Averaging the integral representations of the <Fourier partial sums> gives
$$
\sigma_n(f,x)=\frac1\pi\int_{\mathbb T}
\left(\frac1n\sum_{k=0}^{n-1}D_k(x-t)\right)f(t)\,dt.
$$
The finite trigonometric sum is
$$
\sum_{k=0}^{n-1}\sin\left(k+\frac12\right)u
=\frac{\sin^2(nu/2)}{\sin(u/2)}.
$$
Since $D_k(u)=\sin((k+1/2)u)/(2\sin(u/2))$, the <Fejér kernel> is therefore
$$
\boxed{F_n(u)=\frac1n\sum_{k=0}^{n-1}D_k(u)
=\frac1{2n}\frac{\sin^2(nu/2)}{\sin^2(u/2)}}
$$
and
$$
\boxed{\sigma_n(f,x)=\frac1\pi
\int_{\mathbb T}F_n(x-t)f(t)\,dt}.
$$

The displayed square shows that $F_n\geq0$. Every <Fourier partial sum> preserves the constant function, so $\sigma_n(1)=1$. Substituting $f=1$ in the integral formula gives
$$
\frac1\pi\int_{\mathbb T}F_n(t)\,dt=1.
$$
Nonnegativity then yields
$$
\boxed{\frac1\pi\int_{\mathbb T}|F_n(t)|\,dt=1}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 318 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-318/1/a/solution