<tex> x \ne 0 : p(x) = p(|x| \cdot \frac{x}{|x|}) = |x| \cdot p(\frac{x}{|x|}) \begin{matrix} \le c_2|x| \\ \ge c_1|x| \end{matrix} </tex>
<tex> |p(y) - p(x)| \le p(x - y) = </tex> разложим по базису <tex> p( \sum_{k = 1}^m |y_k - x_k|l_k ) \le </tex>
<tex> \le \sum |y_k - x_k|p(l_k) \le </tex> КБШ <tex> \sqrt{ \sum|x_k - y_k|^2 } \sqrt{\sum p(l_k)^2} = |x - y| \sqrt{\sum p(l_k)^2} </tex>
<tex> \forall \epsilon > 0 \ \exists \delta = \frac{\epsilon}{\sqrt{\sum p(l_k)^2}} </tex>
<tex> \forall y : |x - y| < \delta </tex> верно <tex> |p(x) - p(y)| < \epsilon </tex>
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