<p>Let <InlineEquation ID="IEq1"> <EquationSource Format="TEX">\(f: \mathbb {R} \rightarrow \mathbb {R}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>f</mi> <mo>:</mo> <mi mathvariant="double-struck">R</mi> <mo stretchy="false">→</mo> <mi mathvariant="double-struck">R</mi> </mrow> </math></EquationSource> </InlineEquation>. The Heaviside function <i>H</i> is defined by <Equation ID="Equ7"> <EquationSource Format="TEX">\( H(x) = {\left\{ \begin{array}{ll} 1 \quad (x &gt; 0)\\ 0 \quad (x \le 0). \end{array}\right. } \)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mi>H</mi> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <mfenced open="{"> <mrow> <mtable> <mtr> <mtd columnalign="left"> <mrow> <mn>1</mn> <mspace width="1em" /> <mo stretchy="false">(</mo> <mi>x</mi> <mo>&gt;</mo> <mn>0</mn> <mo stretchy="false">)</mo> </mrow> </mtd> </mtr> <mtr> <mtd columnalign="left"> <mrow> <mrow /> <mn>0</mn> <mspace width="1em" /> <mo stretchy="false">(</mo> <mi>x</mi> <mo>≤</mo> <mn>0</mn> <mo stretchy="false">)</mo> <mo>.</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> </mrow> </math></EquationSource> </Equation>In this paper, we study the composite functional equation <Equation ID="Equ8"> <EquationSource Format="TEX">\(\begin{aligned} f(x + f(y)) = (x + f(y)) G_f (x), \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <mi>f</mi> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo>+</mo> <mi>f</mi> <mrow> <mo stretchy="false">(</mo> <mi>y</mi> <mo stretchy="false">)</mo> </mrow> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo>+</mo> <mi>f</mi> <mrow> <mo stretchy="false">(</mo> <mi>y</mi> <mo stretchy="false">)</mo> </mrow> <mo stretchy="false">)</mo> </mrow> <msub> <mi>G</mi> <mi>f</mi> </msub> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>where <Equation ID="Equ9"> <EquationSource Format="TEX">\(\begin{aligned} G_f (x) = f(H(x) + f(x) - x) + f(H(1 - x) + f(1 - x) - (1 - x)). \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <msub> <mi>G</mi> <mi>f</mi> </msub> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <mi>f</mi> <mrow> <mo stretchy="false">(</mo> <mi>H</mi> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>+</mo> <mi>f</mi> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>-</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>+</mo> <mi>f</mi> <mrow> <mo stretchy="false">(</mo> <mi>H</mi> <mrow> <mo stretchy="false">(</mo> <mn>1</mn> <mo>-</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>+</mo> <mi>f</mi> <mrow> <mo stretchy="false">(</mo> <mn>1</mn> <mo>-</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>-</mo> <mrow> <mo stretchy="false">(</mo> <mn>1</mn> <mo>-</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo stretchy="false">)</mo> </mrow> <mo>.</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>Define the function <InlineEquation ID="IEq2"> <EquationSource Format="TEX">\(I: \mathbb {R} \rightarrow \mathbb {R}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>I</mi> <mo>:</mo> <mi mathvariant="double-struck">R</mi> <mo stretchy="false">→</mo> <mi mathvariant="double-struck">R</mi> </mrow> </math></EquationSource> </InlineEquation> by setting <Equation ID="Equ10"> <EquationSource Format="TEX">\( I(x) = {\left\{ \begin{array}{ll} x \quad (x \in \mathbb {Z})\\ 0 \quad (x \in \mathbb {R} \setminus \mathbb {Z}). \end{array}\right. } \)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mi>I</mi> <mrow> <mo stretchy="false">(</mo> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>=</mo> <mfenced open="{"> <mrow> <mtable> <mtr> <mtd columnalign="left"> <mrow> <mi>x</mi> <mspace width="1em" /> <mo stretchy="false">(</mo> <mi>x</mi> <mo>∈</mo> <mi mathvariant="double-struck">Z</mi> <mo stretchy="false">)</mo> </mrow> </mtd> </mtr> <mtr> <mtd columnalign="left"> <mrow> <mrow /> <mn>0</mn> <mspace width="1em" /> <mo stretchy="false">(</mo> <mi>x</mi> <mo>∈</mo> <mi mathvariant="double-struck">R</mi> <mo lspace="0.15em" rspace="0.15em" stretchy="false">\</mo> <mi mathvariant="double-struck">Z</mi> <mo stretchy="false">)</mo> <mo>.</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </mfenced> </mrow> </math></EquationSource> </Equation>We prove that <InlineEquation ID="IEq3"> <EquationSource Format="TEX">\(f = I\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>f</mi> <mo>=</mo> <mi>I</mi> </mrow> </math></EquationSource> </InlineEquation> holds in the case <InlineEquation ID="IEq4"> <EquationSource Format="TEX">\(f(1) \ne 0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>f</mi> <mo stretchy="false">(</mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>≠</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>. We can regard this fact as a characterization of <InlineEquation ID="IEq5"> <EquationSource Format="TEX">\(\mathbb {Z}.\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi mathvariant="double-struck">Z</mi> <mo>.</mo> </mrow> </math></EquationSource> </InlineEquation></p>

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A characterization of the set of all integers by a composite functional equation

  • Soichi Ikeda

摘要

Let \(f: \mathbb {R} \rightarrow \mathbb {R}\) f : R R . The Heaviside function H is defined by \( H(x) = {\left\{ \begin{array}{ll} 1 \quad (x > 0)\\ 0 \quad (x \le 0). \end{array}\right. } \) H ( x ) = 1 ( x > 0 ) 0 ( x 0 ) . In this paper, we study the composite functional equation \(\begin{aligned} f(x + f(y)) = (x + f(y)) G_f (x), \end{aligned}\) f ( x + f ( y ) ) = ( x + f ( y ) ) G f ( x ) , where \(\begin{aligned} G_f (x) = f(H(x) + f(x) - x) + f(H(1 - x) + f(1 - x) - (1 - x)). \end{aligned}\) G f ( x ) = f ( H ( x ) + f ( x ) - x ) + f ( H ( 1 - x ) + f ( 1 - x ) - ( 1 - x ) ) . Define the function \(I: \mathbb {R} \rightarrow \mathbb {R}\) I : R R by setting \( I(x) = {\left\{ \begin{array}{ll} x \quad (x \in \mathbb {Z})\\ 0 \quad (x \in \mathbb {R} \setminus \mathbb {Z}). \end{array}\right. } \) I ( x ) = x ( x Z ) 0 ( x R \ Z ) . We prove that \(f = I\) f = I holds in the case \(f(1) \ne 0\) f ( 1 ) 0 . We can regard this fact as a characterization of \(\mathbb {Z}.\) Z .