# An integral and function inequality proof

1. Sep 26, 2007

### brh2113

1. The problem statement, all variables and given/known data
Let f be a function such that |f(u)-f(v)|$$\le|u-v|$$ for all u and v in an interval [a,b]. Assume that f is integrable on [a,b]. Prove that

|$$\int_a^bf(x)dx$$ - (b-a)f(a)|$$\leq\frac{(b-a)^{2}}{2}$$

That's the absolute value of the difference of the integral of f(x) from a to b and (b-a)f(a). The subtraction and absolute value signs aren't quite coming out right, but it should be good enough to interpret.

Once I prove this, I have to prove the more general statement that for any c in [a,b],

|$$\int_a^bf(x)dx$$ - (b-a)f(c)|$$\leq\frac{(b-a)^{2}}{2}$$

2. Relevant equations

Standard integration formulas. I might be able to throw the triangle equality somewhere to break up those absolute value differences.

3. The attempt at a solution

I've realized that the given information implies that |m|$$\leq1$$ because
m = $$\frac{|f(u)-f(v)|}{|u-v|}$$ and |u-v| is less than or equal to |f(u)-f(v)|.

$$\frac{(b-a)^{2}}{2}$$ also looks like the formula for a triangle that has a height equal to its base length. So I think that if the slope were 1, then the integral would be equal to the area of the triangle. If the slope is less than 1, then I think the integral is less than the area of the triangle, because it won't reach the same height at point b that the triangle will reach. Using this, I started working with f(x) = (1)x + c, where c is the y intercept (I used c instead of b to avoid confusion). This hasn't gotten me very far, though, and I'm having trouble putting the logic into precise enough mathematical language to call it a proof. Any advice?

2. Sep 26, 2007

### NateTG

Well, let's say we have $f(x)$, and $[a,b]$ as described. If you can you find (nice) functions that are upper and lower bounds for $f(x)$, then you might be able to use those instead of this nebulous $f(x)$.

3. Sep 26, 2007

### brh2113

I've noticed now that (b-a)f(a) removes the area under the rectangle generated from 0 to the y-intercept from a to b. Can I then rewrite the integral - (b-a)f(a) as just the integral of the function without a y-intercept; namely, as $$\int_a^b(mx)dx$$?

Then I get that $$\frac{(b^{2}-a^{2)}}{2}$$$$\leq\frac{(b-a)^{2}}{2}$$.

I'm not sure if this is the right way to go, though.

From here it follows that $${(b+a)}$$ $$\leq$$ $${(b-a)}$$, which isn't always true, so I seem to have either made a mistake or gone down the completely wrong path.

4. Sep 27, 2007

### NateTG

I'd try to attack this problem with:
$$f(a)-(x-a) \leq f(x) \leq f(a)+(x-a)$$
$$\int f(a)-(x-a) dx \leq \int f(x) dx \leq \int f(a)+(x-a) dx$$

5. Sep 27, 2007

### brh2113

Thank you for your help. I managed to solve it by starting with the given information and integrating that to work toward the statement I wanted to prove.