f(x) can't be continuous on [a,b]

let f(x) be defined on [a,b]

Continuity

Def 1)

Lim f(x) requires neighborhood in (a,b) → 0.

Lim f(a) not defined so f(x) not continuous at a.

Def 2)

Lim f(x) requires neighborhood in [a,b] → 0.

If x ≠ a, same as Def 1). If x = a, definition of continuity differs from x ≠ a.

f(x) not continuous at a because you can’t change definition of continuity for x ≠a and x = a (in mid-stream).

Same applies for derivatives.

It becomes an issue if you require conditions at a boundary determine conditions in the interior.

Continuous Functions on Closed Regions

reference post 1.

Functions defined on a closed region can only be continuous and have continuous derivatives at the boundary if they are defined beyond the closed region. After all operations are performed and solutions found, the functions outside the closed region can be ignored if they are irrelevant.

Explanation:

The essence of the general problem condenses to a consideration of three common versions of “The Fundamental Theorem of Calculus.”

I) If f is continuous on [a,b] it has an anti-derivative (F’(x)=f(x)) on [a,b].

II) If f is continuous on (a,b) it has an anti-derivative (F’(x)=f(x)) on (a,b).

III) If f is continuous on (c,d) and (a,b) is in (c,d), f has an antiderivative on (a,b).

I) & II) generalized is the case for generalized divergence theorems and differential equations defined on a closed “volume” with boundary conditions, but in these cases the functions and derivatives can’t be continuous on the “surface” of the closed “volume.”

It turns out that left and right sided functions work even though they don’t satisfy the requirements of continuity as explained in post 1 above:

One sided functions are tangents at the boundary which implies the function could be extended slightly beyond the boundary along the tangent “plane” or normal to the tangent (directional derivative) which turns cases I) and II) into case III. Same for higher order derivatives with function at boundary extended in such a way as to make the higher order derivatives exist and be continuous at boundary.

So cases I and II turn out to be right if you explain why.

Textbook authors don’t explain it. They use the general version of I) or II). I give Kellog’s (Potential Theory) version of the Divergence Theorem because it is typical and uses cases I) and II) (generalized), and some authors defer to his derivation.

“The divergence theorem involves two things, a certain region, or portion of space, and a vector field, or set of three functions X, Y, Z of x,y,z, defined in this region.”….. “As to the field (X,Y,Z), we shall assume that its components and partial derivatives of the first order are continuous within and on the boundary of N.”

Re: f(x) can't be continuous on [a,b]

Put this here so as not to clutter up post 2.

Example: Find area under step function f(x) = 1 for x ε [a,b] (or (a,b)) and 0 otherwise.

You can’t do it using the unexplained Fundamental Theorem of Calculus because an anti-derivative (F’(x) = f(x)) doesn’t exist at a and b.

F(x) = x has right and left derivatives at a and b so f(x) can be extended slightly left and right tangentially and in that case the Fundamental Theorem of Calculus gives ʃf(x)dx = F(1) – F(0) = 1.

This nay seem trivial, but, for ex, the question of the directional derivative of f(x) on the boundary of a closed volume when f(x) is defined only in or on the closed volume isn’t. If the closed volume is in a fluid with a velocity field, deriving the divergence theorem is no problem.

Re: f(x) can't be continuous on [a,b]

Consider

ʃ_{a}^{b }f '(x)dx = lim {[f(a+dx) – f(a)]/dx}dx + ʃ_{a+dx}^{b-dx }f '(x)dx +{[f(b-dx) – f (b)]/dx}dx

= f(b)-f(a)

The left and right hand terms after the first = sign are the same whether f '(x) is continuous in c<a<b<d or in [a,b] with one-sided derivatives.

It then seems reasonable that ʃ f ' over [a,b] depends only on end points even for continuity defined by one-sided derivatives. Presumably a similar demonstration applies for generalized divergence theorem, “volume" interval to “surface” integral when f is only defined inside and on the boundary of a closed domain and continuous there with one-sided partial derivatives. But I would like to see it.

I was convinced by Courant who defines a function as continuous on [a,b] if f(x) approaches a limit for all x in [a,b] with a neighborhood in [a,b], ie, one sided derivatives. Also for 2d and hence presumably for higher dimensions.

My apologies to Rudin even though he never mentions one-sided-derivatives and is impossible to remember.