Solution.
By definition, the n n n f f f
a k = ∫ T e − 2 π i k x f ( x ) d x a_k = \int_\T e^{-2\pi ikx} f\p{x} \,\diff{x} a k = ∫ T e − 2 πik x f ( x ) d x for n ∈ Z n \in \Z n ∈ Z s n ( x ) = ∑ k = − n n a k e 2 π i k x s_n\p{x} = \sum_{k=-n}^n a_k e^{2\pi ikx} s n ( x ) = ∑ k = − n n a k e 2 πik x
a k e 2 π i k x = ( ∫ T e − 2 π i k y f ( y ) d y ) e 2 π i k x = ∫ T e 2 π i k ( x − y ) f ( y ) d y = ( e 2 π i k y ∗ f ) ( x ) . a_ke^{2\pi ikx}
= \p{\int_\T e^{-2\pi iky} f\p{y} \,\diff{y}}e^{2\pi ikx}
= \int_\T e^{2\pi ik\p{x-y}} f\p{y} \,\diff{y}
= \p{e^{2\pi iky} * f}\p{x}. a k e 2 πik x = ( ∫ T e − 2 πik y f ( y ) d y ) e 2 πik x = ∫ T e 2 πik ( x − y ) f ( y ) d y = ( e 2 πik y ∗ f ) ( x ) . Thus, if D n ( x ) = ∑ k = − n n e 2 π i k x D_n\p{x} = \sum_{k=-n}^n e^{2\pi ikx} D n ( x ) = ∑ k = − n n e 2 πik x
s n ( x ) = ( D n ∗ f ) ( x ) . s_n\p{x} = \p{D_n * f}\p{x}. s n ( x ) = ( D n ∗ f ) ( x ) . Notice that we have a geometric sum, so
D n ( x ) = ∑ k = − n n ( e 2 π i x ) k = e − 2 π i n x ∑ k = 0 2 n ( e 2 π i x ) k = e − 2 π i n x e 2 π i ( 2 n + 1 ) x − 1 e 2 π i x − 1 = e 2 π i ( n + 1 ) x − e − 2 π i n x e 2 π i x − 1 ⋅ e − i π x e − i π x = e π i ( 2 n + 1 ) x − e − π i ( 2 n + 1 ) x e π i x − e − π i x = sin ( ( 2 n + 1 ) π x ) sin ( π x ) . \begin{aligned}
D_n\p{x}
= \sum_{k=-n}^n \p{e^{2\pi ix}}^k
&= e^{-2\pi inx}\sum_{k=0}^{2n} \p{e^{2\pi ix}}^k \\
&= e^{-2\pi inx} \frac{e^{2\pi i\p{2n+1}x} - 1}{e^{2\pi ix} - 1} \\
&= \frac{e^{2\pi i\p{n+1}x} - e^{-2\pi inx}}{e^{2\pi ix} - 1} \cdot \frac{e^{-i\pi x}}{e^{-i \pi x}} \\
&= \frac{e^{\pi i\p{2n+1}x} - e^{-\pi i\p{2n+1}x}}{e^{\pi ix} - e^{-\pi ix}} \\
&= \frac{\sin\p{\p{2n+1}\pi x}}{\sin\p{\pi x}}.
