Should the function f to be continuous for applying MVTI or not?

[michaeldoe]

I have found the following proof of remainder term for midpoint rule of integration:

and I'm trying to understand the part of it where author is applying MVTI to composition of functions $f''(\xi_i(x))$ and move it out of integral sign as $f''(\xi_i)$. If we solve Taylor's series for this composition we get $f''(\xi_i(x)) = \frac{2(f(x) - f(c_i) - f'(c_i)(x - c_i))}{(x - c_i)^2}$ which have the problem at $c_i$. The composition $f''(\xi_i(x))$ is discontinuous here. $\frac{0}{0}$ case. So, why is it correct to apply MVTI?

I should say that I'm learning math by myself, as a hobby. And what I've found then ... I tired of controversial information in different resources actually, so I'm here.

In a different resources I have seen different definitions of the Mean Value Theorem for Integrals. Say we have $\int_{a}^{b} f(x) g(x) dx$. In some resources it's written as neither function $f(x)$ nor $g(x)$ should be continuous to apply this theorem, but both $f(x)$ and $g(x)$ should be integrable and $g(x)$ shouldn't to change sign. Example of such definitions are typically can be found in Russian literature like Fikhtengol'ts Calculus, in RU Wikipedia page of this theorem and also some teachers saying different definitions of it.

And in some resources I have seen different definition which said that $f(x)$ should be continuous strictly and $g(x)$ should be integrable and shouldn't change sign. For example EN Wikipedia page of this theorem, OpenStax Calculus book, Stewart Calculus and others.

In this case, in the proof the $f(x)$ is $f''(\xi_i(x))$ which is not continuous by its definition (fraction above) and $g(x) = (x - c_i)^2$.

Well, should $f(x)$ be strictly continuous in order to apply MVT-I theorem and move this $f''(\xi_i(x))$ out of integral sign as $f''(\xi_i)$ or not finally?

 

[Office_Shredder]

Bullet point 2 mentions f being in $C^2$ which means its second derivative is continuous. It's unclear to me if that's an assumption for this entire section or not. It might help to post the entirety of what is being done so we know what assumptions you're starting with.

 

[michaeldoe]

Bullet point 2 mentions f being in $C^2$ which means its second derivative is continuous. It's unclear to me if that's an assumption for this entire section or not. It might help to post the entirety of what is being done so we know what assumptions you're starting with.

The screenshot I have attached is actually what is being done (the whole proof, if that's what you meant).

Also, yesterday I have read more and it seems I have understood the actual difference between two versions of MVTI for weaker (integrability only) and stronger case.

When $f$ is not continuous (only integrable) we get some "average number" as a result, say, $\lambda$ which is not necessary can be attained by $f$ at some particular point. So we can't say $\lambda = f(c)$ in this case.

While with continuity we have intermediate value property and we can say $\lambda = f(c)$ for some $c$.

But it's still doesn't answer why it's valid to move discontinuous $f''(\xi_i(x))$ out of integral sign as $f''(\xi_i)$ for some $\xi_i$. Perhaps because $f''$ itself is continuous and therefore have intermediate value property?

My concern is that while second derivative of $f$ itself is continuous by assumption, that's not the function we have under interegral sign, but $f''(\xi_i(x))$.

 

[fresh_42]

We want to apply the MVTI. This says, if $F(x)$ is a continuous function, $G\, : \,[a,b]\longrightarrow \mathbb{R}$ is either $G\geq 0$ or $G\leq 0$ then there is a $\xi\in [a,b]$ such that
$$
\int_a^b F(x)G(x)\,dx=F(\xi)\int_a^bG(x)\,dx\,.
$$
Since $f''\in C^2$ and we want to apply it to $F=f''\circ\xi$ and $G(x)=(x-c_i)^2.$ Hence, we need that $f''\circ \xi$ is continuous, and we need to consider $\xi.$ If it is continuous, then the statement in your book is exactly the MVTI. But is $\xi(x)$ really a function of $x$?

I think it is just an unfortunate notation. What is $f''(\xi(x))$? I would have written it as $f''(\xi ; x)$ where $\xi$ indicates a parameter and $x$ is the variable. $f''(\xi ; x)$ is still continuous in $x$ so we don't have $F=f''\circ \xi$ but actually $F=f''(\xi ;x)$ and the MVTI should be
$$
\int_{x_i}^{x_{i+1}} f''(\xi, x)(x-c_i)^2\,dx=f''(\xi;\xi')\int_{x_i}^{x_{i+1}} (x-c_i)^2\,dx
$$
where $\xi,\xi'\in [x_i,x_{i+1}].$ It all depends on how $F$ looks like and shouldn't it be something like $\xi'' \cdot (x-c_i)^3$?

 

[michaeldoe]

What is f″(ξ(x))?

$f''(\xi(x))$ comes from Taylor's theorem with the Lagrange form of the remainder. This theorem is actually what is used in the original post to prove/derive the error term for the Midpoint Rule (a numerical method of integration):

$$
\frac{h^2(b-a) f''(\xi)}{24}
$$

The value $\xi$ depends on $x$ and lies between $(c_i, x)$ or $(x, c_i)$. As far as I know, $\xi(x)$ itself is not continuous (it's even not explicitly known).

I actually found this proof here: https://math.stackexchange.com/ques...e-error-for-the-midpoint-rule/4327333#4327333

It's a classical method of proof using Taylor's theorem with the Lagrange form of the remainder. I also found the same proof in a classical book on analysis, but I used the one from Math Stack Exchange for convenience.

However, neither this proof nor the one in the book justifies the use of the Mean Value Theorem for Integrals (MVTI) on $f''(\xi(x))$. The book I'm referring to is in Russian, so I'm not sure it makes sense to include screenshots of the proof here.

-----

Edit:

The one justification I've found used argument that $f''$ itself is continuous so by using Intermediate Value Property we move out of integral sign not $f''(\xi_i(x))$ but original second derivative $f''$. Because $\xi_i(x)$ takes argument $x$ which lies in $[x_i, x_{i + 1}]$ subinterval and returns value from the same subinterval, and original second derivative of a function (which is continuous by assumption) is continuous on this subinterval.

But I can't connect it.