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.

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## 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:**

## FAQ

**Q1: Is log(x) uniformly continuous on (a, b) where both a and b are finite and positive.**

Answer: log x is uniformly continuous on [a, b] as it is closed and bounded. So log x is uniformly continuous on (a, b).

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.