I have been studying Analysis for a while now, It has always confused me what ε and 𝛿 actually mean. For example in the definition below:

Definition 5.2 (pointwise and uniform convergence I)

Let {$f_n$} be a sequence of functions $f_n : <c,d> → \mathbb{R}$, and let $f : <c,d>→ \mathbb{R}$ be another function.

(i) The sequence {$f_n$} converges pointwise on $<c,d>$ to $f$ if the following holds: For all $x ∈ <c,d>$ and for all $ε > 0$ there exists $N = N(x, ε) ∈ N$ such that $n ≥ N$ implies $|f_n(x) − f(x)| < ε.$

(ii) The sequence {$f_n$} converges uniformly on $<c,d>$ to $f$, if for all $ε > 0$ there exists $N = N(ε) ∈ N$ such that $n ≥ N$ implies $|f_n(x) − f(x)| < ε$ for all $x ∈ <c,d>$.

I've always thought of $\epsilon$ as some real number, and $\delta$ as some other real number. I'd guess that $\epsilon$ and $\delta$ change in context in each situation they are used, but I am struggling to understand what this context is.

For example, in the above, is $\epsilon$ simply representing a real number? If it is, then how is N a function of epsilon? (I struggle to see this intuitively) following from this, I get that n is simply the nth term in the sequence of functions, so, in both (i) and (ii) it's saying that some natural number exists s.t. any term in the sequence with a bigger or equal 'index' implies that the sequence of functions converges to each function.

Also, I get that the difference between pointwise convergence and Uniform convergence is that pointwise only 'requires', one 'point' to converge to its function, but uniform convergence requires all points to converge, Hence I half understand that N is a function of $\epsilon$ for pointwise convergence and for uniform convergence, N is both a function of $\epsilon$ and $x$. However, I still struggle to picture what these N actually mean. Maybe I am missing the point somewhere. If someone could try and help me that'd be great.


A lot of these calculus definitions are of the type "as long as the argument to the function is constrained to be close enough to a given value, then the value of the function either vary relatively little (in the case of continuity) or is close to some given value (in the case of limits and convergence)". In this context, yes, indices of a sequence are arguments, and $\infty$ is one of the possible values that arguments can "get close to".

The $\epsilon$-$\delta$ (or some times $\epsilon$-$N$) machinery just formalises these notions so that we can actually confirm what, in a given case, is "close enough", and "varying little".

For instance, in pointwise convergence, a sequence $f_n$ of functions converge pointwise to $f$ if, for any $x$, and any given bound $\epsilon$, you can go far enough out in the sequence of functions so that the difference between $f_n(x)$ and $f(x)$, as real numbers, is smaller than $\epsilon$.

How far out in the sequence you have to go (denoted by $N$), may depend on $\epsilon$ (the narrower bound you set, the further out in the sequence you have to go to stay within the bound), and it may depend on $x$, since the elements here are functions, after all, and while it may take three steps for $f_n(0)$ to get close to $f(0)$, it could take a thousand steps for $f_n(100)$ to get close to $f(100)$. We write these dependences as $N(\epsilon,x)$.

Example: $f_n(x)=\frac xn$ converges pointwise to $f(x)=0$. We see this because for any $x$, and any bound $\epsilon$, we can pick large enough $n$ that $\left|\frac xn\right|<\epsilon$. For instance, any $n>\frac{|x|}\epsilon$ will fulfill this.

Uniform convergence, on the other hand, means that the function converges everywhere simultaneously, in done sense. In other words, given an $\epsilon$-bound, we require that it's possible to go far enough out in the sequence so that no matter which $x$ we pick, the difference between $f_n(x)$ and $f(x)$ is smaller than $\epsilon$.

Example 1: the previous example is not uniformly convergent, because given an $\epsilon$-bound, no matter how large an $n$ we pick, there are going to be values of $x$ that make $\left|\frac xn\right|$ too large.

Example 2: $f_n(x)=x-\frac 1n$ converge uniformly to $f(x)=x$. That's because given any $\epsilon$-bound, we can choose $n$ large enough that no matter what $x$ is, $|f_n(x)-f(x)|$ is smaller than $\epsilon$. Seeing as $|f_n(x)-f(x)|=\frac 1n$, picking $n$ larger than $\frac1\epsilon$ will work.


We use $\epsilon$ and $\delta$ to make concepts which are otherwise hard to explain precise. Of course $\epsilon$ and $\delta$ are real numbers and usually $\epsilon$ is positive and small while $\delta$ depends on $\epsilon$. The concept of limit is a good example of how $\epsilon$ and $\delta$ come to play their role. If you want $f(x)$ gets as close as you wish to $L$ you say for every positive $\epsilon$ I like to make $|f(x)-L|<\epsilon$. Now this does not happen without paying a price.The price is making making the $|x-a|$ small enough. We say there is a $\delta$ such that if $|x-a|<\delta$ then $|f(x)-L|<\epsilon.$ A smaller $\epsilon$ means more precision so you may have to pay a higher price by choosing a smaller $\delta$ that is, $\delta$ normally depends on $\epsilon$.


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