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Let $f : [0,1] \to \mathbb{R}$ be a continuous function that vanishes at $x = 1.$ Show that there exists a sequence of polynomials, each vanishing at $x = 1$, which converges to $f$ uniformly on $[0,1].$

It feels like a Stone-Weierstrass question, but after looking over the Stone-Weierstrass proof several times I am not sure if it truly applies.

My second thought is Arzela-Ascoli - namely that the family of functions you consider are the collection of polynomials, call this collection $P$, that vanish at $x = 1.$ My problem then becomes that this family is not uniformly bound over the interval, and I do not know if we can assert that a sequence of these polynomials converges to a specific $f,$ I think Arzela-Ascoli only proves that a uniformly convergent sequence in $P$ exists.

Anyone have any insight? Thanks in advance.

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Another trick may be the following one: take a sequence of polynomials $q_1(x),q_2(x),\ldots$ that uniformly approximate $g(x)=\frac{f(x)}{1-x}$ over $(0,1)$ then consider $p_n(x)=(1-x)q_n(x)$. It is very easy to show that the sequence $p_1(x),p_2(x),\ldots$ meets the given constraints.

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  • $\begingroup$ Hello Jack. Is it true that there is a sequence of polynomials that uniformly approximates $g$? What if $f$ is slower than $x \mapsto 1-x$, so that $g(x) \to \infty$ as $x\to 1$? An example would be: $f(x) = \sqrt{1-x}$. $\endgroup$ – user00000 Aug 24 '16 at 11:54
  • $\begingroup$ @user00000: well, that is an issue, but we may circumvent it by first approximating (for instance, by convolution) the original function with a differentiable one that vanishes at $x=1$, hence it is not really restrictive to assume that the $g$ function above is continuous over $(0,1)$. Thanks for pointing that out. Out of curiosity, are you a somewhat special user? That string of zeroes in your user name striked me :D $\endgroup$ – Jack D'Aurizio Aug 24 '16 at 11:56
  • $\begingroup$ However, how can we approximate the original function with a differentiable one which vanishes at $x =1$ without resorting to polynomials? $\endgroup$ – user00000 Aug 24 '16 at 12:10
  • $\begingroup$ @user00000: by considering the convolution (en.wikipedia.org/wiki/Convolution) with a smooth kernel. The trick is that the convolution between a $C^0$ and a $C^1$ function is a $C^1$ function, and if the kernel is concentrated enough, that convolution also gives a uniform approximation. $\endgroup$ – Jack D'Aurizio Aug 24 '16 at 12:12
  • $\begingroup$ Ah, excuse me, I just saw your updated comment now. Convolution would have never hit my mind. And no, I am not by any means a special user! :) $\endgroup$ – user00000 Aug 24 '16 at 12:16
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Take the Stone Weierstrass polynomials $\{p_n\}$, and modify them. Let $p_n(1) = a_n$ and $q_n(x) = p_n(x) - a_n$. Since $a_n = p_n(1) \rightarrow f(1) = 0$, $\{q_n\}$ are convergent to $f$

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  • $\begingroup$ Are $p_n$'s the Bernstein polynomials? $\endgroup$ – Merkh Aug 21 '16 at 18:23
  • $\begingroup$ @Merkh anything converging uniformly to $f$ on $[0,1]$ will do.. Bernstein or not $\endgroup$ – user290300 Aug 21 '16 at 18:24
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As already noted, if $p_n$ is any sequence of polynomials converging uniformly to $f$ then $q_n=p_n-p_n(1)$ works. Another way to look at this is the version of S-W for $C_0(X)$, where $X$ is locally compact Hausdorff; you have $f\in C_0([0,1))$, and that version of S-W says that the polynomials vanishing at $1$ are dense in $C_0([0,1))$.

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