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{{Math proof|title=Proof|proof=
The first part results by using the [[chain rule]] for differentiating both sides of the equation <math>f(s\mathbf x ) = s^k f(\mathbf x)</math> with respect to <math>s,</math> and taking the limit of the result when {{mvar|s}} tends to {{math|1}}.
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The solutions of this [[linear differential equation]] have the form <math>g(s)=g(1)s^k.</math>
Therefore, <math display="block"> f(s \mathbf{x}) = g(s) = s^k g(1) = s^k f(\mathbf{x}),</math> if {{mvar|s}} is sufficiently close to {{math|1}}. If this solution of the partial differential equation would not be defined for all positive {{mvar|s}}, then the [[functional equation]] would allow to prolongate the solution, and the partial differential equation implies that this prolongation is unique. So, the ___domain of a maximal solution of the partial differential equation is a linear cone, and the solution is positively homogeneous of degree {{mvar|k}}. <math>\square</math>
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As a consequence, if <math>f : \R^n \to \R</math> is continuously differentiable and homogeneous of degree <math>k,</math> its first-order [[partial derivative]]s <math>\partial f/\partial x_i</math> are homogeneous of degree <math>k - 1.</math>
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