Question

Assume that \(f(x)\) is a continuous function defined on the interval \(a \leq x \leq b\). Suppose it is known that $$ \int_{x_{1}}^{x_{2}} f(x) d x=0 $$ for all choices of \(x_{1}\) and \(x_{2}\) satisfying \(a \leq x_{1} \leq x_{2} \leq b .\) Prove that \(f(x)=0, a \leq x \leq b\). [Hint: You can use a contradiction argument; that is, you can assume that the hypotheses hold but that the conclusion is false. For example, assume that \(f(c)>0\) at some point \(c, a0\) such that \((c-\delta, c+\delta)\) lies within \((a, b)\) while at the same time \(f(x)>\frac{1}{2} f(c)\) for all \(x\) in \((c-\delta, c+\delta) .\) Show that this fact leads to a contradiction.]

Short answer

Question: Prove that if a continuous function has an integral equal to zero in any subinterval within a given interval, the function must be equal to zero in the entire interval. Answer: The function cannot have any positive or negative function values within the given interval, as this would create a contradiction when considering the continuity and properties of the integral. Therefore, the function must be equal to 0 on the entire interval.

Step-by-step solution

Assumptions

Assume that there is a point \(c\) in the interval \((a, b)\) where \(f(c)>0\). Also, since the function is given to be continuous on \([a, b]\), it means that the function is continuous at point \(c\) as well.

Continuity and an auxiliary subinterval

Due to the continuity of \(f(x)\) at point \(c\), we can find a \(\delta>0\) such that for any \(x\) within the interval \((c-\delta, c+\delta)\), \(f(x)>\frac{1}{2}f(c)\). It is important to notice that \((c-\delta, c+\delta)\) lies within the interval \((a, b)\).

Integral within the subinterval

Now let's consider the integral of \(f(x)\) between \(x_1\) and \(x_2\) with \(x_1=c-\delta\) and \(x_2=c+\delta\): $$ \int_{c-\delta}^{c+\delta} f(x) dx. $$

Contradiction argument

We know that \(f(x)>\frac{1}{2}f(c)\) for all \(x\) in the interval \((c-\delta, c+\delta)\). Therefore: $$ \int_{c-\delta}^{c+\delta} f(x) dx > \int_{c-\delta}^{c+\delta} \frac{1}{2}f(c) dx. $$ Then, because \(\frac{1}{2}f(c)\) is a constant: $$ \int_{c-\delta}^{c+\delta} \frac{1}{2}f(c) dx = \frac{1}{2}f(c)\int_{c-\delta}^{c+\delta} dx=\frac{1}{2}f(c)(2\delta). $$ Thus, we have $$ \int_{c-\delta}^{c+\delta} f(x) dx > \frac{1}{2}f(c)(2\delta), $$ which implies that the integral on the left-hand side is positive. However, we are given that the integral of \(f(x)\) on any subinterval within \((a, b)\) is equal to zero. This creates a contradiction, as we have found a subinterval where the integral is strictly greater than zero when we assumed the existence of a point \(c\) with a positive function value. In conclusion, there cannot exist a point \(c\) with a positive function value. Using a similar argument, we can show that there cannot exist a point with a negative function value. Therefore, \(f(x)=0\) for all \(x\) in the interval \([a,b]\).

Key concepts

Continuous Function

A continuous function is one that does not have any breaks, jumps, or points of discontinuity. In simpler terms, if you were to draw the graph of a continuous function, you could do so without lifting your pencil from the paper. This property is crucial when discussing functions like the one in our exercise, where the function is defined to be continuous on the interval \(a \leq x \leq b\).

Continuity guarantees the behavior of the function around any point within its domain. If the function takes on a certain value at some point, continuity ensures that its values at nearby points will be close to that value. In the given exercise, if there's a point \(c\) where \(f(c)>0\), continuity implies that there's a delta-neighborhood \( (c-\delta, c+\delta) \) where \(f(x)\) stays above half of \(f(c)\). This means we have a consistent interval where the function maintains positive values, which is essential for constructing the contradiction argument.

Contradiction Argument

The contradiction argument is a logical method used to prove a statement by assuming that the contrary is true, and then showing that this assumption leads to a logical inconsistency. It's like saying, 'let's assume this is false, and see if things start to make sense or not.' If they do not, we must conclude that our initial assumption was incorrect.

In the context of our problem, the contradiction argument begins by assuming that there exists some \(c\) in the interval \(a < c < b\) where \(f(c) > 0\), despite the condition that integral from \(x_1\) to \(x_2\) of \(f(x)\) dx equals zero for any \(x_1\) and \(x_2\) within \(a \leq x_1 \leq x_2 \leq b\). The argument takes advantage of the function's continuity by pinpointing a subinterval around \(c\) where \(f(x)\) would be necessarily positive; hence the whole integral should be greater than zero, blatantly contradicting the given condition of the integral being zero. Thus, our original assumption—that there is a positive value of \(f(c)\)—cannot be true.

Definite Integral Properties

Definite integrals have several key properties that can be leveraged to solve problems like the one in our exercise. One such property is that the definite integral over an interval sums up the area under the curve of a function between two points on the x-axis. If the function is positive, this area will also be positive; if the function is negative, the area is taken to be negative.

Another critical property is linearity, which means that if we integrate a constant, we would end up with the product of that constant and the length of the interval. Applying this to our problem, the integral of \( (1/2)f(c) \) over the subinterval \( [c-\delta, c+\delta] \) should yield a positive result if \(f(c) > 0\), contrasting with the zero integral requirement for any subinterval within \( [a,b] \). This apparent inconsistency is precisely what's exploited in the contradiction argument to establish that \(f(x)\) must be zero everywhere on the interval.