# Physical Interpretation of uniformly continuous function.

I have been reading material on uniformly continuous functions. And going through the problems where we have to prove that a function is not uniformly continuous or otherwise.

A function defined on an interval I is said to be uniformly continuous on I if to each $$\epsilon$$ there exists a $$\delta$$ such that

$$|f(x_1)-f(x_2)| < \epsilon$$, for arbitrary points $$x_1, x_2$$ of I for which $$|x_1-x_2|<\delta$$.

Now I understand the above definition in mathematical terms and able to apply this definition to solve problems. But I don't understand why this definition was introduced, how does uniformly continuous functions look like ? If given a graph of a function, how can I tell if it is a uniformly continuous function ? When I imagine continuous functions, I have this picture in mind as to how they look like but for uniformly continuous functions, I can't think of any picture.

• Well, the wiki. page uniform continuity has the history and nice visualizes. I would look there.
– user853982
Commented Feb 14, 2021 at 11:28

For uniformly continuous functions, there is for each $$\varepsilon >0$$ a $$\delta >0$$ such that when we draw a rectangle around each point of the graph with width $$2\delta$$ and height $$2\varepsilon$$ , the graph lies completely inside the rectangle.
For functions that are not uniformly continuous, there is an $$\varepsilon >0$$ such that regardless of the $$\delta >0$$, when we draw a $${\displaystyle 2\varepsilon \times 2\delta }$$ rectangle around a point of the graph, there are points of the graph directly above or below the rectangle. There might be midpoints where the graph is completely inside the rectangle but this is not true for every midpoint.
My favorite example is $$f(x) = \frac{1}{1-x}$$ on the interval $$(0,1)$$. The nearer $$x$$ is to $$1$$ the steeper the graph, so the larger a $$\delta$$ you need for each given $$\epsilon$$.