Answers & Hints for Selected Exercises 663
EXERCISE SET 5.4
- For [-3, 1], choose 6 = c:/17. For (-2, 2), choose 6 = c:/ 13.
- Let Xo E A and € > 0. Since f is uniformly continuous on A, 36 > 0 3
Vx, y E A, Ix - YI < 6 :::::} lf(x) - f(y)I < c:. Thus, Vx E A, Ix - xol < 6 :::::}
lf(x) - fxo)I < c:. That is , f is continuous at x 0. - Vx, y E IR, lf(x) - f(y)I = l(7x - 8) - (7y - 8)1 = 7lx -yl. For c: > 0, choose
6 = c:/7. Then Ix - xol < 6:::::} lf(x) - f(y)I = 7lx -yl < 7 · c:/7 = c:. - Vx, YE [1, +oo), lf(x) - f(y)I = j ~ - t j = lx,,;yl :::; Ix - YI·
- In Section 5.1 we showed that Vx, y E IR, I sin x - sin YI :::; Ix - YI.
- All but (e) are uniformly continuous on the given interval.
- Suppose f, g are uniformly continuous on A. Let c: > 0. Then 3 6 1 , 6 2 > 0 3
Ix - YI < 61:::::} lf(x) - f(y)I < c:/2 and Ix - YI < 62:::::} lg(x) - g(y)I < c:/2.
Choose 6 = min{61, 62}. Then Vx, y EA, Ix - xril < 6:::::}
lf(x) + g(x) - (f(y) + g(y))I:::; lf(x) - f(y)I + lg(x) - g(y)I < ~ + ~ = c:. - As shown in Section 5.1, sinx and cosx are continuous everywhere. Thus, by
Thm. 5.1.13, tanx = ~~~~ and secx = co~x are continuous wherever cosx =f. 0.
Hence, they are continuous on ( -~, ~). But they are not uniformly continuous
on ( -~, ~) since they are unbounded there. - Let c: > 0. Then :JN E IR 3 x;:::: N:::::} lf(x) - LI < c:/4. Thus, x, y;:::: N:::::}
lf(x) - f(y)I :::; lf(x) - LI+ IL - f(y)I < c:/2.
Since f is continuous on [a, NJ, it is uniformly continuous there. So 36 >
0 3 Vx, y E [a, NJ, Ix - YI < 6:::::} lf(x) - f(y)I < c:/2.
Thus, for all x, y 2 a and Ix - YI < 6,
(i) if x, y E [a, NJ, then lf(x) - f(y)I < c:;
(ii) if x , y E (N, oo), then lf(x) - f(y)I < c:;
(iii) if x E [a, NJ and y E (N, oo ), then
lf(x) - f(y)I :::; lf(x) - f(N)I + lf(N) - f(y)I < ~ + ~ = c:.
:. f is uniformly continuous on [a, oo). - Let c: > 0. Then
361 > 0 3 Vu,v E f(A), lu-vl < 61:::::} lg(u)-g(v)I < c:, and
362 > 0 3 Vx, y EA, Ix - YI < 62:::::} lf(x) - f(y)I < 61
:::::} lg(f(x))-g(f(y))I <c:::::} l(gof)(x)-(gof)(y)I <c:.
Therefore, g o f is uniformly continuous on A.
- By Cor. 5.1.15, f is continuous on [O, 1], so it is uniformly continuous there.
But Vx, y E [O, 1], l~=:I = v'x!Vil' which is unbounded on [O, l]. Hence,
~ K > O 3 l~=yY71 :::; K. :. f cannot satisfy a Lipschitz condition on [O, l].