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Differentiable at endpoints?

Does differentiation only work on open sets?

Admittedly, there are some questions and answers as to why a function defined on a closed interval is not differentiable on the interval's endpoints. But I find no answer as to why, in spite of this, continuity can be defined on an interval's endpoints.

For differentiability, the intuition is that the neighborhood of $x$ that allows $\lim_{x \rightarrow c}\frac{f(x) - f(c)}{x - c} = L\quad$ must (or is this our definition) be populated from both the left ($x < c\;$) and the right ($x > c\;$). Hence, differentiability isn't defined on endpoints of an interval.

Why does such a requirement not exist for continuity?

A function $f$ is continuous at $c$ if $$ \forall \epsilon, \exists \delta, |x - c| < \delta \rightarrow |f(x) - f(c)| < \epsilon$$

Continuity seems not to care whether points appear left or right of $c$ -- just that, if they exist in some chosen neighborhood, they satisfy the condition above.

Here is an example of this "dichotomy"

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Muno
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  • There is another reason the mean value theorem is formulated as such: so that it applies to functions like $f(x) = \sqrt{1-x^2}$ over $[-1, 1]$. – MathematicsStudent1122 Apr 23 '16 at 01:15
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    Part of the answer is that differentiability relies on the algebraic structure of $\mathbb{R}$, whereas continuity just relies on the distance function: $$d(x,y) = |y-x|.$$ Another viewpoint that while the closed interval $[a,b]$ is a perfectly good topological space, its not a smooth manifold. Contrast this with the open interval $(a,b)$, which is both a topological space, and a smooth manifold. – goblin GONE Apr 23 '16 at 01:37
  • @goblin uh, well it's a smooth manifold with boundary. – Matt Samuel Apr 23 '16 at 02:17
  • @goblin er, nothing dummied down for someone working through "elementary" real analysis!? – Muno Apr 23 '16 at 02:20
  • @MattSamuel, which is not a smooth manifold :) the point really being that given a function $f : X \rightarrow \mathbb{R}$ where $X$ is a "smooth manifold with boundary," there's no definition of "$f$ is differentiable." You have to restrict $f$ to the non-boundary part to formulate this notion. – goblin GONE Apr 23 '16 at 02:23
  • @Muno, sorry, I don't really know how to dumb it down. But if you look up the phrase "metric space", it will all start to make a bit more sense methinks. – goblin GONE Apr 23 '16 at 02:47
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    Can you think of any fundamental difference between "permissible" and "necessary'? A function is certainly PERMITTED to be differentiable at the endpoints of the interval in the MVT; if the function is, in fact, differentiable at the end points, that doesn't make the MVT invalid or untrue. –  Apr 23 '16 at 02:51
  • @mathguy Ah -- so differentiability at an endpoint is a valid concept (just, in this case, not necessary for the proof). If differentiability at an endpoint is possible, doesn't that invalidate what goblin and MattSamuel say, then? – Muno Apr 23 '16 at 03:54
  • @Muno It's not quite that simple. Read the answers/comments on this post: http://math.stackexchange.com/questions/126176/differentiablility-over-open-intervals?rq=1 – MathematicsStudent1122 Apr 23 '16 at 08:39

2 Answers2

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Not all mathematicians forbid taking a derivative at the endpoint of an interval.

Consider the question Derivative at Endpoint, which provides a passage from Rudin's Principles of Mathematical Analysis according to which a function defined only on a closed interval may have derivatives at the endpoints of that interval.

You can also examine the answers to derivative on endpoints.

The choice to allow or disallow derivatives at endpoints depends on technical details of how you define a derivative in the first place. For example, if you use a definition that computes left and right derivatives separately and then requires them to be equal in order have and ordinary derivative, differentiation is clearly impossible unless the function is defined on both sides of the point where you wish to differentiate it. But it is not necessary to define differentiation that way.

With regard to the examples you cited in the question:

The fact that someone may say in some piece of mathematics that if $f$ is differentiable on $(a,b)$ then a certain fact is true, it does not mean that $f$ cannot be differentiable at $a$ or $b.$ It merely means that we do not need $f$ to be differentiable at $a$ or $b$ in order to reach the desired conclusion. Putting unnecessary conditions in a theorem is generally a bad idea, since in those cases where the condition isn't met you are unable to use the theorem, whereas if you had omitted the unnecessary condition you might find all the other conditions are satisfied and the theorem applies.

Conversely, when we say that if some other fact is true then $f$ is differentiable on $(a,b),$ we may mean only that sometimes $f$ can fail to have a derivative at $a$ or $b$ even though the given fact is true. It might even just mean that it was too much bother (and not worthwhile enough) to prove that $f$ is differentiable at the endpoints. It is certainly not warranted to conclude from such a statement that $f$ cannot be differentiable at $a$ or $b.$

David K
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I know this is an old post but anyway, I think your confusion comes from you thinking that the function in question is only defined on the closed interval. Take the modulus function as an obvious example, call it f. it's defined on the whole of R the set of real numbers. now consider the closed interval I = [0,1]. Then here is an example that is continuous on I but only differentiable on (0,1). (i.e (0,1)= [0,1]-{0,1}). This is because no limit exit of the chord-slope function of f near 0 but limit f exits near 0 and f(0)= lim f as x tends to 0 = 0. For an silly (trivial) example of a function differentiable on a closed interval say g(x)=x*x defined on [0, infinity) then for all c in I= [2,4] g(c) is both differentiable and (hence) continuous on I. Note there must be 'elbow-room' on either side of any point, c for the limit of g at c to exist, although for a limit we don't-care what the value is of g(x) at c or even if exists at all! I.e. g(c) may be undefined yet the limit exists.

continuity is weaker but necessary than that of differentiability. However the opposite is not true.

differentiability (at a point c) is a local property in which a neighbourhood $N = \left( {c - \delta ,c + \delta } \right) \subset I$ of c has the property that for every x in $N = \left| {x - c} \right| < r \cap I$ is differentiable at $x$ so is only defined on interior points and therefore not differentiable at end points of a closed interval. While continuity in any interval use the property for any given epsilon positive there exists delta positive s.t. for all $x \in I$ satisfying $\left| {x - a} \right| < \delta $ guarantees that $\left| {f(x) - f(a)} \right| < \varepsilon $ so this covers point end point $a$. note that if $f$ is continuous on$I$ is continuous at each point of $I$. This can also be rephrased in terms to limits of a sequence.

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