\end{aligned} D n ( x ) = k = − n ∑ n ( e 2 πi x ) k = e − 2 πin x k = 0 ∑ 2 n ( e 2 πi x ) k = e − 2 πin x e 2 πi x − 1 e 2 πi ( 2 n + 1 ) x − 1 = e 2 πi x − 1 e 2 πi ( n + 1 ) x − e − 2 πin x ⋅ e − iπ x e − iπ x = e πi x − e − πi x e πi ( 2 n + 1 ) x − e − πi ( 2 n + 1 ) x = sin ( π x ) sin ( ( 2 n + 1 ) π x ) . Moreover, by Hölder's inequality, ∥ s n ∥ L ∞ ≤ ∥ D n ∥ L 1 ∥ f ∥ L ∞ \norm{s_n}_{L^\infty} \leq \norm{D_n}_{L^1} \norm{f}_{L^\infty} ∥ s n ∥ L ∞ ≤ ∥ D n ∥ L 1 ∥ f ∥ L ∞ ∥ f ∥ L ∞ < ∞ \norm{f}_{L^\infty} < \infty ∥ f ∥ L ∞ < ∞ f f f ∥ D n ∥ L 1 \norm{D_n}_{L^1} ∥ D n ∥ L 1
First, notice that sin ( π x ) \sin\p{\pi x} sin ( π x ) [ 0 , 1 2 ] \br{0, \frac{1}{2}} [ 0 , 2 1 ] t ∈ [ 0 , 1 ] t \in \br{0, 1} t ∈ [ 0 , 1 ]
sin ( π t 2 ) ≥ t sin ( π 2 ) = t ⟹ sin ( π x ) ≥ 2 x for x ∈ [ 0 , 1 2 ] . \sin\p{\pi \frac{t}{2}} \geq t\sin\p{\frac{\pi}{2}} = t
\implies \sin\p{\pi x} \geq 2x
\quad\text{for } x \in \br{0, \frac{1}{2}}. sin ( π 2 t ) ≥ t sin ( 2 π ) = t ⟹ sin ( π x ) ≥ 2 x for x ∈ [ 0 , 2 1 ] . Thus,
∥ D n ∥ L 1 = ∫ − 1 / 2 1 / 2 ∣ sin ( ( 2 n + 1 ) π x ) sin ( π x ) ∣ d x = 2 ∫ 0 1 / 2 ∣ sin ( ( 2 n + 1 ) π x ) sin ( π x ) ∣ d x ( evenness of ∣ sin ( π x ) ∣ ) ≤ 2 ∫ 0 1 / 2 ∣ sin ( ( 2 n + 1 ) π x ) 2 x ∣ d x = 1 2 ∫ 0 ( 2 n + 1 ) / 2 ∣ sin ( π x ) x ∣ d x ( x ↦ ( 2 n + 1 ) x ) ≤ 1 2 ∫ 0 n + 1 ∣ sin ( π x ) x ∣ d x = 1 2 ∑ k = 0 n ∫ k k + 1 ∣ sin ( π x ) x ∣ d x ≤ 1 2 ∫ 0 1 ∣ sin ( π x ) x ∣ d x + 1 2 ∑ k = 1 n 1 k ≤ 1 2 ∫ 0 1 ∣ sin ( π x ) x ∣ d x + 1 2 ( 1 + ∫ 1 n 1 x d x ) ≤ C log n , \begin{aligned}
\norm{D_n}_{L^1}
&= \int_{-1/2}^{1/2} \abs{\frac{\sin\p{\p{2n+1}\pi x}}{\sin\p{\pi x}}} \,\diff{x} \\
&= 2\int_0^{1/2} \abs{\frac{\sin\p{\p{2n+1}\pi x}}{\sin\p{\pi x}}} \,\diff{x}
&& (\text{evenness of } \abs{\sin\p{\pi x}}) \\
&\leq 2\int_0^{1/2} \abs{\frac{\sin\p{\p{2n+1}\pi x}}{2x}} \,\diff{x} \\
&= \frac{1}{2} \int_0^{\p{2n+1}/2} \abs{\frac{\sin\p{\pi x}}{x}} \,\diff{x}
&& (x \mapsto \p{2n+1}x) \\
&\leq \frac{1}{2} \int_0^{n+1} \abs{\frac{\sin\p{\pi x}}{x}} \,\diff{x} \\
&= \frac{1}{2} \sum_{k=0}^n \int_k^{k+1} \abs{\frac{\sin\p{\pi x}}{x}} \,\diff{x} \\
&\leq \frac{1}{2} \int_0^1 \abs{\frac{\sin\p{\pi x}}{x}} \,\diff{x} + \frac{1}{2} \sum_{k=1}^n \frac{1}{k} \\
&\leq \frac{1}{2} \int_0^1 \abs{\frac{\sin\p{\pi x}}{x}} \,\diff{x} + \frac{1}{2} \p{1 + \int_1^n \frac{1}{x} \,\diff{x}} \\
