# Is this an equivalent definition of uniform continuity?

The motivation for this definition is that a main characteristic of functions defined on an interval which are continuous but not uniformly continuous is that the largest possible $\delta$ required to keep $\,f(x)\,$ within $\varepsilon$ of $\,f(x_0)\,$ can be made arbitrarily small by choosing the appropriate $x_0$. Consider, for example, $\,f(x) = \tan x\,$ in $\,\left(\dfrac{\pi}{2} - \varepsilon_0, \dfrac{\pi}{2}\right)$ for $\varepsilon_0$ small. It is continuous everywhere in that interval but we are forced to choose smaller and smaller $\delta$ as $\,x_0 \to \dfrac{\pi}{2}$.

Let $\,f:I \to \mathbb{R}$ be a continuous real function. Consider some $\,x_0 \in I\,$ and fix $\,\varepsilon \in \mathbb{R_+}$. Define $P = \left\lbrace\delta \mid \forall \,x \in I \ , \left\lvert x-x_0\right\rvert < \delta \implies \left\lvert \,f(x) - f(x_0)\right\rvert < \varepsilon\right\rbrace$. Let $$\delta_{x_0}^* = \begin{cases} \sup P, & P \ \text{ bounded above} \\ 1, & \text{otherwise} \end{cases}$$ We say $\,f$ is uniformly continuous if for all $\,\varepsilon \in \mathbb{R_+}, \:\inf\left\lbrace\delta_{x_0}^* \mid x_0 \in I \right\rbrace \in \mathbb{R_+}$.

It should be noted that the choice $1$ is arbitrary and $P$ is non empty by hypothesis. I'm not sure if this is equivalent, and if so how can one prove it is equivalent to the standard definition?

• Yes, this is right, though you've made the key point -- that the infimum is positive rather than zero -- look like mere notation. One direction of the equivalence is obvious: picking that infimum as your delta works for all $x$. In the other direction, it's easy to get a contradiction. – symplectomorphic Apr 18 '16 at 6:47

Old, but might as well. Note $\mathbb{R}_{+} = (0, \infty)$ in the original question.

$$\text{My definition} \implies \text{Normal Definition}$$

Choose $\delta_0 \stackrel{\rm def}{=} \frac 12\inf\{\delta_{x_0}^{*} | x_0 \in I\}$ as our uniform $\delta$.

Now, let $x,y \in I$ by arbitrary such that $|x-y| < \delta_0$ and consider $\delta_x^{*}$. If $\sup P = +\infty$ so that $\delta_x^{*} = 1$, the set $P$ of $\delta$'s which work for $x$ is unbounded, so $\delta_0$ necessarily works. Hence, we have $|f(x) - f(y)| < \epsilon$. On the other hand, if $\delta_x^{*} = \sup P < +\infty$, $\delta_x^{*}$ is the supremum of the $\delta$'s which work for $x$, and $0<\delta_0<\delta_x^{*}$, hence $\delta_0$ works here again. So $|f(x) - f(y)| < \epsilon$.

$$\text{Normal definition} \implies \text{My Definition}$$

Suppose that some uniform $\delta_0>0$ works on all of $I$ and that $\inf\{\delta_{x_0}^{*} | x_0 \in I\} = 0$. Without loss of generality, suppose $\delta_0<1$ so the piecewise definition is not a problem. We will deduce a contradiction.

Since the infimum is zero, there is a $\delta_{x_0}^{*}$ such that $0<\delta_{x_0}^{*}<\delta_0$. But if for all $y \in I$,

$|x_0 - y| < \delta_0 \Longrightarrow |f(x_0) - f(y)| < \epsilon$, we have $\delta_0 \in P$, $\delta_{x_0}^{*} = \sup P$ and $0<\delta_{x_0}^{*}<\delta_0$ which yields a contradiction.