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Why Read This Book? - Index of

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7.7 Uniform Continuity 237<br />

Next, let’s split up the fraction in the right-hand side <strong>of</strong> Eq. (7.36) and apply<br />

the triangle inequality. Also, notice that x, y > 0 implies that 1/(x + 1),1/(y + 1),<br />

x/(x + 1), and y/(y + 1) are all less than 1. So we have<br />

�<br />

�<br />

�<br />

�<br />

xy + x + y �<br />

�<br />

�(x<br />

+ 1)(y + 1) � ≤<br />

�<br />

�<br />

�<br />

�<br />

xy �<br />

�<br />

�(x<br />

+ 1)(y + 1) � +<br />

�<br />

�<br />

�<br />

�<br />

x �<br />

�<br />

�(x<br />

+ 1)(y + 1) � +<br />

�<br />

�<br />

�<br />

�<br />

y �<br />

�<br />

�(x<br />

+ 1)(y + 1) �<br />

� � � �<br />

�<br />

= �<br />

x � �<br />

� �<br />

y �<br />

�<br />

�(x<br />

+ 1) � �(y<br />

+ 1) � +<br />

� � � �<br />

�<br />

�<br />

x � �<br />

� �<br />

1 �<br />

�<br />

�(x<br />

+ 1) � �(y<br />

+ 1) �<br />

� � � �<br />

�<br />

+ �<br />

y � �<br />

� �<br />

1 �<br />

�<br />

�(y<br />

+ 1) � �(x<br />

+ 1) � ≤ 3<br />

(7.37)<br />

Having arrived at |f(x) − f(y)| ≤ 3 |x − y|, we are ready to write a pro<strong>of</strong>.<br />

Let ɛ>0 be given, and let δ = ɛ/3. Then for any x, y ≥ 0 such that |x − y| 0 such that |f(x) − f(y)| ≤ m |x − y| for all<br />

x, y ∈ A, then f is uniformly continuous on A.<br />

If y �= x, the hypothesis condition <strong>of</strong> Exercise 7.7.3 is equivalent to<br />

�<br />

�<br />

�<br />

�<br />

f(x) − f(y) �<br />

�<br />

� x − y � ≤ m (7.39)<br />

which means that f has a bound on the slopes <strong>of</strong> lines through any two points on<br />

its graph. Loosely speaking, if the steepness <strong>of</strong> f (as measured by slopes <strong>of</strong> secant<br />

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