Calculus problem involving Mean-value Theorem and Riemann integrable functions

[CatWhisperer]

Homework Statement

Let ##f:[a,b] \rightarrow R## be a differentiable function. Show that if ##P = \{ x_0 , x_1 , ... , x_n \}## is a partition of ##[a,b]## then $$L(P,f')=\sum_{j=1}^n m_j \Delta x_j \leq f(b) - f(a)$$ where ##m_j=inf \{ f'(t) : t \in [x_{j-1} , x_j ] \}## and ##\Delta x_j = x_j - x_{j-1}## for each ##1 \leq j \leq n##. Hint: Use the Mean-value Theorem.

Homework Equations

Mean-value Theorem:
If ##f## is continuous on ##[a,b]## and differentiable on ##(a,b)## then there exists a ##c \in (a,b)## such that $$f'(c) = \frac{f(b)-f(a)}{b-a}$$

The Attempt at a Solution

##f## is differentiable on ##[a,b]##, so it must also be continuous on ##[a,b]##, so by the mean value theorem there exists a ##c \in [a,b]## such that $$f'(c) = \frac{f(b)-f(a)}{b-a}$$ $$\sum_{j=1}^n \Delta x_j = b-a$$ $$\Leftrightarrow \sum_{j=1}^n m_j \Delta x_j = (b-a)\sum_{j=1}^n m_j \leq f(b) - f(a)$$ $$\Leftrightarrow \sum_{j=1}^n m_j \leq \frac{f(b) - f(a)}{b-a} = f'(c)$$ (where ##f'(c)## is described by the MVT)

which is about as far as I have gotten. I would really appreciate any assistance. I don't really know how I am supposed to bring MVT into this; how I have done so here doesn't really seem to help me.

Thanks very much in advance.

 

[CatWhisperer]

It just occurred to me that I can't necessarily assume I can factor out ##(b-a)## unless I know that all ##\Delta x## are of equal width... which I do not. The only thing I can think to do in that case is: $$L(P,f') = \sum_{j=1}^n m_j \Delta x_j \leq f'(b) - f'(a)$$ but I am still stumped as to how I use MVT here, and how this relates to the original inequality.

 

[pasmith]

Apply the mean value theorem not to [itex][a,b][/itex], but to each [itex][x_{j-1},x_j][/itex].

 

[CatWhisperer]

Apply the mean value theorem not to [itex][a,b][/itex], but to each [itex][x_{j-1},x_j][/itex].

Following your suggestion, I have made a further attempt at solving the problem:

##f## is differentiable and continuous on ##[a,b]## so ##f## must also be differentiable and continuous on every ##\Delta x=[x_{j-1},x_j]##. Thus, there exists a ##c \in [x_{j-1},x_j]## such that $$f'(c)=\frac{f(x_j)-f(x_{j-1})}{\Delta x_j}$$ $$\Leftrightarrow \Delta x_j=\frac{f(x_j)-f(x_{j-1})}{f'(c)}$$ $$\Leftrightarrow \sum_{j=1}^n m_j\frac{f(x_j)-f(x_{j-1})}{f'(c)}\leq f(b)-f(a)$$ Now, I am not sure about this next part.

We know that ##m_j=inf\{ f'(x):x\in [x_{j-1},x_j]\}##, so if we choose ##x=c## then we get ##m_j=inf\{ f'(c):c\in [x_{j-1},x_j]\}## which gives $$\sum_{j=1}^n [f(x_j)-f(x_{j-1})]\leq f(b)-f(a)$$ but I don't really know if that's right, or how to explain if it is.

Is anyone able to give me further guidance here? Thanks in advance.

 

[LCKurtz]

Following your suggestion, I have made a further attempt at solving the problem:

##f## is differentiable and continuous on ##[a,b]## so ##f## must also be differentiable and continuous on every ##\Delta x=[x_{j-1},x_j]##. Thus, there exists a ##c \in [x_{j-1},x_j]## such that $$f'(c)=\frac{f(x_j)-f(x_{j-1})}{\Delta x_j}$$

You don't get the same ##c## on each interval. You could say there exists ##c_j \in [x_{j-1},x_j]## such that$$
f'(c_j)=\frac{f(x_j)-f(x_{j-1})}{\Delta x_j}$$

The rest of your argument has gone off the tracks. The next thing you should write down is the sum you are trying to work with:$$
L(P,f') = \sum_{j=1}^n m_j\Delta x_j$$Now think about how ##f'(c_j)## compares with ##m_j##.

