I wish to determine all functions f of bounded variation on [0,1] such that f ( x ) + ( T V ( f [ 0 , x ] ) ) 1 / 2 = 1 for all x ∈ [ 0 , 1 ] and ∫ 0 1 f ( x ) d x = 1 / 3.Here T V ( f [ a , b ] ) denotes the total variation of f on [a,b].So looking at what needs to be done, it is clear to me that the function f(x) = 1/3 satisfies the integral equation however, I have no intuition on how to proceed with the equation that involves the total variation or how to bring it all together.

Question

I wish to determine all functions   f of bounded variation on [0,1] such that  f  (  x  )  +  (  T  V  (      f          [      0      ,      x      ]        )      )          1              /            2        =  1    for all   x  ∈  [  0  ,  1  ]    and        ∫    0    1    f  (  x  )    d  x  =  1      /    3.Here   T  V  (      f          [      a      ,      b      ]        ) denotes the total variation of   f on [a,b].So looking at what needs to be done, it is clear to me that the function f(x) = 1/3 satisfies the integral equation however, I have no intuition on how to proceed with the equation that involves the total variation or how to bring it all together.

odmeravan5c · Accepted Answer

In these cases it is a good idea to start plugging particular values and see what happens: here x=0 tells us that f(0)=1, because       f          [      0      ,      0      ]       has 0 variation.Another useful fact is that   T  V  (      f          [      0      ,      a      ]        )  ≤  T  V  (      f          [      0      ,      b      ]        ) for   a  ≤  b. Then, from your first equation you have   f  (  a  )  =  1  −  T  V  (      f          [      0      ,      a      ]        )  ≥  1  −  T  V  (      f          [      0      ,      b      ]        )  ≥  f  (  b  ), so f is non-increasing.For non-increasing functions, the total variation simplifies to   T  V  (      f          [      0      ,      x      ]        )  =  f  (  0  )  −  f  (  x  )  =  1  −  f  (  x  ) and your first equation becomes  (  1  −  f  (  x  )      )          1              /            2        =  1  −  f  (  x  )  ,which tells us that   f  (  x  )  ∈  {  0  ,  1  } for all   x  ∈  [  0  ,  1  ].EDIT: Since f is non-increasing and is either 0 or 1, the only way to satisfy the integral equation is having   f  (  x  )  =  1  ,  ∀  x  &amp;lt;  1      /    3 and   f  (  x  )  =  0  ,  ∀  x  &amp;gt;  1      /    3. The value at x=1/3 does not change the integral, so both f(1/3)=0 and f(1/3)=1 are possible. So your problem has 2 solutions.

I wish to determine all functions f of bounded variation on [0,1] such that f ( x

Answered question

Answer & Explanation

New Questions in Pre-Algebra