An integral and function inequality proof

[brh2113]

Homework Statement

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}

Homework Equations

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

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?

 

[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).

 

[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.

 

[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

 

[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.