&\leq C\log{n},
\end{aligned} ∥ D n ∥ L 1 = ∫ − 1/2 1/2 ∣ ∣ sin ( π x ) sin ( ( 2 n + 1 ) π x ) ∣ ∣ d x = 2 ∫ 0 1/2 ∣ ∣ sin ( π x ) sin ( ( 2 n + 1 ) π x ) ∣ ∣ d x ≤ 2 ∫ 0 1/2 ∣ ∣ 2 x sin ( ( 2 n + 1 ) π x ) ∣ ∣ d x = 2 1 ∫ 0 ( 2 n + 1 ) /2 ∣ ∣ x sin ( π x ) ∣ ∣ d x ≤ 2 1 ∫ 0 n + 1 ∣ ∣ x sin ( π x ) ∣ ∣ d x = 2 1 k = 0 ∑ n ∫ k k + 1 ∣ ∣ x sin ( π x ) ∣ ∣ d x ≤ 2 1 ∫ 0 1 ∣ ∣ x sin ( π x ) ∣ ∣ d x + 2 1 k = 1 ∑ n k 1 ≤ 2 1 ∫ 0 1 ∣ ∣ x sin ( π x ) ∣ ∣ d x + 2 1 ( 1 + ∫ 1 n x 1 d x ) ≤ C log n , ( evenness of ∣ sin ( π x ) ∣ ) ( x ↦ ( 2 n + 1 ) x ) for C C C n n n n n n
∥ s n ∥ L ∞ log n ≤ C ∥ f ∥ L ∞ log n log n = C ∥ f ∥ L ∞ . \frac{\norm{s_n}_{L^\infty}}{\log{n}}
\leq \frac{C\norm{f}_{L^\infty}\log{n}}{\log{n}}
= C\norm{f}_{L^\infty}. log n ∥ s n ∥ L ∞ ≤ log n C ∥ f ∥ L ∞ log n = C ∥ f ∥ L ∞ . Let ε > 0 \epsilon > 0 ε > 0 T \T T f f f p ( x ) = ∑ j = 1 N c j e 2 π i j x p\p{x} = \sum_{j=1}^N c_j e^{2\pi ijx} p ( x ) = ∑ j = 1 N c j e 2 πij x ∥ f − p ∥ L ∞ < ε \norm{f - p}_{L^\infty} < \epsilon ∥ f − p ∥ L ∞ < ε
b k = ∫ T e − 2 π i k x p ( x ) d x = ∑ j = 1 N c j ∫ 0 1 e − 2 π i ( j − k ) x d x . b_k
= \int_\T e^{-2\pi ikx} p\p{x} \,\diff{x}
= \sum_{j=1}^N c_j \int_0^1 e^{-2\pi i\p{j-k}x} \,\diff{x}. b k = ∫ T e − 2 πik x p ( x ) d x = j = 1 ∑ N c j ∫ 0 1 e − 2 πi ( j − k ) x d x . By orthogonality, b k = c j b_k = c_j b k = c j j = k j = k j = k 0 0 0 t n t_n t n N N N
∥ t n ∥ L ∞ log n = ∣ ∑ k = 1 n b k e 2 π i k x ∣ log n ≤ ∑ k = 1 N ∣ b k ∣ log n → n → ∞ 0. \frac{\norm{t_n}_{L^\infty}}{\log{n}}
= \frac{\abs{\sum_{k=1}^n b_k e^{2\pi ikx}}}{\log{n}}
\leq \frac{\sum_{k=1}^N \abs{b_k}}{\log{n}}
\xrightarrow{n\to\infty} 0. log n ∥ t n ∥ L ∞ = log n ∣ ∣ ∑ k = 1 n b k e 2 πik x ∣ ∣ ≤ log n ∑ k = 1 N ∣ b k ∣ n → ∞ 0. We then see
lim sup n → ∞ ∥ s n ∥ L ∞ log n ≤ lim sup n → ∞ ( ∥ s n − t n ∥ L ∞ log n + ∥ t n ∥ L ∞ log n ) ≤ C ∥ f − p ∥ L ∞ < ε . \limsup_{n\to\infty}\,\frac{\norm{s_n}_{L^\infty}}{\log{n}}
\leq \limsup_{n\to\infty}\,\p{\frac{\norm{s_n - t_n}_{L^\infty}}{\log{n}} + \frac{\norm{t_n}_{L^\infty}}{\log{n}}}
\leq C\norm{f - p}_{L^\infty}
< \epsilon. n → ∞ lim sup log n ∥ s n ∥ L ∞ ≤ n → ∞ lim sup ( log n ∥ s n − t n ∥ L ∞ + log n ∥ t n ∥ L ∞ ) ≤ C ∥ f − p ∥ L ∞ < ε . Sending ε → 0 \epsilon \to 0 ε → 0