# Thread: Integral proof problem

1. ## Integral proof problem

Prove that, if f is continuous on [a,b], and

$\displaystyle \int_{a}^{b} |f(x)| dx = 0$

then $\displaystyle f(x)=0$ for all x in [a,b].

A hint provided was to prove this via contradiction.

2. Hello, Runty!

I have an intuitive (graphic) proof . . .

Prove that, if $\displaystyle f(x)$ is continuous on $\displaystyle [a,b]$, and: .$\displaystyle \int_{a}^{b} |f(x)| dx \:=\: 0$
. . then $\displaystyle f(x)\,=\,0$ for all $\displaystyle x \in [a,b].$

A hint provided was to prove this via contradiction.
Suppose $\displaystyle f(x)$ is not equal to 0 for all $\displaystyle x \in [a,b].$
. . Then $\displaystyle f(x)$ is not the $\displaystyle x$-axis.

There are three cases:

. . (1) $\displaystyle f(x)$ is above the x-axis on the interval $\displaystyle [a,b]$
. . (2) $\displaystyle f(x)$ is below the x-axis on the inverval $\displaystyle [a,b].$
. . (3) $\displaystyle f(x)$ crosses the x-axis on the interval $\displaystyle [a,b].$

For case (1), we have this graph:
Code:
|
|           *
|         *:|
|       *:::|
|     *:::::|
|   *:::::::|
|   |:::::::|
- - | - + - - - + - -
|   a       b
$\displaystyle \int^b_a|f(x)|\,dx$ represents the area under the curve,
. . . which is evidently not zero.

For case (2), we have this graph:
Code:
|
|   a       b
- - + - + - - - + - -
|   |:::::::|
|   *:::::::|
|     *:::::|
|       *:::|
|         *:|
|           *
$\displaystyle \int^b_a|f(x)|\,dx$ represents the area above the curve,
. . . which also is not zero.

For case (3), we have this graph for $\displaystyle f(x)$:
Code:
|
|   *
|     *
|       *
|         *
- - + - + - - - * - - + - - -
|   a         *   b
|               *
|                 *

The the graph of $\displaystyle |f(x)|$ looks like this:
Code:
|
|   *
|   |:*           *
|   |:::*       *:|
|   |:::::*   *:::|
- - | - + - - - * - - - + - - -
|   a               b
|
$\displaystyle \int^b_a|f(x)|\,dx$ represents the area under the curve,
. . . which again is not zero.

[This may not be considered a a satisfactory proof.]

3. Originally Posted by Soroban
Hello, Runty!

I have an intuitive (graphic) proof . . .

Suppose $\displaystyle f(x)$ is not equal to 0 for all $\displaystyle x \in [a,b].$
. . Then $\displaystyle f(x)$ is not the $\displaystyle x$-axis.

There are three cases:

. . (1) $\displaystyle f(x)$ is above the x-axis on the interval $\displaystyle [a,b]$
. . (2) $\displaystyle f(x)$ is below the x-axis on the inverval $\displaystyle [a,b].$
. . (3) $\displaystyle f(x)$ crosses the x-axis on the interval $\displaystyle [a,b].$

For case (1), we have this graph:
Code:
|
|           *
|         *:|
|       *:::|
|     *:::::|
|   *:::::::|
|   |:::::::|
- - | - + - - - + - -
|   a       b
$\displaystyle \int^b_a|f(x)|\,dx$ represents the area under the curve,
. . . which is evidently not zero.

For case (2), we have this graph:
Code:
|
|   a       b
- - + - + - - - + - -
|   |:::::::|
|   *:::::::|
|     *:::::|
|       *:::|
|         *:|
|           *
$\displaystyle \int^b_a|f(x)|\,dx$ represents the area above the curve,
. . . which also is not zero.

For case (3), we have this graph for $\displaystyle f(x)$:
Code:
|
|   *
|     *
|       *
|         *
- - + - + - - - * - - + - - -
|   a         *   b
|               *
|                 *

The the graph of $\displaystyle |f(x)|$ looks like this:
Code:
|
|   *
|   |:*           *
|   |:::*       *:|
|   |:::::*   *:::|
- - | - + - - - * - - - + - - -
|   a               b
|
$\displaystyle \int^b_a|f(x)|\,dx$ represents the area under the curve,
. . . which again is not zero.

[This may not be considered a a satisfactory proof.]

It's a good proof, but it needs to be solved via calculus equations. I'm afraid I can't really use it.

4. Are you allowed to use this definition of continuity of at c ?

$\displaystyle \forall \epsilon > 0 \: \exists \eta > 0 \:\forall x \in [a,b] \:|x-c| < \eta \Rightarrow |f(x) - f(c)| < \epsilon$

If so suppose that f is not equal to 0 for all x in [a,b]

This means that exists c in [a,b] such that f(c) is not equal to 0

Let $\displaystyle \epsilon = \frac{|f(c)|}{2}$

5. Originally Posted by Runty
Prove that, if f is continuous on [a,b], and

$\displaystyle \int_{a}^{b} |f(x)| dx = 0$

then $\displaystyle f(x)=0$ for all x in [a,b].

A hint provided was to prove this via contradiction.
Originally Posted by running-gag
Are you allowed to use this definition of continuity of at c ?

$\displaystyle \forall \epsilon > 0 \: \exists \eta > 0 \:\forall x \in [a,b] \:|x-c| < \eta \Rightarrow |f(x) - f(c)| < \epsilon$

If so suppose that f is not equal to 0 for all x in [a,b]

This means that exists c in [a,b] such that f(c) is not equal to 0

Let $\displaystyle \epsilon = \frac{|f(c)|}{2}$
Are you trying to prove this using Riemann-Stieltjes integration concepts?

6. Originally Posted by Drexel28
Are you trying to prove this using Riemann-Stieltjes integration concepts?
No, that was not mentioned.

7. Originally Posted by Runty
Prove that, if f is continuous on [a,b], and $\displaystyle \int_{a}^{b} |f(x)| dx = 0$ then $\displaystyle f(x)=0$ for all x in [a,b].
This is a well known theorem: If $\displaystyle g$ is continuous on $\displaystyle [a,b]$ and $\displaystyle \left( {\exists p \in [a,b]} \right)\left[ {g(p) > 0} \right]$ then $\displaystyle \left( {\exists [c,d] \subseteq [a,b]} \right)\left[ {\left( {\forall x \in [c,d],\;g(x) > 0} \right)} \right]$.

Apply that theorem to $\displaystyle |f|$. Note that $\displaystyle \int_a^b {\left| f \right|dx} = \int_a^c {\left| f \right|dx} + \int_c^d {\left| f \right|dx} + \int_d^b {\left| f \right|dx} > 0$.