Fourier partial-sum operator norm equals the Lebesgue constant

Put $L_N={\int }_0^1\mathrm{\mid }{D}_N(t)\mathrm{\mid }dt$. The convolution formula gives $\mathrm{\mid }{S}_{N}f(x)\mathrm{\mid }\le L_N\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $x$. Finite Fourier sums are linear and continuous as functions of $x$, so both maps in the statement are bounded linear maps and $\mathrm{\parallel }{T}_{N,x_0}\mathrm{\parallel }\le \mathrm{\parallel }{S}_N\mathrm{\parallel }\le L_N$.

For $N\ge 1$ and $0\le j\le N-1$, set $a_j=(j+1\mathrm{/}6)\mathrm{/}(2N+1)$ and $b_j=(j+5\mathrm{/}6)\mathrm{/}(2N+1)$. These disjoint intervals lie in $(0,1\mathrm{/}2)$. On them $\mathrm{\mid }\sin ((2N+1)\pi t)\mathrm{\mid }\ge 1\mathrm{/}2$ and $0<\sin (\pi t)\le \pi t$, so $L_N\ge \frac{1}{2\pi }{\sum }_{j=0}^{N-1}{\int }_{a_j}^{b_j}dt\mathrm{/}t$. Each integral is at least $(b_j-a_j)\mathrm{/}{b}_j=\frac{2\mathrm{/}3}{j+5\mathrm{/}6}\ge \frac{2}{3(j+1)}$.

For $\delta >0$ define $f_{\delta }(u)=D_N(x_0-u)\mathrm{/}\max \{\delta ,\mathrm{\mid }{D}_N(x_0-u)\mathrm{\mid }\}$. The denominator is positive, so this is a real continuous periodic function with norm at most one, also admissible in the complex space. Writing $a=\mathrm{\mid }{D}_N(x_0-u)\mathrm{\mid }$, we have $0\le a-a^2\mathrm{/}\max \{\delta ,a\}\le \delta$. Changing variables in the periodic integral yields $0\le L_N-T_{N,x_0}{f}_{\delta }\le \delta$. Thus $\mathrm{\parallel }{T}_{N,x_0}\mathrm{\parallel }\ge L_N-\delta$ for every $\delta >0$, proving both norm identities. Zeros of the kernel cause no discontinuity in this test.

Consequently $L_N\ge \frac{1}{3\pi }{\sum }_{j=1}^{N}1\mathrm{/}j\ge \frac{1}{3\pi }{\int }_1^{N+1}dt\mathrm{/}t=\frac{1}{3\pi }\log (N+1)$. For $N=0$, $D_0=1$ and both norms equal one by the already proved identities (the test $f=1$ attains the value). This covers the initial index and proves the asserted unboundedness.