Is there an operation that intensifies fractions

Is there an operation that can be used on fractions (greater than zero and less than 1) that, given a value larger than 0.5 returns a larger fraction (but still less than 1), and given a value smaller than 0.5 returns an even smaller fraction (but greater than 0)? I.e., given 0.7 you will get 0.9, and given 0.2 you will get 0.1 or something like that.
The catch is that it has to be just an operation, not an algorithm (ie you cant check if the value you were given is larger than 0.5, etc).

Is this possible?

f(x) = −1/3 x³ + 1/2 x² + 5/6 x

Draw line y = x from point (0,0) to point (1,1), which passes through point (1/2, 1/2)
Now we need a curve that passes through the 3 points above, but that is below line for x between 0 and 1/2, and is above line when x is between 1/2 and 1

What we want to find is cubic function that passes through these 3 points and that has point of inflection at (1/2, 1/2).

So we need f''(1/2) = 0
f''(x) = (2x - 1) or (−2x + 1)

f''(x) > 0 for 0 < x < 1/2
f''(x) < 0 for 1/2 < x < 1

f''(x) = −2x + 1

f'(x) = −x² + x + a

f(x) = −1/3 x³ + 1/2 x² + ax + b

Since we want f(0) = 0 ----> b = 0
We also want f(1) = 1
−1/3 + 1/2 + a = 1
1/6 + a = 1
a = 5/6

f(x) = −1/3 x³ + 1/2 x² + 5/6 x

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Now we check that f(1/2) = 1/2
f(1/2) = −1/3 (1/8) + 1/2 (1/4) + 5/6 (1/2)
. . . . . = −1/24 + 3/24 + 10/24
. . . . . = 12/24
. . . . . = 1/2

So f(x) passes through point (0,0), (1/2, 1/2), and (1,1)
It has point of inflection at (1/2, 1/2) since f''(1/2) = 0
It is below line y = x for 0 < x < 1/2, since f''(x) > 0 (f concave up) for these points
It is above line y = x for 1/2 < x < 1, since f''(x) < 0 (f concave down) for these points

We can check different values on the interval.