@@ -7,7 +7,7 @@
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\begin{enumerate}
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\begin{enumerate}
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\item For each $f \in C([0, 1]; \complex)$ and $\eps > 0$, there exists $C > 0$ such that
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\item For each $f \in C([0, 1]; \complex)$ and $\eps > 0$, there exists $C > 0$ such that
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\[
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\[
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|f(t) - f(x)| \le \eps + cg_x(t)
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|f(t) - f(x)| \le \eps + Cg_x(t)
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\]
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\]
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for all $x, t \in [0, 1]$.
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for all $x, t \in [0, 1]$.
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@@ -37,7 +37,7 @@
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so
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so
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\[
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\[
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|f(x) - f(x)| \le \eps \left\lceil \abs{\frac{x - t}{\delta}} \right\rceil \le \eps + C(s - t)^2 = \eps + Cg_x(t)
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|f(t) - f(x)| \le \eps \left\lceil \abs{\frac{x - t}{\delta}} \right\rceil \le \eps + C(x - t)^2 = \eps + Cg_x(t)
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\]
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\]
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(3): Let $f \in C([0, 1]; \complex)$ and $\eps > 0$. By (2), there exists $C > 0$ such that
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(3): Let $f \in C([0, 1]; \complex)$ and $\eps > 0$. By (2), there exists $C > 0$ such that
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@@ -48,7 +48,7 @@
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for all $n \in \natp$, where
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for all $n \in \natp$, where
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\begin{align*}
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\begin{align*}
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\sup_{x \in [0, 1]}|\phi_n(g_x)(x)| &\le \sup_{x \in [0, 1]}|\phi_n(t^2 - 2tx + x^2)(x)| \\
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\sup_{x \in [0, 1]}|\phi_n(g_x)(x)| &\le \sup_{x \in [0, 1]}|\phi_n(t^2 - 2tx + x^2)(x)| \\
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&\le \sup_{x \in [0, 1]}|\phi_n(t^2)(x) - \phi_n(tx)(x) + x^2| \\
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&\le \sup_{x \in [0, 1]}|\phi_n(t^2)(x) - 2\phi_n(tx)(x) + x^2| \\
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&\le \normn{\phi_n(t^2) - t^2}_u + 2\sup_{x \in [0, 1]}\normn{\phi_n(xt) - xt}_u
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&\le \normn{\phi_n(t^2) - t^2}_u + 2\sup_{x \in [0, 1]}\normn{\phi_n(xt) - xt}_u
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\end{align*}
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\end{align*}
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@@ -98,7 +98,7 @@
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\begin{align*}
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\begin{align*}
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\phi_n(t^2) &= \sum_{k = 1}^n \frac{n!}{k!(n-k)!}\frac{k^2}{n^2}t^k(1 - t)^{n-k} \\
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\phi_n(t^2) &= \sum_{k = 1}^n \frac{n!}{k!(n-k)!}\frac{k^2}{n^2}t^k(1 - t)^{n-k} \\
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&= \sum_{k = 1}^n \frac{(n-1)!}{(k-1)!(n-k)!}\frac{k}{n}t^k(1 - t)^{n-k} \\
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&= \sum_{k = 1}^n \frac{(n-1)!}{(k-1)!(n-k)!}\frac{k}{n}t^k(1 - t)^{n-k} \\
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&= t \sum_{k = 0}^{n-1} \frac{(n-1)!}{(k-1)!(n-k)!}\frac{k+1}{n}t^k(1 - t)^{n-k} \\
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&= t \sum_{k = 0}^{n-1} \frac{(n-1)!}{(k-1)!(n-k-1)!}\frac{k+1}{n}t^k(1 - t)^{n-k} \\
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&= t \phi_{n-1}\paren{\frac{(n-1)t + 1}{n}} = t^2 + \frac{t - t^2}{n}
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&= t \phi_{n-1}\paren{\frac{(n-1)t + 1}{n}} = t^2 + \frac{t - t^2}{n}
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\end{align*}
|
\end{align*}
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Reference in New Issue
Block a user