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If there is a sequence of odd polynomials <math> {p_n(x)} </math> with <math>p_n \rightarrow f </math>, then <math> f(0) = 0 </math>. | If there is a sequence of odd polynomials <math> {p_n(x)} </math> with <math>p_n \rightarrow f </math>, then <math> f(0) = 0 </math>. | ||
Since <math>p_n(x)</math> are odd polynomials, then <math> p_n(0) = 0. </math>. Then <math>f(0) = \lim_{n \rightarrow \infty} p_n(0) = 0</math> | Since <math>p_n(x)</math> are odd polynomials, then <math> p_n(0) = 0. </math>. Then <math>f(0) = \lim_{n \rightarrow \infty} p_n(0) = 0</math> | ||
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| + | <math>(\Leftarrow)</math> | ||
| + | If <math>f \in C([0,1])</math>, and <math> f(0) = 0 </math>, then let <math> g(x) = f(x)</math> if <math>x \in [0,1]</math>, and <math>g(x) = - f(-x)</math> if <math>x \in [-1,0]</math>. In this case <math>g</math> is an odd function and <math>g \in C([-1,1])</math>. Then there exists a sequence of polynomials <math>g_n(x)</math> such that <math>g_n(x) \rightarrow g </math> uniformly on [-1,1]. Let <math> p_n(x) = \frac{1}{2}(g_n(x)-g_n(-x))</math>, then <math>p_n(x)</math> is an odd polynomial. and <math>\lim_{n \rightarrow \infty}p_n(x) = \lim_{n \rightarrow \infty} \frac{1}{2}(g_n(x)-g_n(-x)) = \frac{1}{2}(g(x)-g(-x)) = g(x)</math>. Then<math> p_n(x) \rightarrow f(x) </math> on <math>[0,1]</math>. | ||
Latest revision as of 09:02, 10 July 2008
$ (\Rightarrow) $ If there is a sequence of odd polynomials $ {p_n(x)} $ with $ p_n \rightarrow f $, then $ f(0) = 0 $. Since $ p_n(x) $ are odd polynomials, then $ p_n(0) = 0. $. Then $ f(0) = \lim_{n \rightarrow \infty} p_n(0) = 0 $
$ (\Leftarrow) $
If $ f \in C([0,1]) $, and $ f(0) = 0 $, then let $ g(x) = f(x) $ if $ x \in [0,1] $, and $ g(x) = - f(-x) $ if $ x \in [-1,0] $. In this case $ g $ is an odd function and $ g \in C([-1,1]) $. Then there exists a sequence of polynomials $ g_n(x) $ such that $ g_n(x) \rightarrow g $ uniformly on [-1,1]. Let $ p_n(x) = \frac{1}{2}(g_n(x)-g_n(-x)) $, then $ p_n(x) $ is an odd polynomial. and $ \lim_{n \rightarrow \infty}p_n(x) = \lim_{n \rightarrow \infty} \frac{1}{2}(g_n(x)-g_n(-x)) = \frac{1}{2}(g(x)-g(-x)) = g(x) $. Then$ p_n(x) \rightarrow f(x) $ on $ [0,1] $.
