Integral proof problem

**Runty** · January 22nd 2010, 08:37 AM

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

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

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

A hint provided was to prove this via contradiction.

**Soroban** · January 22nd 2010, 10:41 AM

Hello, Runty!

I have an intuitive (graphic) proof . . .

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

A hint provided was to prove this via contradiction.

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

There are three cases:

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

For case (1), we have this graph:

Code:

        |
        |           *
        |         *:|
        |       *:::|
        |     *:::::|
        |   *:::::::|
        |   |:::::::|
    - - | - + - - - + - -
        |   a       b

$\int^b_a|f(x)|\,dx$ represents the area under the curve,
. . . which is evidently not zero.

For case (2), we have this graph:

Code:

        |
        |   a       b
    - - + - + - - - + - -
        |   |:::::::|
        |   *:::::::|
        |     *:::::|
        |       *:::|
        |         *:|
        |           *

$\int^b_a|f(x)|\,dx$ represents the area above the curve,
. . . which also is not zero.

For case (3), we have this graph for $f(x)$ :

Code:

        |
        |   *
        |     *
        |       *
        |         *
    - - + - + - - - * - - + - - -
        |   a         *   b
        |               *
        |                 *

The the graph of $|f(x)|$ looks like this:

Code:

        |
        |   *
        |   |:*           *
        |   |:::*       *:|
        |   |:::::*   *:::|
    - - | - + - - - * - - - + - - -
        |   a               b
        |

$\int^b_a|f(x)|\,dx$ represents the area under the curve,
. . . which again is not zero.

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

**Runty** · January 22nd 2010, 10:50 AM

Originally Posted by Soroban

Hello, Runty!

I have an intuitive (graphic) proof . . .

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

There are three cases:

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

For case (1), we have this graph:

Code:

        |
        |           *
        |         *:|
        |       *:::|
        |     *:::::|
        |   *:::::::|
        |   |:::::::|
    - - | - + - - - + - -
        |   a       b

$\int^b_a|f(x)|\,dx$ represents the area under the curve,
. . . which is evidently not zero.

For case (2), we have this graph:

Code:

        |
        |   a       b
    - - + - + - - - + - -
        |   |:::::::|
        |   *:::::::|
        |     *:::::|
        |       *:::|
        |         *:|
        |           *

$\int^b_a|f(x)|\,dx$ represents the area above the curve,
. . . which also is not zero.

For case (3), we have this graph for $f(x)$ :

Code:

        |
        |   *
        |     *
        |       *
        |         *
    - - + - + - - - * - - + - - -
        |   a         *   b
        |               *
        |                 *

The the graph of $|f(x)|$ looks like this:

Code:

        |
        |   *
        |   |:*           *
        |   |:::*       *:|
        |   |:::::*   *:::|
    - - | - + - - - * - - - + - - -
        |   a               b
        |

$\int^b_a|f(x)|\,dx$ 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.

**running-gag** · January 22nd 2010, 11:02 AM

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

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

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

**Drexel28** · January 22nd 2010, 12:09 PM

Originally Posted by Runty

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

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

then $f(x)=0$ 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 ?

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

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

Are you trying to prove this using Riemann-Stieltjes integration concepts?

**Runty** · January 22nd 2010, 12:55 PM

Originally Posted by Drexel28

Are you trying to prove this using Riemann-Stieltjes integration concepts?

No, that was not mentioned.

**Plato** · January 22nd 2010, 02:47 PM

Originally Posted by Runty

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

This is a well known theorem: If $g$ is continuous on $[a,b]$ and $\left( {\exists p \in [a,b]} \right)\left[ {g(p) > 0} \right]$ then $\left( {\exists [c,d] \subseteq [a,b]} \right)\left[ {\left( {\forall x \in [c,d],\;g(x) > 0} \right)} \right]$ .

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

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