There’s another post asking for the motivation behind uniform continuity. I’m not a huge fan of it since the top-rated comment spoke about local and global interactions of information, and frankly I just did not get it.

Playing with the definition, I want to say uniform continuity implies there’s a maximum “average rate of change”. Not literally a derivative, but the rate of change between two points is bounded in the domain. I’m aware that this is essentially Lipschitz continuity, and that Lipschitz implies uniform. This implies there’s more to uniform continuity than just having a bounded average rate of change.

And also, how is it that $ f(x)=x$ is uniform yet $f(x)f(x)=g(x)=x^2$ is not? I understand why it isn’t, I can prove it. But I just don’t understand the motivation and importance of uniform continuity.

**Answer**

The real “gist” of continuity, in its various forms, is that it’s the “property that makes calculators and measurements useful”. Calculators and measurements are fundamentally *approximate* devices which contain limited amounts of precision. Special functions, like those which are put on the buttons of a calculator, then, if they are to be useful, should have with them some kind of “promise” that, if we only know the input to a limited amount of precision, then we will at least know the output to *some* useful level of precision as well.

Simple continuity is the weakest form of this. It tells us that if we want to know the value of a target function $f$ to within some tolerance $\epsilon$ at a target value $x$, but using an approximating value $x’$ with limited precision instead of the true value $x$ to which we may not have access or otherwise know to unlimited precision, i.e. we want

$$|f(x) – f(x’)| < \epsilon$$

then we will be able to have that *if* we can make our measurement of $x$ suitably accurate, i.e. we can make that

$$|x – x’| < \delta$$

for some $\delta > 0$ which may or may not be the same for every $\epsilon$ and $x$.

Uniform continuity is stronger. It tells us that not only do we have the above property, *but* in fact the *same* $\delta$ threshold on $x’$‘s accuracy will be sufficient to get $\epsilon$ worth of accuracy in the approximation of $f$ *no matter what $x$ is*. Basically, if the special function I care about is uniform continuous, and I want 0.001 accuracy, and the max $\delta$ required for that is, say, 0.0001, by measuring to that *same* tolerance I am assured to *always* get 0.001 accuracy in the output *no matter what $x$ I am measuring*. If, on the other hand, it were the case that the function is merely continuous but not uniformly so, I could perhaps measure at one value of $x$ with 0.0001 accuracy and that accuracy would be sufficient to get 0.001 accuracy in the function output, but if I am measuring at another, such a tolerance might give me only 0.5 accuracy – terrible!

Lipschitz continuity is *even better*: it tells us that the max error in approximating $f$ is *proportional* to that in approximating $x$, i.e. $\epsilon \propto \delta$, so that if we make our measurement 10 times more accurate, say (i.e. one more significant figure), we are assured 10 times more accuracy in the function (i.e. gaining a significant figure in the measurement lets us gain one in the function result as well).

And in fact, all the functions (that are real-analytic, not combinatorial functions like nCr and what not) on your real-life calculator are at least *locally* Lipschitz continuous, so that while this proportionality factor (effectively, *absolutely* how many sig figs you get for a given number of such in the input) may not be the same everywhere, you can still be assured that in relative terms, adding 10x the precision to your measurements, i.e. one more significant figure, will always make the approximation (however good or not it actually is) returned by your calculator 10x more accurate, i.e. also to one more significant figure.

And to top it all off, all these forms of continuity – at least in their *local* variants, that is, over any bounded interval – are implied by differentiability.

**Attribution***Source : Link , Question Author : Spencer Kraisler , Answer Author : The_Sympathizer*