# A proof problem.

1. Jun 23, 2007

### MathematicalPhysicist

let f(x) be continuously differentiable in [0,infinity) such that the derivative f'(x) is bounded. suppose that the integral $$\int_{a}^{\infty}|f(x)|dx$$ converges. prove that f(x)->0 when x->infinity.

now, here's what i did:
f'(x) is bounded then for every x>=0 |f'(x)|<=M for some M>0.
now bacause the integral $$\int_{a}^{\infty}|f(x)|dx$$ converges for some a>0, then by cauchy criterion we have that for every e>0 there exists B such that for every b1>b2>B>0 $$\int_{b2}^{b1}|f(x)|dx<e$$ now we can use the inequality: $$|\int_{b2}^{b1}f(x)dx|<=\int_{b2}^{b1}|f(x)|dx<e$$.
now, the crucial point here is that i need to show that for every e>0 there exists M'>0 such that for every x>M' |f(x)|<e.
now i think i can use the mean value theorem for integral here, so there exists a point x in (b2,b1) such that $$\int_{b2}^{b1}f(t)dt=f(x)(b1-b2)$$ and then we have that for every x greater than B
|f(x)|(b1-b2)<e so |f(x)|<e/(b1-b2) but this doesnt work cause b1 and b2 aren't constants, so perhaps here i should use that the derivative is bounded, so by lagrange theorem we have that: |f(b1)-f(b2)|<=M(b1-b2) but still i don't see how to connect both of these theorems to prove this statement.
any hints?

2. Jun 23, 2007

### StatusX

If the derivative is bounded, it means the function can't changes that quickly near a local max. So you can give a lower bound for the integral over a region in terms of the sup of f(x) in that region.

3. Jun 23, 2007

### MathematicalPhysicist

how does a lower bound will help me here?
i think i need to prove it by epsilon delta, unless there's another way?

4. Jun 23, 2007

### StatusX

Trust me, you want a lower bound. It will help you show that if the integral of |f(x)| is small enough, then sup{|f(x)|} must be small too.

5. Jun 23, 2007

### MathematicalPhysicist

ah, ok i understand.
but still i dont see how to use here the boudedness of the derivative, i mean: |f'(x)|<=M for some M>0, so bacause f(x) is continuous it is bounded so if sup(f(x))=max(f(x) in some closed interval, what to do from here?

6. Jun 23, 2007

### StatusX

The idea is that the function can only fall from its max so quickly. Find the function that does this the fastest, and you know the actual function is at least as big.

7. Jun 24, 2007

### MathematicalPhysicist

sorry but i don't see how to implement it here, i mean if we have a closed interval then sup=max in this interval and the function is not greater than this value for every x in the interval now if i were to use the boudedness, i would get by the mvt that |f(b1)-f(b2)|<=M(b1-b2) but f(x)<=max(f(x))
|f(x)|<=|f(x)-f(b2)|+|f(b2)-f(b1)|+|f(b1)|<=M(x-b2)+M(b1-b2)+max(f(x)) but still stuck.