Spectral Theories, II, p. 278
On keeps the notations of the preceding exercise. (a) Let f : C → C f:\mathbf{C}\rightarrow \mathbf{C} f : C → C be an element of E 1 / 2 \mathscr{E}_{1/2} E 1/2 such that the restriction of f f f to R \mathbf{R} R belongs to L 2 ( R ) L^2(\mathbf{R}) L 2 ( R ) . For every z ∈ C − Z z\in \mathbf{C}-\mathbf{Z} z ∈ C − Z , we have
f ( z ) = f(z) = f ( z ) = sin(π π z \pi \pi z π π z )∑ n ∈ Z ( − 1 ) n f ( n ) z − n \sum_{n\in\mathbf{Z}}(-1)^n\frac{f(n)}{z-n} ∑ n ∈ Z ( − 1 ) n z − n f ( n ) .
(Use the Paley-Wiener theorem and formula a) of Exercise 1.)
(b) Let f : C → C f:\mathbf{C}\rightarrow \mathbf{C} f : C → C be an element of E 1 / 2 \mathscr{E}_{1/2} E 1/2 such that f f f is bounded on R \mathbf{R} R . For every z ∈ C − Z z\in \mathbf{C}-\mathbf{Z} z ∈ C − Z , we have
f ( z ) = f ( 0 ) + f(z) =f(0) + f ( z ) = f ( 0 ) + sin(π π z \pi \pi z π π z )( f ′ ( 0 ) + ∑ n ∈ Z − { 0 } ( − 1 ) n z f ( n ) − f ( 0 ) n ( z − n ) ) (f'(0) +\sum_{n\in\mathbf{Z}-\{0\}}(-1)^nz\frac{f(n) -f(0)}{n(z-n)}) ( f ′ ( 0 ) + ∑ n ∈ Z − { 0 } ( − 1 ) n z n ( z − n ) f ( n ) − f ( 0 ) )
(apply the preceding question to the function defined by z ↦ ( f ( z ) − f ( 0 ) ) / z z\mapsto (f(z)-f(0))/z z ↦ ( f ( z ) − f ( 0 )) / z for z ≠ 0 z\not= 0 z = 0 and which maps 0 to f ′ ( 0 ) f'(0) f ′ ( 0 ) ).
(c) Let f f f be an element of E 1 / 2 \mathscr{E}_{1/2} E 1/2 bounded on R \mathbf{R} R . Then f ′ f' f ′ is bounded on R \mathbf{R} R and we have
supx ∈ R ∣ f ′ ( x ) ∣ ⩽ 2 _{x\in\mathbf{R}}|f'(x)|\leqslant 2 x ∈ R ∣ f ′ ( x ) ∣ ⩽ 2 supx ∈ R ∣ f ( x ) ∣ _{x\in\mathbf{R}}|f(x)| x ∈ R ∣ f ( x ) ∣
(“Bernstein inequality”; reduce to majorizing ∣ f ′ ( 1 / 2 ) ∣ |f'(1/2)| ∣ f ′ ( 1/2 ) ∣ and apply the preceding formula).
(d) Let f f f be an element of E 1 / 2 \mathscr{E}_{1/2} E 1/2 such that the restriction of f f f to R \mathbf{R} R belongs to L 2 ( R ) L^2(\mathbf{R}) L 2 ( R ) . Let u u u and v v v be the 1-periodic functions on R \mathbf{R} R defined by
u ( t ) = f ^ ( t ) + f ^ ( t − 1 ) , v ( t ) = 2 i π ( t f ^ ( t ) + ( t − 1 ) f ^ ( t − 1 ) )
u(t) =\widehat{f}(t) +\widehat{f}(t-1),v(t) = 2i\pi (t\widehat{f}(t) + (t-1)\widehat{f}(t-1))
u ( t ) = f ( t ) + f ( t − 1 ) , v ( t ) = 2 iπ ( t f ( t ) + ( t − 1 ) f ( t − 1 ))
for 0 ⩽ t < 1 0\leqslant t <1 0 ⩽ t < 1 . For almost every t ∈ [ − 1 , 1 ] t\in [-1,1] t ∈ [ − 1 , 1 ] , we have
f ^ ( t ) = ( 1 − ∣ t ∣ ) u ( t ) + ( 2 i π ) − 1 \widehat{f}(t) = (1- |t|)u(t) + (2i\pi )^{-1} f ( t ) = ( 1 − ∣ t ∣ ) u ( t ) + ( 2 iπ ) − 1 s(t t t )v ( t ) v(t) v ( t ) ,
and the coefficients of the Fourier series of u u u and v v v are f ( − n ) f(-n) f ( − n ) and f ′ ( − n ) f'(-n) f ′ ( − n ) , respectively, for n ∈ Z n\in \mathbf{Z} n ∈ Z .
(e) Let f ∈ E 1 / 2 f\in \mathscr{E}_{1/2} f ∈ E 1/2 as in the preceding question. We have for every z ∈ C z\in \mathbf{C} z ∈ C the formula
f ( z ) = ( f(z) =( f ( z ) = ( sin(π π z \pi \pi z π π z ))2 { ( ∑ n ∈ Z f ( n ) ( z − n ) 2 + ∑ n ∈ Z f ′ ( n ) z − n } ^2\{(\sum_{n\in\mathbf{Z}}\frac{f(n)}{(z-n)^2}+\sum_{n\in\mathbf{Z}}\frac{f'(n)}{z-n}\} 2 {( ∑ n ∈ Z ( z − n ) 2 f ( n ) + ∑ n ∈ Z z − n f ′ ( n ) } ,
where the series converges symmetrically in O ( C ) \mathscr{O}(\mathbf{C}) O ( C ) , that is to say, uniformly on the compact subsets of C \mathbf{C} C . (Use the formulas of question f) of Exercise 1.)
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