By Kuttler K.

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**Example text**

Since AB = I it follows B is one to one since otherwise, there would exist, x = 0 such that Bx = 0 and then ABx = A0 = 0 = Ix. Therefore, from what was just shown, B is also onto. In addition to this, A must be one to one because if Ay = 0, then y = Bx for some x and then x = ABx = Ay = 0 showing y = 0. Now from what is given to be so, it follows (AB) A = A and so using the associative law for matrix multiplication, A (BA) − A = A (BA − I) = 0. But this means (BA − I) x = 0 for all x since otherwise, A would not be one to one.

Suppose φ is strictly increasing on (a, b) , a similar argument holding for φ strictly decreasing on (a, b) . If x > a, then pick y ∈ (a, x) and from the above, φ (y) < φ (x) . Now by continuity of φ at a, φ (a) = lim φ (z) ≤ φ (y) < φ (x) . x→a+ Therefore, φ (a) < φ (x) whenever x ∈ (a, b) . Similarly φ (b) > φ (x) for all x ∈ (a, b). This proves the lemma. 2. 4 Let f : (a, b) → R be one to one and continuous. Then f (a, b) is an open interval, (c, d) and f −1 : (c, d) → (a, b) is continuous. Proof: Since f is either strictly increasing or strictly decreasing, it follows that f (a, b) is an open interval, (c, d) .

2. 4, the lower sums on the interval [0, 1] are all equal to zero while the upper sums are all equal to one. 3. Let f (x) = 1 + x2 for x ∈ [−1, 3] and let P = −1, − 31 , 0, 21 , 1, 2 . Find U (f, P ) and L (f, P ) for F (x) = x and for F (x) = x3 . 4. Show that if f ∈ R ([a, b]) for F (x) = x, there exists a partition, {x0 , · · · , xn } such that for any zk ∈ [xk , xk+1 ] , n b f (x) dx − a f (zk ) (xk − xk−1 ) < ε k=1 n This sum, k=1 f (zk ) (xk − xk−1 ) , is called a Riemann sum and this exercise shows that the Riemann integral can always be approximated by a Riemann sum.

### Topics in analysis by Kuttler K.

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