<p>We describe a universal procedure that allows to obtain continuous solutions of inequalities of the form <Equation ID="Equ18"> <EquationSource Format="TEX">$\begin{aligned} a_1f(\alpha _1x+(1-\alpha _1)y)+\cdots +a_nf(\alpha _nx+(1-\alpha _n)y)\le 0,\;x,y\in I, \end{aligned}$</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <msub> <mi>a</mi> <mn>1</mn> </msub> <mi>f</mi> <mrow> <mo stretchy="false">(</mo> <msub> <mi>α</mi> <mn>1</mn> </msub> <mi>x</mi> <mo>+</mo> <mrow> <mo stretchy="false">(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>α</mi> <mn>1</mn> </msub> <mo stretchy="false">)</mo> </mrow> <mi>y</mi> <mo stretchy="false">)</mo> </mrow> <mo>+</mo> <mo>⋯</mo> <mo>+</mo> <msub> <mi>a</mi> <mi>n</mi> </msub> <mi>f</mi> <mrow> <mo stretchy="false">(</mo> <msub> <mi>α</mi> <mi>n</mi> </msub> <mi>x</mi> <mo>+</mo> <mrow> <mo stretchy="false">(</mo> <mn>1</mn> <mo>-</mo> <msub> <mi>α</mi> <mi>n</mi> </msub> <mo stretchy="false">)</mo> </mrow> <mi>y</mi> <mo stretchy="false">)</mo> </mrow> <mo>≤</mo> <mn>0</mn> <mo>,</mo> <mspace width="0.277778em" /> <mi>x</mi> <mo>,</mo> <mi>y</mi> <mo>∈</mo> <mi>I</mi> <mo>,</mo> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>where <InlineEquation ID="IEq1"> <EquationSource Format="TEX">$I\subset {\mathbb {R}}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>I</mi> <mo>⊂</mo> <mi mathvariant="double-struck">R</mi> </mrow> </math></EquationSource> </InlineEquation> is an interval, <InlineEquation ID="IEq2"> <EquationSource Format="TEX">$f:I\rightarrow {\mathbb {R}}$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>f</mi> <mo>:</mo> <mi>I</mi> <mo stretchy="false">→</mo> <mi mathvariant="double-struck">R</mi> </mrow> </math></EquationSource> </InlineEquation> is an unknown function and <InlineEquation ID="IEq3"> <EquationSource Format="TEX">$a_i\in {\mathbb {R}},$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>a</mi> <mi>i</mi> </msub> <mo>∈</mo> <mi mathvariant="double-struck">R</mi> <mo>,</mo> </mrow> </math></EquationSource> </InlineEquation> <InlineEquation ID="IEq4"> <EquationSource Format="TEX">$\alpha _i\in [0,1]$</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>α</mi> <mi>i</mi> </msub> <mo>∈</mo> <mrow> <mo stretchy="false">[</mo> <mn>0</mn> <mo>,</mo> <mn>1</mn> <mo stretchy="false">]</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> are given numbers. After presenting and explaining all necessary theoretical details, we provide a computer program that solves inequalities of such type.</p>

Solving linear functional inequalities with a computer

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2025年9月9日

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Chisom Prince Okeke,
Tomasz Szostok

摘要

We describe a universal procedure that allows to obtain continuous solutions of inequalities of the form $\begin{aligned} a_1f(\alpha _1x+(1-\alpha _1)y)+\cdots +a_nf(\alpha _nx+(1-\alpha _n)y)\le 0,\;x,y\in I, \end{aligned}$ $\begin{matrix} a_{1} f (α_{1} x + (1 - α_{1}) y) + \dots + a_{n} f (α_{n} x + (1 - α_{n}) y) \leq 0, x, y \in I, \end{matrix}$ where $I\subset {\mathbb {R}}$ $I \subset R$ is an interval, $f:I\rightarrow {\mathbb {R}}$ $f : I \to R$ is an unknown function and $a_i\in {\mathbb {R}},$ $a_{i} \in R,$ $\alpha _i\in [0,1]$ $α_{i} \in [0, 1]$ are given numbers. After presenting and explaining all necessary theoretical details, we provide a computer program that solves inequalities of such type.