Uniform continuity? How do I prove this?

Question

Show f=1/x² is unif. cont. on [1, ∞) but not unif.cont. on (0, ∞).

The interval (0, ∞) does not include 0.

Answer 1

First I'll prove a lemma which will be the key step in proving f is uniformly continuous on [1, ∞).

Lemma: For x, y in [1, ∞), it's true that (x + y)/x^2y^2 <= 2.

Proof: Since x >=1 and y >=1, then (xy)^2 >= x and (xy)^2 >= y. Since (xy)^2 is bigger than both x and y, then it's bigger than their average. So (xy)^2 >= (x + y)/2. Dividing by (xy)^2 and multiplying by 2 gives 2 >= (x + y)/(xy)^2, which is what I wanted to show.

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Now I'll show 1/x^2 is uniformly continuous on [1, ∞).

Let ε > 0. Choose δ = 2ε. Then if |x - y| < δ, we have

|f(x) - f(y)| = |(1/x^2) - (1/y^2)| = |(y^2 - x^2)/(x^2y^2)| = (|y - x||y + y|)/(x^2y^2) = δ|y+x|/(x^2y^2) = δ(x + y)/(x^2y^2) <= 2δ = ε.

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Now I'll show 1/x^2 is not uniformly continuous on (0, ∞).

Let δ > 0. Choose x so small that x < 1 and x < 2δ. I will show that |x - (x/2)| < δ, but |f(x) - f(x/2)| > 1.

Since x < 2δ, then x/2 < δ, so x - (x/2) < δ, so |x - (x/2)| < δ.

However, we have |f(x) - f(x/2)| = |1/(x^2) - 1/((x/2)^2)| = |1/x^2 - 1/(x^2 / 4)| = |1/x^2 - 4/x^2| = 3/x^2. This is greater than 1 since x < 1.

Since no matter how small I choose δ, I can find two points (x and x/2) less than δ apart such that f(x) and f(x/2) are more than 1 apart, then f is not uniformly continuous on (0, ∞).

Answer 2

Themathemagician has given you an excellent explanation. I'll give you some more insight.

We know f is uniformly continuous on an interval I if, and only if, for every sequence x_n in I that converges to some x (not necessarily in I), the image sequence f(x_n) is convergent.

Let x_n = 1/n. Then, x_n is a sequence in (0, oo) that converges to 0. It's image sequence is f(x_n) = 1/(x_n)^2) = n^2, which clearly goes to infinity. Therefore, the condition for uniform continuity is not satisfied on (0, oo), so that f is not uniformly continuous on this set.

To show f is uniformy continuous on [a, oo), a>0, we can use the following well known theorem: If f is differentiable on an interval I, then f is Lipschitz on I, if and only if, its derivative f' is bounded. If w = infimum {|f(x)| : x is in I}, then, for every x and y in I, we have |f(x) - f(y)| <= w|x - y|. It's immediate that being lipschitz implies uniform continuity.
In our case, f(x) = 1/(x^2), so that f'(x) = -2/(x^3). Since x>=a >0, |f|' attains a global minimum at x=a, which is 2/(a^3). Therefore, |f(x) - f(y)| <= 2/(a^3).|x - y|, so that f is Lipschitz and, therefore, uniformly continuous on I. Since, in our case, a=1, we have |f(x) - f(y| <= 2|x - y|. This the same conclusion Themathemagician got using just algebra.

When a function is differentiable, it's often easier to prove it's uniformly continous considering it's derivative than using algebra.

Answer 3

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Answer 4

The function is not continuous over (0, ∞) because the function is undefined at zero.