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I'm reading through my professor's class notes for Fourier Analysis and there is a remark that I can't verify.

He notes that:

$$[\sigma_n (f)](x) - f(x) = \frac1{2\pi}\int_{-\pi}^\pi [f(x+t) - f(x)]K_n(t)\;\mathrm dt.$$

Note that here $K_n$ is the Fejer kernel of $f$, and the factor of $2\pi$ comes from the fact that we scaled the $L_1$ norm by this factor which I have noticed is not always done.

I tried using the fact that $\sigma_n (f)$ is equal to $f*K_n$, which let me obtain:

$$[\sigma_n (f)](x) - f(x) = \frac1{2\pi}\int_{-\pi}^\pi f(x+t)K_n (t)\;\mathrm dt - f(x)$$

I can't see how to get the $K_{n}$ inside the integral, let alone attached to the factor of $K_{n}$. Thanks for the patience, I just am learning how to use this site. :)

Note that the definition of $\sigma_n (f)$ is $\frac1{n}\sum\limits_{j=0}^{n-1}S_j (f)$, where $S_j (f) = \sum\limits_{m=-k}^k\widehat{f}(m) e^{imt}$ is the $j$-th partial sum of the Fourier series of $f$.

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it's difficult to read. Put the formulas between two "$$" Can you add it in the question? – Davide Giraudo Sep 30 '11 at 18:37
Ok. I put all my subsequent comments into the OP with proper formatting. Sorry I didn't realize I could edit my post at first. Now I seem to have the hang of inserting tex code too. Thank you. – roo Sep 30 '11 at 19:00
The Fejer Kernel is integrable, so you can write $f(x)=c\int_{-\pi}^{\pi}f(x)K_n(t)dt$. Most likely $c=\frac{1}{2\pi}=\left( \int_{-\pi}^{\pi} K_n(t)dt\right)^{-1}$ but I haven't done the calculations. – Jose27 Sep 30 '11 at 19:15
I have never heard of this fact. I guess I'll just have to accept it and plough onward.... – roo Sep 30 '11 at 19:27
This is literally the first thing stated immediately after the definition of both the Fejer Kernel and the Cesaro mean. So I would understand that there is very little machinery available to verify this statement. – roo Sep 30 '11 at 19:44
up vote 3 down vote accepted

You simply use the fact that $\frac{1}{2 \pi} \int_{-\pi}^{\pi} K_n(t) \mathrm{d} t=1$. To see this, note the closed form expression for the Fejer kernel:

$$ K_n(t) = \frac{1}{n} \sum_{k=0}^{n-1} D_k(t) $$ where $D_k(t) = \sum_{n=-k}^{k} \exp(i t n)$. Now:

$$ \frac{1}{2\pi}\int_{-\pi}^\pi K_n(t) \mathrm{d} t = \frac{1}{2 \pi n} \int_{-\pi}^\pi \sum_{k=0}^{n-1} D_k (t) \mathrm{d} t = \frac{1}{2 \pi n} \sum_{k=0}^{n-1} \int_{-\pi}^\pi D_k (t) \mathrm{d} t = \frac{1}{2 \pi n} \sum_{k=0}^{n-1} (2 \pi) = 1 $$ where we used $\int_{-\pi}^\pi D_k (t) \mathrm{d} t = \sum_{n=-k}^{k} \int_{-\pi}^\pi {\rm e}^{i t n} \mathrm{d} t = \sum_{n=-k}^{k} ( 2\pi \delta_{n,0} ) = 2 \pi$.

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Thank you Sasha. I literally facepalmed when I clued into the fact that $f(x)$ is independant of the variable of integration $t$. (Please forgive me I'm on cold meds today.) Of course $f(x)$ can move freely in and out of the integral. I now see why Jose27 was so brief: there really isn't much to prove. Thanks to both of you. – roo Sep 30 '11 at 20:47

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