# Calculus problem involving Mean-value Theorem and Riemann integrable functions

1. Sep 29, 2014

### CatWhisperer

1. The problem statement, all variables and given/known data

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.

2. Relevant 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 $$f'(c) = \frac{f(b)-f(a)}{b-a}$$
3. 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.

2. Sep 29, 2014

### 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.

Last edited: Sep 29, 2014
3. Sep 29, 2014

### pasmith

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

4. Sep 29, 2014

### CatWhisperer

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.

5. Sep 30, 2014

### LCKurtz

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 $m_j$.

6. Sep 30, 2014

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

7. Sep 30, 2014

### LCKurtz

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.

Last edited: Sep 30, 2014
8. Oct 1, 2014

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

?

9. Oct 1, 2014

### LCKurtz

$c_j$ 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 $f'(c_j)$ ?

Last edited: Oct 2, 2014