How do I find all continuous functions $f:[a,b]\to \mathbb{R}$ such that $$ \int_{a}^x f(t) dt = \int_{x}^b f(t) dt$$ , for all $x \in (a,b)$

This problem is from old qualifyings of real analysis , it does not look too bad but still I struggled quit a good time. I tried to define some functions to work out but non of them did the job.

Any little hint would be appreciated.

  • 4
    $\begingroup$ Taking a derivative might help. $\endgroup$ – muaddib Jun 1 '15 at 23:41
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    $\begingroup$ do not we need $f$ to be differentiable? $\endgroup$ – mp100 Jun 1 '15 at 23:42
  • $\begingroup$ Derivative of the integral <--> Fundamental Theorem of Calculus $\endgroup$ – muaddib Jun 1 '15 at 23:43
  • $\begingroup$ @mp100 Not to differentiate $F\colon x\mapsto =\int_a^x f$, which is differentiable (even $C^1$). $\endgroup$ – Clement C. Jun 1 '15 at 23:43
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    $\begingroup$ The only function satisfying that requirement I got is a zero function. I do not think that we can have such non zero function for arbitrary $x$. $\endgroup$ – mp100 Jun 2 '15 at 0:04

Hint: $\int_a^x = \int_a^b - \int_x^b = \ldots$.


If a = b, then any function works. If a < b, then f is a constant function equal to 0. Really, for any c < d from [a, b] consider three integrals:

I1 from a to c, I2 from c to d, I3 from d to b.

Obviously I1 + I2 = I3 and I1 = I2 + I3 by the property of f. This means I2 is 0. Because it is true for any c, d, integral from a to x is always 0 for any x from [a, b]. Taking its derivative by x becomes 0, and it is also f.


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