Proving ( f(x y) f(x) f(y) ) Implies ( f(x) cx )
Proving ( f(x y) f(x) f(y) ) Implies ( f(x) cx )
In this article, we delve into the mathematical proof demonstrating that the function ( f ) adheres to the equation ( f(x y) f(x) f(y) ) for all ( x ) and ( y ) if and only if ( f(x) cx ) for some constant ( c ). We will explore the steps and underlying properties that lead to this conclusion.
Identifying the Functional Equation
Our starting point is the functional equation given by:
[ f(x y) f(x) f(y) ]
This equation is known as Cauchy's functional equation. To begin the proof, we need to understand the implications of this equation for different values of ( x ) and ( y ).
Checking for Basic Properties
We first set ( y 0 ). This gives us:
[ f(x cdot 0) f(x) f(0) implies f(0) f(x) f(0) ]
If ( f(0) eq 0 ), then we can divide both sides by ( f(0) ), resulting in ( 1 f(x) ), which contradicts the general form of ( f ). Hence, we conclude that:
[ f(0) 0 ]
Considering Rational Inputs
Beyond the basic property, we consider rational values for ( x ). First, let us set ( x y 1 ).
[ f(1 cdot 1) f(1) f(1) implies f(1) f(1)^2 ]
This implies either ( f(1) 1 ) or ( f(1) 0 ).
Next, let us set ( x 1 ) and ( y frac{1}{n} ) where ( n ) is a positive integer:
[ fleft(1 cdot frac{1}{n}right) f(1) fleft(frac{1}{n}right) implies fleft(frac{1}{n}right) f(1) fleft(frac{1}{n}right) implies 1 f(1) quad text{(since ( fleft(frac{1}{n}right) eq 0 ))} ]
Thus, ( f(1) 1 ). Our functional equation now simplifies to:
[ fleft(x frac{1}{n}right) f(x) fleft(frac{1}{n}right) implies fleft(x frac{1}{n}right) f(x) frac{1}{f(n)} ]
By induction, we can show that ( fleft(frac{1}{n}right) frac{1}{f(n)} ). This implies:
[ fleft(frac{m}{n}right) m fleft(frac{1}{n}right) frac{m}{f(n)} frac{m}{n} ]
Show that ( fleft(frac{m}{n} xright) frac{m}{n} f(x) )
Now, let ( r frac{m}{n} ) where ( m ) and ( n ) are integers. We can show that:
[ fleft(frac{m}{n} xright) frac{m}{n} f(x) ]
This can be done using induction or by manipulating the functional equation.
Extend to Real Inputs
To extend the result to all real numbers, we use the density of rational numbers in the reals. If ( f ) is continuous or bounded on some interval, then:
[ text{If } fleft(r xright) r f(x) text{ for all rational } r text{, then it holds for all real } r. ]
Conclusion
Since ( f(1) ) can be denoted as ( c ) where ( c f(1) ), we can express ( f(x) ) as:
[ f(x) c x text{ for some constant } c. ]
Thus, we have shown that if ( f(x y) f(x) f(y) ) for all ( x ) and ( y ), then ( f(x) c x ) for some constant ( c ).
Note: The above result is not always true without additional conditions such as continuity or measurability. For example, if ( f ) is a measurable function, the result holds. Similarly, if ( f ) is bounded, the result also holds.
The additivity of ( f ) implies that ( f(N x) N f(x) ) for every real ( x ) and each natural number ( N ). This is because:
[ f(x) f(1 x) 1 f(x) ]
For ( x 1 ), it gives:
[ f(N) N f(1) c N text{ where } c f(1) ]
For ( x frac{1}{N} ), we have:
[ f(1) f(N frac{1}{N}) N fleft(frac{1}{N}right) implies fleft(frac{1}{N}right) frac{c}{N} ]
Thus, for any rational number ( x frac{m}{n} ) where ( m ) and ( n ) are positive integers, we have:
[ fleft(frac{m}{n}right) m fleft(frac{1}{n}right) m frac{c}{n} frac{m}{n} c ]
Assuming ( f ) is continuous on ( mathbb{R} ), let ( x lim q_n ) where ( q_n ) is a rational sequence converging to ( x ):
[ f(x) fleft(lim q_nright) lim f(q_n) lim c q_n c left(lim q_nright) c x ]
Therefore, ( f(x) c x ) for all real ( x ).
Note: A function may be continuous on the set of rational numbers ( mathbb{Q} ) but discontinuous on the set of real numbers ( mathbb{R} ), such as ( f(x) 1 ) if ( x ) is rational, and ( 0 ) if ( x ) is irrational.