If integral equals zero, then function equals zero

In summary: N such that f(x)>d/2, then it follows that f is continuous at x. However, this does not prove that f is also continuous at every point in N. For example, if x is not in N, then f(x) may still be greater than zero.
  • #1
jdz86
21
0

Homework Statement



Let f : [a,b] [tex]\rightarrow[/tex] [tex]\Re[/tex] be continuous and assume f [tex]\geq[/tex] 0. Prove that if [tex]\int_{[a,b]}[/tex]f = 0 then f = 0.

Homework Equations



Nothing really. If relevant, mean value theorem was discussed in earlier problems, so I'm not sure if it fits though.

The Attempt at a Solution



I tried using the MVT and stating that if all that was mentioned above holds true then you can say f(c)(b-a) = 0, then solving b = a. From there I couldn't go anywhere. Pretty sure I wasn't on the right track to begin with in the first place anyway.
 
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  • #2
Intuitively the problem is clear i would guess. SInce f>=0 then what it menas is that from a to b there is no area accumulated at all, so the only possibility in this case is for f to be the zero function(f=0).

Now, to formally prove it i would suggest starting from the definition of the definite integral in terms of Riman sums would help.
 
  • #3
Suppose that f is not identically zero, then a lower bound for the integral is...
 
  • #4
Well the upper bound, sup(f) = b , and the lower bound, inf(f) = a

And with Riemann Sums, I don't see how you can add up the partitions and take the Upper Sum and Lower Sum of a function that you don't know.

I just can't see where they fit into the proof.
 
  • #5
No, sup(f) is not b, nor is inf(f) equal to a. In fact you don't know what inf and sup are.

You have also failed to use or mention the fact that f is continuous.

Suppose that f is not identically 0, i.e. there exists a c in (a,b) with f(c)=d>0.

f is continuous. What does that say about f in small enough neighbourhood of c?
 
  • #6
f converges to c??
 
  • #7
No, no, no. If f(c)=d>0 and f is continuous, then there is an interval N=(c-delta,c+delta) around x=c such that for x in N, f(x)>d/2, for example. If you can tell me why, then I'll tell you why the integral must be greater than zero.
 
  • #8
With your example f(x)>d/2, since f(c)=d where you have x=c so then f(x)=d. And d/2<d. But I still can't see the relation to my original problem.
 
  • #9
The relation to your original problem is to show that if f(x)>0 at any point then the integral must be greater than zero. Because if f is continuous and positive at any point, then it is positive on an interval around that point. That forces the integral to be nonzero. That's what people have been trying to tell you. Hence, f(x) is equal to zero everywhere. Now can you add the details?
 
  • #10
Haha, I must've just stared at your statement for about 2 minutes, then it clicked. So f would have to be zero everywhere because if it was greater than zero than it's integral would have to be greater than zero as well. Thanks. Now I just have to put everything together.
 
  • #11
jdz86 said:
Haha, I must've just stared at your statement for about 2 minutes, then it clicked. So f would have to be zero everywhere because if it was greater than zero than it's integral would have to be greater than zero as well. Thanks. Now I just have to put everything together.

You still haven't fully(formaly) justified, as Dick asked you to do, as to why f(x)>0 on some interval N, if the function f(x) is nonzero at any point whatsoever...
 
  • #12
Well if the function is non-zero at every point and continuous, then when taking the value of the function at some point along a given interval, that point will be non-zero as well since every other point along the function is non-zero. Is that what you were asking??
 
  • #13
jdz86 said:
Well if the function is non-zero at every point and continuous, then when taking the value of the function at some point along a given interval, that point will be non-zero as well since every other point along the function is non-zero. Is that what you were asking??

No, not really. What i am saying is that, you haven't shown why f(x)>0, if say,(like Dick already explained it really well) at some point c,
f(c)=d>0. In other words, if at some single point f is greater than zero, then you have to show, why will there exist an interval
N=(c-delta,c+delta), such that for every point x in this interval N, f(x)>0?
 
  • #14
Apologies for bringing up an old thread but I am attempting to solve a similar problem...

I can understand the workings of the proof that has been discussed above, but am having difficulties in actually writing it formally.

In particular, I am having trouble trying to prove why if at some single point f is greater than zero, then there must exist an interval N=(c-delta,c+delta), such that for every point x in this interval N, f(x)>0?

Clearly, by the epsilon-delta definition of continuity, if there is a point f(c)=d>0 then there must be points within epsilon either side of f(c)=d. Can we then just increase our value epsilon (by definition of continuity, we can take any value for epsilon) until all x such that f(x) is non-zero is within delta either side of c? Hence, there would then be an interval N=(c-delta,c+delta) such that for every point x in this interval N, f(x)>0.

Does this work? How do I then show that the integral of f(x) between c-delta and c+delta is greater than 0?

Thanks very much for your help!
 
  • #15
You don't have to get that involved with it. Just choose [tex]\epsilon < d[/tex] to begin with.
 
  • #16
I don't know how I didn't see that before...thanks!

How do I then go on to show that the integral of f(x) between c-delta and c+delta (and therefore the whole function between a and b) is greater than 0?
 
  • #17
I messed up, you don't want to take [tex]\epsilon < d[/tex] but some fraction of d. 1/2 is convenient.

then [tex]f > d/2[/tex]. But [tex]c - \delta < x < c + \delta[/tex] gives you an interval for which [tex]f(x) > d/2[/tex]
 
  • #18
THe idea is relatively simple. If f is a continuous function on an interval (a,b), and if there is a point c in that interval such that f(c)=d>0, then we have to show that there must be some neighborhood Q of c such that for all x in Q f(x)>0.

It is not difficult to construct a sequence {x_n} of members of (a,b) each distinct from c, and converging to c, hence c is an accumulation point of (a,b).

Now, given that f is continuous, for every epsilon greater than zero and also for e=d/2 there should exist some delta such that anytime |x-c|<delta |f(x)-d|<d/2.

so: -d/2<f-d<d/2=> 0<d/2<f<3d/2, for all x in (c-delta, c+delta).


Then to show that the integral is not zero, go to the definition of the integral using rieman sums. when you divide the interval [a,b] into n subintervals take (c-delta, c+delta) as one of your intervals.
 

1. What does it mean if the integral of a function is equal to zero?

If the integral of a function is equal to zero, it means that the area under the curve of the function is zero. This could indicate that the function is symmetric about the x-axis, or that the positive and negative areas cancel each other out.

2. Can a function be equal to zero if its integral is not equal to zero?

Yes, it is possible for a function to equal zero without its integral being equal to zero. This is because the integral of a function represents the area under the curve, while the function itself represents the value at a specific point on the curve.

3. Does a zero integral always mean that the function is identically zero?

No, a zero integral does not always mean that the function is identically zero. It is possible for a function to have a zero integral, but still have non-zero values at certain points on the curve.

4. Are there any other conditions for a function to equal zero besides having a zero integral?

Yes, there are other conditions for a function to equal zero. For example, the function could have a root or zero at a specific point, or it could have a horizontal asymptote at y=0.

5. How can the statement "if integral equals zero, then function equals zero" be proven?

This statement can be proven using the Fundamental Theorem of Calculus, which states that if a function f(x) is continuous on the interval [a,b] and its derivative f'(x) exists on that interval, then the definite integral of f'(x) from a to b is equal to f(b) - f(a). Therefore, if the integral of a function is equal to zero, then f(b) - f(a) = 0, which means that f(b) = f(a). Since f(b) and f(a) are both equal to zero, this proves that the function is equal to zero.

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