f( x+ y) f( x− y) − = 4 xy = ( x+ y) −( x− y) x+ y x− y 2 2 Mathematical Database f( x+ y) 2 f( x− y) 2 − ( x + y) = −( x− y ) for all x, y . x+ y x− y Thus f( x) x 2 − x is a constant. Let f( x) x 2 − x = k , then 3 f ( x) x kx = + . Remark. In this question, we have a symmetric condition. By us**in**g the symmetry, we reduce the equation to a **one**-**variable** functional equation. This is a useful technique for symmetric functional **equations**. Example 2.3. If f : → satisfies 2 f ( x + f( y)) = y+ xf( x) for all xy∈ , , f**in**d f ( x ). Solution. Putt**in**g x = 0 , then f ( f( y)) = y. Thus we have, (2.1) f y xf x f f x f y x f y 2 2 ( + ( )) = ( ( + ( ))) = + ( ) Now replace x by f ( x ), 2 f ( y+ ( f( x) f( f( x))) = ( f( x)) + f( y) . Remember that f ( f( y)) = y, so (2.2) f y xf x f x f y 2 ( + ( )) = ( ( )) + ( ) Compar**in**g (2.1) and (2.2), we have (2.3) ( ) 2 2 f ( x) = x Now replace y by f ( y ) **in** the orig**in**al equation, we have 2 f ( x + y) = f( y) + xf( x) . Squar**in**g both sides, we have ( ) 2 2 2 2 2 2 2 4 2 ( x + y) = f( x + y) = f( y) + x f( x) + 2 xf( x) f( y) = x + y + 2 xf( x) f( y) From this, we have (2.4) xy = f( x) f( y) From (2.3), we have f ( x ) = x or f ( x) =− x. If f ( x ) = x, from (2.4) we have f ( x ) = x for x∈R . Page 6 of 10

If f ( x ) =−x, from (2.4) we have f ( x) = − x for x∈R . Mathematical Database Therefore f ( x ) = x or f ( x) =− x. Remark. In this question, we replace (or substitute) **variable**s by other th**in**gs. (we replaced x by f ( x ) **in** the question). This is a very useful technique. Some common replacements and substitutions **in**clude: replac**in**g x by f ( x ); replac**in**g x by f ( f( x )); substitut**in**g x = 0; substitut**in**g x = y = 0; substitut**in**g x = 1, etc. Example 2.4. If f : → + satisfies **1.** f ( xf ( y)) = yf ( x) for all xy , ∈ 2. f( x) → 0 as x →∞, f**in**d f ( x ). + Solution. First, let us show that f ( x ) is surjective. Let x ⎛ ⎛ x ⎞ ⎞ x y = , then f ⎜xf ⎜ ⎟⎟ = f ( x) = x . Therefore f ( x ) is surjective, i.e. for all y f ( x) ⎝ ⎝ f( x) ⎠⎠ f( x) there exists an x such that f ( x) Assume = y. f( y ) = **1.** Putt**in**g x = 1, then f (1) = f(1 f( y)) = yf(1) . So y = 1 and f (1) = **1.** Now putt**in**g x = y , then f ( xf ( x)) = xf ( x) . Thus xf( x ) is a fixed po**in**t for the function. Let us show two th**in**gs about fixed po**in**t now. (i) If a, b are fixed po**in**ts of f, then f ( ab) = f ( af ( b)) = bf ( a) = ab . Thus ab is also a fixed (ii) po**in**t. ⎛ 1⎞ ⎛1 ⎞ ⎛1⎞ If a is a fixed po**in**t of f, then 1 = f(1) = f ⎜ai ⎟= f ⎜ f( a) ⎟= af ⎜ ⎟ ⎝ a⎠ ⎝a ⎠ ⎝a⎠ . Thus ⎛1⎞ 1 f ⎜ ⎟ = ⎝a⎠ a , which means 1 a is also a fixed po**in**t. Now, if xf ( x ) > 1 , then (( ( )) n n f xf x ) = ( xf ( x)) . As n , n n f (( xf ( x)) ) = ( xf ( x)) . This contradicts with condition 2. →∞ [ ] xf ( x) n →∞ while Page 7 of 10