 0 such that for all x in [C -d, C + d], | f (x) |<M. Now since C -d < C, and sup K = C, there exists a x in K such that C -d < x £ C. That means f is bounded on [a, x] say by N and so f is bounded on [a, C + d] =[a, x]È[C -d, C + d] by max (M, N).

I.e., for all x in [a, C + d], | f (x) |< max (M, N ). Therefore, C + d Î K. Hence C + d £ C. contradicting d > 0. Hence C = b. (This means that for any x such that a £ x < b, f is bounded on [a, x] . Why? The reason is as follows. There exists a y in K such that a £ x < y £ b and f is bounded on [a, y] and so f is bounded on [a, x].)

Now we observe that by the continuity of f at b, for a fixed choice of an e > 0, we have that there exists a d > 0 such that for any x in [a, b] with b - d < x £ b, we get f (b) - e < f (x) < f (b) + e. Hence f is bounded on [x, b] by max (| f (b) - e| , | f (b) + e|) for some x in K with b - d < x £ b since b = C = sup K . Therefore, f is bounded on [a, x] say by M₁ . Therefore, f is bounded on [a, b] = [a, x]È [x, b] by max ((| f (b) - e| , | f (b) + e|, M ₁ ). This completes the proof.

Theorem 2 (Extreme Value Theorem). A continuous function on a closed and bounded interval attains its supremum and infimum. That is, if f : [a, b] ® R is continuous, then there exist c and d in [a, b] such that f (c) = supremum f ([a, b]) and f (d) = infimum f ([a, b]).

Proof. Again if a = b, we have nothing to prove. Assume a < b. Consider the range f ([a, b]). Let M = supremum f ([a, b]). Suppose M does not lie in the range, i.e., there does not exist c in [a, b] such that f (c) = M. Then f (x) < M for all x in [a, b]. That means M - f (x) > 0 . Since f is continuous on [a, b], the function defined by h(x) = M - f (x) for x in [a, b] is also continuous and h(x) > 0 for all x in [a, b]. Therefore, since h(x) is non-zero, g(x) = 1/h(x) is also continuous on [a, b]. Hence by Theorem 1, g is bounded and so is bounded above. Hence the sup of its range exists by the completeness property of R. Let K = supremum g([a, b]). Because h(x) > 0 for all x in [a, b], g(x) > 0 for all x in [a, b]. Thus K > 0 and so 0 < g(x) £ K for all x in [a, b]. Hence h(x) = 1/g(x)³ 1/K for all x in [a, b]. That means M - f (x)³ 1/K and hence f (x) £ M - 1/K. for all x in [a, b]. Therefore, M - 1/K ³ M since M = supremum f ([a, b]) and so 1/K £ 0 contradicting K > 0.

We conclude that there exists c in [a, b] such that f (c) = M.

Similarly if there does not exists d in [a, b] such that f (d) = N = infimum f ([a, b]), we can arrive at a contradiction. We use h(x) = f (x) - N > 0. Then g(x) = 1/h(x) > 0 for all x in [a, b] and g is continuous on [a, b]. Let L = supremum g([a, b]) > 0. L exists by Theorem 1. Thus 0 < g(x) £ L for all x in [a, b]. Therefore, h(x) = 1/g(x) ³ 1/L for all x in [a, b]. That means f (x) - N ³ 1/L and hence f (x) ³ N + 1/L for all x in [a, b]. Therefore, N + 1/L £ N

since N = infimum f ([a, b]), contradicting 1/L > 0. Hence there exists a d in [a, b] such that f (d) = infimum f ([a, b]). (See Figure 1 below for illustration.)

Theorem 3 Intermediate Value Theorem. Suppose f : [a, b] ® R is continuous. If g is an intermediate value between f (a) and f (b), i.e. either f (a) £ g £ f (b) or f (b) £ g £ f (a), then there exists c in [a, b] such that f (c) = g .

Proof. Again if a = b, we have nothing to prove. Assume a < b. Without loss of generality we may assume that f (a) < f (b). If g = f (a) or f (b), we have nothing to prove. Now take any g such that f (a) < g < f (b). Then define g : [a, b] ® R by g(x) = f (x) - g for x in [a, b]. Then g is a continuous function, g(a) < 0 and g(b) > 0. We are going to find a point k in [a, b] such that g(k) = 0. We do this by using the completeness property of the real numbers R. Let F = { x Î [a, b]: g(x) < 0}. Then F ¹ Æ since a Î F because g(a) < 0. Obviously F is bounded above by b. Hence by the completeness property of R, supremum of F exists. Let k = sup F. Since g is continuous at a and g(a) < 0, there exists d > 0 such that for all x with a £ x < a + d < b , g(x) < 0. ( Take e = - g(a)/2. By continuity of g at a, there exists d₁ > 0 such that for all x in [a, b] with a £ x < a + d₁ , |g(x) - g(a)| < - g(a)/2 or 3g(a)/2 < g(x) < g(a)/2 < 0. Take d = min(d₁ , (b -a)/2). ) This means k ³ a + d ' > a for any d ' with 0 < d ' < d. Therefore, k > a. Thus a < k £ b. Now by the continuity of g at b and the fact that g(b) > 0, there exists d₂ > 0 such that for all x with a < b - d₂ < x £ b , g(x) > 0. That means for any k with b - d₂ < k < b, k Ï F and consequently any k with b - d₂ < k < b is an upper bound for F . Thus k = sup F £ b - d₂ < b. Hence a < k < b.

We now claim that g(k) = 0. That is f (k) = g.

Suppose g(k) < 0. Then by the continuity of g at k, there exists d₃ > 0 such that for any x with a < k - d₃ £ x £ k + d₃ < b, we have g(x) < 0. This means k + d₃ Î F . Thus k + d₃ £ sup F = k, and d₃ £ 0 contradicting d₃ > 0. Hence g(k) ³ 0. Similarly if g(k) > 0, then by the continuity of g at k, there exists d₄ > 0 such that for any x with a < k - d₄ £ x £ k + d₄ < b, we have g(x) > 0. Then any x in [a, b] and g(x) < 0 would imply that x < k - d₄. Thus k - d₄ is an upper bound for F and hence k £ k - d₄ giving d₄ < 0 contradicting d₄ > 0. Hence g(k) = 0. We now take c = k and f (c) = g.

If f (a) < f (b), then multiply by -1, we get - f (a) > - f (b). Replace f above by - f , g by - g and the proof proceed exactly the same manner as above to obtain a c in [a, b] such that - f (c) = - g and that is the same as f (c) = g . This completes the proof.

Theorem 4. (Compactness Theorem). If f : [a, b] ® R is a continuous function defined on a closed and bounded interval [a, b], then its range is also a closed and bounded interval.

Proof. By the Extreme Value Theorem, Theorem 2 above, the range of f , f ( [a, b]) Í [inf f ( [a, b]), sup f ( [a, b])] = [ f (c), f (d)] for some c and d in [a, b]. But by the Intermediate Value Theorem, Theorem 3 above, [ f (c), f (d)] Í f ( [a, b]). Hence f ( [a, b]) = [ f (c), f (d)] and [ f (c), f (d)] is a closed and bouinded interval. That means the range f ( [a, b]) is a closed and bounded interval. ( See figure 3 below.)

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