 

[CatWhisperer]

Okay so I think I understand this a lot better now. I drew a picture to understand all the parts of the problem and how they relate to one another and have attempted to write the proof, however I can't be sure if my explanations are adequate. Any feedback is much appreciated:

##f## is differentiable on ##[a,b]##, so ##f## is differentiable on each ##[x_{j-1},x_j]##, so ##f## is continuous on each ##[x_{j-1},x_j]##. Thus, by the Mean-value Theorem for differentiation, there exists a ##c\in [x_{j-1},x_j]## such that $$f'(c)=\frac{f(x_j)-f(x_{j-1})}{\Delta x_j}$$ $$\Leftrightarrow f'(c)\Delta x_j=f(x_j)-f(x_{j-1})$$ $$\Leftrightarrow \sum_{j=1}^n f'(c)\Delta x_j=\sum_{j=1}^n f(x_j)-f(x_{j-1})$$ for each ##c\in [x_{j-1},x_j]##.

As ##\Delta x_j\rightarrow 0##, ##m_j\rightarrow f'(c)##, thus $$\sum_{j=1}^n m_j\Delta x_j\leq \sum_{j=1}^n f'(c_j)\Delta x_j$$ (this part I am not sure if I need to include additional explanation. I proved it to myself visually but that wouldn't be an acceptable method for this type of problem).

As ##\Delta x_j\rightarrow 0## $$\sum_{j=1}^n f(x_j)-f(x_{j-1})\rightarrow f(b)-f(a)$$ Thus, $$\sum_{j=1}^n f'(c_j)\Delta x_j\leq f(b)-f(a)$$ Therefore $$\sum_{j=1}^n m_j\Delta x_j\leq \sum_{j=1}^n f'(c_j)\Delta x_j\leq f(b)-f(a)$$ $$\Leftrightarrow L(P,f')=\sum_{j=1}^n m_j\Delta x_j\leq f(b)-f(a)$$

 

[LCKurtz]

Okay so I think I understand this a lot better now.

I don't think so. It's still very confused. There is nothing in this problem about ##\Delta x_j\rightarrow 0##.

Here's what I suggested before:
You could say there exists ##c_j\in[x_{j−1},x_j]## such that
##f′(c_j)=\frac{f(x_j)−f(x_{j−1})}{Δx_j}##The next thing you should write down is the sum you are trying to work with:$$
L(P,f′)=\sum_{j=1}^n m_jΔx_j$$
Now think about how ##f′(c_j)## compares with ##m_j##. Try following that advice without quoting a bunch of extraneous stuff. It only takes about two steps.

 

[CatWhisperer]

Okay...

There exists a ##c_j\in [x_{j-1},x_j]## such that $$f'(c_j)=\frac{f(x_j)-f(x_{j-1})}{\Delta x_j}$$ Now, $$L(P,f')=\sum_{j=1}^n m_j\Delta x_j$$ And $$m_j=inf\{ f'(c_j):c_j\in [x_{j-1},x_j]\}$$ $$\Leftrightarrow m_j=\frac{inf\{ f(x_j)-f(x_{j-1})\} }{\Delta x_j}$$ $$\Leftrightarrow L(P,f')=\sum_{j=1}^n \frac{inf\{ f(x_j)-f(x_{j-1})\} }{\Delta x_j}\Delta x_j=\sum_{j=1}^n inf\{ f(x_j)-f(x_{j-1})\} $$ $$\Leftrightarrow L(P,f')=\sum_{j=1}^n m_j\Delta x_j\leq f(b)-f(a)$$

?

 

[LCKurtz]

Okay...

There exists a ##c_j\in [x_{j-1},x_j]## such that $$f'(c_j)=\frac{f(x_j)-f(x_{j-1})}{\Delta x_j}$$ Now, $$L(P,f')=\sum_{j=1}^n m_j\Delta x_j$$ And $$m_j=inf\{ f'(c_j):c_j\in [x_{j-1},x_j]\}$$

##c_j## is given on each interval by the MVT. You have no choice about it and you can't vary it so that infimum makes no sense. Look again at your definition of ##m_j## in your original post. Like I asked before, how does ##m_j## compare with ##f'(c_j)## ?