Prove that log(x) is Uniformly Continuous on [1, ∞)

The function log x is defined for all positive values of x. Note that log(x) is uniformly continuous on [1, ∞). It is also uniformly continuous on any open interval (a, b) where both a (>0) and b are finite.

log x is Uniformly Continuous

Let us consider the function

f(x) = log(x), x ∈ (0, ∞).

Prove that f(x) is uniformly continuous on [1, ∞).

Proof:

Let us consider two points x, y in [1, ∞). We assume that x < y. So $\dfrac{y}{x}>1$. Using the fact log(1+x) < x for x>0, this implies that

0 < log $\dfrac{y}{x}$ < $\dfrac{y}{x}$ – 1

⇒ log $\dfrac{y}{x}$ < $\dfrac{y-x}{x}$

⇒ log $\dfrac{y}{x} \leq y-x$ as x >1 …(∗)

Now, assume that x > y. So $\dfrac{x}{y}>1$. In a similar way as above, we obtain that

0 < log $\dfrac{x}{y} \leq x-y$ …(∗∗)

Combining (∗) and (∗∗), it follows that

| log x – log y | ≤ |x-y| for all x, y ∈ [1, ∞).

This proves that f satisfies the Lipschitz condition. Now, by the fact Lipschitz functions are uniformly continuous we conclude that log(x) is uniformly continuous on [1, ∞).

Remark:

By the same method as above, one can prove that log x is uniformly continuous on [a, ∞) where a > 0. In that case, the Lipschitz constant will be 1/a.

Read Also: Continuous but not Uniformly Continuous: An Example

Show that log x is Uniformly Continuous on (1, 2)

As [1, 2] is a closed and bounded interval, and log x is defined on it, we say that the function f(x) = log x is uniformly continuous on [1, 2].

Observe that (1, 2) ⊂ [1, 2]. So log x is uniformly continuous on the open interval (1, 2).

Related Topics:

• Intermediate Value Theorem
• Fixed Point Theorem

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