1)How does the graph of a funtion of bounded variation behave. 2)Why a bounded function is not always a function of bounded variation.Please explain graphically. 3)What purpose does bounded variation serve.I mean why they are defined. Sorry if the question is already asked. And thank you in advance.
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For 2, think of a trigonometric function with a particular argument. – YoTengoUnLCD Mar 29 '16 at 04:39
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On a interval, the length of the function's curve is finite.
Think of $\sin x^{-1}$ on $]0,1]$.
Let $f$ be of bounded variation. It has the following properties:
It is the difference of two bounded increasing functions.
The left and right limits exists at every point within $f$'s domain.
The discontinuity set of $f$ is countable, and $f$ is thus Riemann integrable.
$f'$ is Lebesgue integrable.
Henricus V.
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Sir this doesnt answer my question.As you.can see I have laid amphasis on the graph of function. – Rayees Ahmad Mar 29 '16 at 04:50
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Will you explain a bit. Isnt the graph of $x\sin\frac{\pi}{x}$ smoth except on a countable set.Why is it not function of bounded variation. – Rayees Ahmad Mar 29 '16 at 04:58
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@RayeesAhmad It is not smooth on any neighbourhood of $0$. Its derivative is unbounded there. – Henricus V. Mar 29 '16 at 05:00
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@RayeesAhmad It depends on your definition of smooth of a graph. Its curve oscillates rapidly around $0$ and cannot be drawn to approximate the function itself. – Henricus V. Mar 29 '16 at 05:05
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Yes sir I have varified it from wolform but what is the significance of defining a function of bounded variation? – Rayees Ahmad Mar 29 '16 at 05:14
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@RayeesAhmad It allows the function to have "good" properties as listed above, and guarentees differentiability at many points. – Henricus V. Mar 29 '16 at 05:31