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# 1. Functional equations in one variable

1. Functional equations in one variable

## If xf ( x ) < 1, then by

If xf ( x ) < 1, then by (ii) we have condition 2. Thus, we must have Remark. 1 xf( x) xf ( x ) = 1. This means Mathematical Database is a fixed point larger than 1. This again contradicts with f( x) 1 = . x The concept of fixed point is introduced in this question. If f ( x) of f. It is not so commonly used but it is also a useful technique. = x, then x is called a fixed point 3. Some famous functional equations In this part, we will introduce some famous functional equations. We may quote them directly in competitions. Theorem 3.1. (Cauchy equation) If f is a continuous function such that f ( x+ y) = f( x) + f( y) for all xy∈ , , then f ( x) = cx where c is a constant. Proof. We may use the technique mentioned in example 2.1. First, put f ( x) y = 1 and let x be a positive integer. Then f ( x+ 1) = f ( x) + c, where c = f(1) . Thus = cx for positive integers x. It is easy to verify that (0) 0 original equation. For negative integers, we replace x by –x and get f = by putting x = y = 0 into the f ( − x+ 1) = f( − x) + c. Thus f ( − x) =−cx for positive integers x. Therefore we conclude that f ( x) = cx for integers x. Let m x = , where m, n are integers. Then we have n ⎛m+ 1⎞ ⎛m⎞ ⎛1⎞ f ⎜ ⎟= f ⎜ ⎟+ f ⎜ ⎟ ⎝ n ⎠ ⎝ n ⎠ ⎝n⎠ , thus ⎛m ⎞ ⎛1 ⎞ f ⎜ ⎟= mf ⎜ ⎟ ⎝ n ⎠ ⎝n⎠ . However, 1 ⎛1⎞ c = f ( ⋅ n ) = nf ⎜ ⎟ n ⎝n⎠ ⎛1 ⎞ c f ⎜ ⎟ = ⎝n⎠ n and we have ⎛1 ⎞ c f ⎜ ⎟ = ⎝n⎠ n . This implies f ( x) = cx for rational numbers x. . So Page 8 of 10

Mathematical Database Now, as f is continuous, we can always bound an irrational number by two rational numbers. (For example, we may bound π by 3, 3.1, 3.14, 3.141 and 4, 3.2, 3.15, 3.142 respectively). Let { xn}, xn∈ be such a sequence with lim x n = x , then by continuity of f, we have n→∞ f ( x) = f(lim x ) = lim f( x ) = lim cx = cx. Therefore f ( x) n n n n→∞ n→∞ n→∞ = cx for all x∈R . Q.E.D. Corollary 3.2. If f is a continuous function and for all xy∈ , , (i) f ( x+ y) = f( x) f( y) then f ( x) = c (ii) f ( xy) = f ( x) + f ( y) then f ( x) = clnx (iii) f ( xy) = f ( x) f ( y) then f ( x) = x where c is a constant. x c Example 3.3. If f :(1, +∞) → is a continuous function such that f ( xy) = xf ( y) + yf ( x) for all1 < xy , ∈ , find f ( x ). Solution. It is equivalent to f ( xy) = f ( x) + f ( y) . If we let xy x y gxy ( ) = gx ( ) + gy ( ) f ( x) gx ( ) = , then the equation becomes x which is simply (ii) of corollary 1.32 (although there is a minor difference between them). So we have g( x) = clnxand f ( x) = xg( x) = cxlnx. Page 9 of 10

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