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Properties of Continuous Functions with Additivity 📂Functions

Properties of Continuous Functions with Additivity

Theorem

  • [1] If a continuous function f:R→Rf : \mathbb{R} \to \mathbb{R} satisfies f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y) for all x,y∈Rx, y \in \mathbb{R} f(x)=f(1)x f(x) = f(1) x
  • [2] If a continuous function g:R→(0,∞)g : \mathbb{R} \to ( 0 , \infty ) satisfies g(x+y)=g(x)g(y)g(x + y) = g(x) g(y) for all x,y∈Rx, y \in \mathbb{R} g(x)=(g(1))x g(x) = \left( g(1) \right)^x

Explanation

The property where addition is preserved across functions as in f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y) is called additivity, and the property where multiplication is preserved is called multiplicativity. gg feels like a mix of additivity and multiplicativity, which is called homomorphism.

Proof

Strategy: First, it’s shown that rational numbers can move in and out of functions, then a sequence of rational numbers converging to xx is created. Since continuity is guaranteed, lim⁡\lim is also utilized as a point that moves in and out of the function.

[1]

Part 1. f(−x)=−f(x)f(-x) = - f(x)

By additivity f(0)=f(0+0)=f(0)+f(0)  ⟹  f(0)=0 f( 0 ) = f( 0 + 0 ) = f(0) + f(0) \implies f(0) = 0 Similarly, by additivity 0=f(0)=f(x+(−x))=f(x)+f(−x)  ⟹  f(−x)=−f(x) 0 = f( 0 ) = f( x + (-x) ) = f(x) + f( -x ) \implies f(-x) = - f(x)

Part 2. f(qx)=qf(x)f(qx) = q f(x)

For n∈Nn \in \mathbb{N} f(nx)=f(x+⋯+x⏟n)=f(x)+⋯+f(x)⏟n=nf(x)  ⟹  f(nx)=nf(x) f(nx) = f \left( \underbrace{x + \cdots + x}_{n} \right) = \underbrace{f(x) + \cdots + f(x)}_{n} = n f(x) \implies f(nx) = nf(x) If we set y:=xn\displaystyle y := {{ x } \over {n}} f(x)=f(ny)=nf(y)=nf(xn)  ⟹  f(xn)=1nf(x) f(x) = f(ny) = n f(y) = n f \left( {{x} \over {n}} \right) \implies f \left( {{x} \over {n}} \right) = {{1} \over {n}} f(x) And according to Part 1, these properties also hold for negative numbers, so for all q∈Qq \in \mathbb{Q} f(qx)=qf(x) f(qx) = q f(x)

Part 3. f(x)=mxf(x) = mx

Defining a sequence of rational numbers converging to x∈Rx \in \mathbb{R}, {qn}n∈N\left\{ q_{n} \right\}_{n \in \mathbb{N}}, based on Part 2 and the continuity in ff

f(x)=f(x⋅1)=f(lim⁡n→∞qn⋅1)=lim⁡n→∞f(qn⋅1)=lim⁡n→∞qnf(1)=f(1)x \begin{align*} f(x) =& f( x \cdot 1 ) \\ =& f( \lim_{n \to \infty} q_{n} \cdot 1 ) \\ =& \lim_{n \to \infty} f( q_{n} \cdot 1 ) \\ =& \lim_{n \to \infty} q_{n} f( 1 ) = f(1) x \end{align*}

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[2]

Part 1. g(−x)=(g(x))−1\displaystyle g(-x) = \left( g (x) \right)^{-1}

Since g(x)≠0g(x) \ne 0 g(0)=g(0+0)=g(0)g(0)  ⟹  g(0)=1 g( 0 ) = g( 0 + 0 ) = g(0) g(0) \implies g(0) = 1 Similarly, 1=g(0)=g(x+(−x))=g(x)g(−x)  ⟹  g(−x)=1g(x) 1 = g( 0 ) = g( x + (-x) ) = g(x) g( -x ) \implies g(-x) = {{1} \over {g(x)}}

Part 2. g(qx)=(g(x))qg(qx) = \left( g(x) \right)^{q}

For n∈Nn \in \mathbb{N} g(nx)=g(x+⋯+x⏟n)=g(x)×⋯×g(x)⏟n=(g(x))n  ⟹  g(nx)=(g(x))n g(nx) = g \left( \underbrace{x + \cdots + x}_{n} \right) = \underbrace{ g(x) \times \cdots \times g(x)}_{n} = \left( g(x) \right)^{n} \implies g(nx) = \left( g(x) \right)^{n} If we set y:=xn\displaystyle y := {{ x } \over {n}} g(x)=g(ny)=(g(y))n=(g(xn))n  ⟹  g(xn)=(g(x))1n g(x) = g(ny) = \left( g(y) \right)^{n} = \left( g \left( {{x} \over {n}} \right) \right)^{n} \implies g \left( {{x} \over {n}} \right) = \left( g(x) \right)^{{1} \over {n}} And according to Part 1, these properties also hold for negative numbers, so for all q∈Qq \in \mathbb{Q} g(qx)=(g(x))q g(qx) = \left( g(x) \right)^{q}

Part 3.

With g(x)=axg(x) = a^x and defining a sequence of rational numbers converging to x∈Rx \in \mathbb{R}, {qn}n∈N\left\{ q_{n} \right\}_{n \in \mathbb{N}}, based on Part 2 and the continuity in gg g(x)=g(x⋅1)=g(lim⁡n→∞qn⋅1)=lim⁡n→∞g(qn⋅1)=lim⁡n→∞g(1)qn=g(1)lim⁡n→∞qn=(g(1))x \begin{align*} g(x) =& g( x \cdot 1 ) \\ =& g( \lim_{n \to \infty} q_{n} \cdot 1 ) \\ =& \lim_{n \to \infty} g( q_{n} \cdot 1 ) \\ =& \lim_{n \to \infty} g( 1 )^{q_{n}} \\ =& g( 1 )^{ \displaystyle \lim_{n \to \infty} q_{n}} \\ =& \left( g(1) \right)^{x} \end{align*}

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