Proving a Continuous Function Fails to Be Zero: A Deep Dive into Theoretical Mathematics

When studying mathematical proofs, one often encounters the notion of a continuous function and theorems that deal with the zeros of such functions. In this article, we will explore a specific scenario where we prove that for a continuous function, the function cannot be identically zero over a certain interval. This proof will not rely on a theorem but rather on the contrapositive method and properties of continuous functions.

Introduction to Continuous Functions

A continuous function is one that behaves smoothly over its domain without any sudden jumps or breaks. This property ensures that small changes in the input result in only small changes in the output. In calculus, continuity is a fundamental concept that is essential for understanding various aspects of mathematics, including integration and differentiation.

Proof by Contrapositive

To prove that a continuous function is non-zero over an interval, we can use the method of proof by contrapositive. This method involves proving the contrapositive statement, which is logically equivalent to the original statement. If we can show that the negation of the conclusion implies the negation of the hypothesis, then we have proven the original statement.

The original statement for our problem is: If a continuous function ( u ) on the interval [0, 1] is zero everywhere, then ( u(x) 0 ) for all ( x in [0, 1] ).

The contrapositive statement is: If there exists a point ( x_0 ) in the interval [0, 1] such that ( u(x_0) eq 0 ), then ( u(x) eq 0 ) for at least one point ( x in [0, 1] ).

Since the contrapositive is logically equivalent to the original statement, proving the contrapositive will suffice to prove the original statement.

Proof Steps

Let's proceed with the proof:

Step 1: Assume the existence of a non-zero point

Assume there is a point ( x_0 in [0, 1] ) such that ( u(x_0) eq 0 ). For simplicity, let's assume ( u(x_0) > 0 ) (the case for ( u(x_0)

Step 2: Use the definition of continuity

By the definition of continuity, for any ( epsilon > 0 ), there exists a ( delta > 0 ) such that ( |u(x) - u(x_0)| 0 ) such that for all ( x in (x_0 - delta, x_0 delta) cap [0, 1] ), we have:

[ |u(x) - u(x_0)| This implies:

[ -frac{u(x_0)}{2} Adding ( u(x_0) ) to all parts of the inequality:

[ u(x_0) - frac{u(x_0)}{2} Simplifying the inequality:

[ frac{u(x_0)}{2} Therefore, ( u(x) > 0 ) for all ( x in (x_0 - delta, x_0 delta) cap [0, 1] ).

Step 3: Conclude the existence of a non-zero interval

Since ( u(x) > 0 ) in the interval ( (x_0 - delta, x_0 delta) cap [0, 1] ), we can assert that there is a non-zero interval around ( x_0 ) where the function ( u ) is positive.

Let ( a ) and ( b ) be the endpoints of this interval such that ( a [ u(x) > 0 ]

Integration and Conclusion

Now, let's consider the integral of ( u(x) ) over the interval [0, 1]. Since ( u(x) > 0 ) in the interval ( [a, b] ), we can write:

[ int_0^1 u(x) , dx geq int_a^b u(x) , dx geq int_a^b frac{u(x_0)}{2} , dx ]

Since ( u(x_0) eq 0 ), we can simplify the integral as:

[ int_a^b u(x) , dx geq int_a^b frac{u(x_0)}{2} , dx frac{u(x_0)}{2} (b - a) ]

Thus, we have:

[ int_0^1 u(x) , dx geq frac{u(x_0)}{2} (b - a) geq 0 ]

Since ( frac{u(x_0)}{2} > 0 ) and ( b - a > 0 ), it follows that:

[ int_0^1 u(x) , dx > 0 ]

Therefore, we have shown that if there exists a point ( x_0 in [0, 1] ) such that ( u(x_0) eq 0 ), then the integral of ( u(x) ) over [0, 1] is positive. This, in turn, proves that ( u(x) ) cannot be identically zero over the interval [0, 1].

In conclusion, if a continuous function ( u ) on the interval [0, 1] is not identically zero, then the integral of ( u(x) ) over [0, 1] is strictly positive. This confirms the original statement that a continuous function cannot be identically zero over a compact interval without any additional theorems.

References and Further Reading

[1] Stewart, J. (2015). Calculus: Early Transcendentals (8th ed.). Cengage Learning.

[2] Apostol, T. M. (1974). Mathematical Analysis (2nd ed.). Addison-Wesley.