Let $f\in C^0([-1,1])$ and $P_n(f)$ its interpolation polynomial at the Chebyshev nodes.

I would be interested to know about any existing results (positive or negative) about the convergence of $P_n(f)$ to $f$ in $L^1([-1,1])$ (only assuming that $f$ is continuous).

The negative results I'm aware of (existence of a continuous function $f$ for which $P_n(f)(x)$ does not converge for any $x$ in $[-1,1]$, from Marcinkiewicz and Grünwald (1936)) do not seem to leave much hope, but still they are not strong enough to rule out convergence in $L^1$. Any additional insight would be welcome!

  • $\begingroup$ If I understand it right what the Chebyshev nodes are, you are just interpolating continuous functions on the unit circle with the property $f(\bar z)=f(z)$ by trigonometric polynomials of degree $n$ over $2n+2$ equidistant nodes. This creates a bounded operator from $C$ to $L^2$ on the circle, so you have convergence in $L^2(\frac{dx}{\sqrt{1-x^2}})$ in the original problem, which is stronger than what you are asking for. Am I misunderstanding something? $\endgroup$ – fedja Jan 10 '20 at 16:07
  • $\begingroup$ Dear fedja, thanks for your answer! Chebyshev nodes are usually defined as $\cos((2k+1)\pi/(2n))$, $k=0,\ldots,n$, or as $\cos(k\pi/n)$, $k=0,\ldots,n$, and in both cases you can indeed view the problem on the circle as you described (but in the second case, which is the one I'm mostly interested in, you only have $2n$ different points on the circle). Could you please elaborate (or give me a reference) regarding the rest of your arguement? (I agree that once we have convergence in $L^2(\frac{dx}{\sqrt{1-x^2}})$ we have convergence in $L^1$) $\endgroup$ – Maxime Jan 10 '20 at 16:58

Here are the details (for the second definition of Chebyshev nodes). We can define the new continuous function $g$ on the circle by $g(z)=f(\frac{z+z^{-1}}2)$ and consider the interpolation by the trigonometric polynomials $Q_n(z)=\sum_{k=0}^n b_k\frac{z^k+z^{-k}}{2}$ (so that $P_n(\frac{z+z^{-1}}{2})=Q_n(z)$) on the $2n$-th roots of unity. The square of the $L^2$ norm of $Q_n$ on the circle is (up to a normalization factor) just $\sum_{k=-n}^n|c_k|^2$ where $c_0=b_0$ and $c_k=\frac {b_{|k|}}2$ for $1\le k\le n$ (so $Q_n(z)=\sum_{k=-n}^n c_kz^k$.

Notice that $z^n= z^{-n}$ on every node $z$, so we can just as well consider the modified polynomial $\widetilde Q_n(z)=\sum_{k=-(n-1)}^{n-1}c_kz^k+b_nz^n$. It will have the same values on the nodes but now all the participating powers of $z$ will be orthogonal with respect to the counting measure on the nodes, so we will have $$ \sum_{k=-(n-1)}^{n-1}|c_k|^2+2(|c_n|^2+|c_{-n}|^2)=\frac 1{2n}\sum_{z:z^{2n}=1}|\widetilde Q_n(z)|^2\\ =\frac1{2n}\sum_{z:z^{2n}=1}|g(z)|^2\le\|g\|_C^2=\|f\|_C^2 $$ and the declared boundedness of the interpolation operator from $C$ to $L^2$ on the circle follows immediately.

The rest is the usual mumbo-jumbo. We have a sequence of linear interpolation operators $I_n:g\mapsto Q_n$ whose norms from $C$ to $L^2$ are uniformly bounded by some constant $M$. Now decompose $g$ into $g_m+h_m$ where $g_m$ is a trigonometric polynomial of degree $m$ and $\|h_m\|\to 0$ as $m\to\infty$ and notice that $I_ng_m=g_m$ for $n>m$, say, so $$ \|g-I_ng\|_{L^2}=\|h_m-I_nh_m\|_{L^2}\le (1+M)\|h_m\|_C. $$ The rest should be clear, but feel free to ask questions if needed :-)

  • $\begingroup$ Thanks for the clarification! Now that I know the answer, and that you made me think about going via the $L^2$ norm, I think this can also be proved by using the $L^2$ convergence of the Chebyshev series (the Chebyshev polynomials form a Hilbert basis for the weigth $\frac{1}{\sqrt{1-x^2}}$), and then the "aliasing" relation between the Chebyshev series and the Chebyshev interpolant. (I'm not saying that this way is any better, maybe just closer to the type of arguments I'm used to) $\endgroup$ – Maxime Jan 10 '20 at 18:34
  • $\begingroup$ @Maxime You should be a bit careful with that idea because we have boundedness $C$ to $L^2$, not $L^2$ to $L^2$, so pure $L^2$ convergence of the series is insufficient and you'll have to invoke some additional ideas as well. On the other hand, you can, of course, rewrite the argument without the trigonometric substitution if you feel like you would prefer it that way. $\endgroup$ – fedja Jan 10 '20 at 19:14

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