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In preparing for a qualifying exam in analysis, I came across the following question that I have been unable to solve.

Prove or disprove:

If $f$ is infinitely differentiable on $[0,1],$ then there exists a sequence of polynomials, $\{g_n\},$ such that for every $k, k=0,1,2,\ldots, g_n^{(k)} \rightarrow f^{(k)}$ as $n\rightarrow \infty,$ uniformly on $[0,1].$

I'm quite sure that this is false. I believe the counterexample will be an infinitely differentiable function $f$ such that the Taylor series of $f$ does not converge to $f.$ One thought I had was trying to show that, if the Taylor series of $f$ converges to $f,$ then any sequence of approximating polynomials must be the partial sums of the Taylor series. However, this doesn't seem to be true. An idea a classmate of mine had was to try to show that if such a sequence exists, then the Taylor series converges. I'm not sure how one would go about proving that. Part of the difficulty seems to be that the polynomials in this sequence don't necessarily have anything to do with one another. That is, it doesn't seem that I can assume they are the partial sums of some series. Any hints/thoughts/counterexamples would be appreciated.

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All sequences are partial sums of some series, namely the series of their differences. The distinction between sequences and series is artificial. –  joriki Nov 17 '11 at 23:42

1 Answer 1

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The Stone-Weierstraß theorem asserts that this is possible for $k=0$. A constructive proof of this is in the Wikipedia article on Bernstein polynomials.

To do it for all $k$, we can use a sort of diagonal argument. For given $n$, approximate $f^{(n)}$ by a polynomial that differs by less than $\delta_n=1/n$ from $f^{(n)}$ everywhere on $[0,1]$. If we integrate this polynomial $n$ times, using the values at $0$ of the corresponding derivatives of $f$ to determine appropriate integration constants, then in each step the integrated error can't accumulate to $\delta_n$ since we're integrating over an interval of length $1$. Thus the first $n$ derivatives of the resulting polynomial $g_n$ are everywhere within $\delta_n$ of the first $n$ derivatives of $f$. Now for given $k$, we just need to skip the first $k$ polynomials up to $g_k$, and from then on $g_n^{(k)}$ converges uniformly to $f^{(k)}$ with error bound $\delta_n$.

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