Vietnam J Math DOI 10.1007/s10013-015-0146-y A Study of the Sequence of Norm of Derivatives (or Primitives) of Functions Depending on Their Beurling Spectrum Ha Huy Bang1 · Vu Nhat Huy2 Received: 22 October 2014 / Accepted: December 2014 © Vietnam Academy of Science and Technology (VAST) and Springer Science+Business Media Singapore 2015 Abstract In this paper, we characterize the behavior of the sequence of norm of derivatives (or primitives) of functions by their Beurling spectrum in Banach spaces The Bernstein inequality for Banach spaces is also obtained Keywords Banach Spaces · Beurling spectrum Mathematics Subject Classification (2010) 26D10 · 46E30 Introduction The following result was proved in [9]: Let ≤ p ≤ ∞ and f (m) ∈ Lp (R), m = 0, 1, 2, Then there always exists the following limit 1/m lim f (m) p m→∞ and lim m→∞ 1/m f (m) p = σf = sup{|ξ | : ξ ∈ supp fˆ}, where fˆ is the Fourier transform of f 1/m The result shows that the behavior of the sequence f (m) p is wholly characterized by the spectrum of f , and it reminds us of the following well-known spectral radius formula in Ha Huy Bang hhbang@math.ac.vn Vu Nhat Huy nhat huy85@yahoo.com Institute of Mathematics, Vietnamese Academy of Science and Technology, 18 Hoang Quoc Viet Street, Cau Giay, Hanoi, Vietnam Department of Mathematics, College of Science, Vietnam National University, 334 Nguyen Trai Street, Thanh Xuan, Hanoi, Vietnam H H Bang, V N Huy the Theory of Banach algebras: Let A be a unital Banach algebra and x ∈ A Then the limit limm→∞ x m 1/m always exists and satisfies xm lim m→∞ = rA (x), 1/m where rA (x) = sup{|λ| : λ ∈ σA (x)} is the spectral radius of x, σA (x) is the spectrum of x which is the set of all λ ∈ C such that x − λ1 is not invertible in A and is the unit in A (see [18, 22]) That is why σf is called the local spectral radius of the differential operator D, and this result was studied and developed by many authors (see [1–12, 15–20, 26–29]) It is natural to ask what will happen when we replace derivatives by integrals To this question, Tuan answered for p = in [29] and we answered in [13] for ≤ p ≤ ∞, and in [14] for a more general case (but with the assumption that the spectrum must be compact) In this paper, we extend the results in [9, 13, 14, 29] to Banach spaces, we also extend the results in [17] to a general case and the Bernstein inequality is also obtained Main Results Let f ∈ L1 (R) and fˆ = Ff be its Fourier transform +∞ fˆ(ξ ) = √ 2π −∞ e−ixξ f (x)dx, and fˇ = F −1 f be its inverse Fourier transform +∞ fˇ(ξ ) = √ eixξ f (x)dx 2π −∞ Let (X, · X ) denote a complex Banach space and let BC(R → X) denote the set of all X valued bounded continuous functions f : R → X For a given function f ∈ BC(R → X), we define f ∞ = sup{ f (t) X : t ∈ R} Then (BC(R → X), · ∞ ) itself is a Banach space We define the derivative Df of f ∈ BC(R → X), as usual, f (s + δ) − f (s) δ Note that for every λ ∈ C \ iR the operator λ − D is invertible Hence, the spectrum of the differential operator is iR Clearly, the inverse of λ − D is a bounded operator on (BC(R → X), · ∞ ) Moreover, Df (s) = lim δ→0 +∞ −∞ ϕ(t)(λ − D)−n f (t)dt = +∞ −∞ (λ − D)−n ϕ(t) f (t)dt for any f ∈ BC(R → X) The convolution ϕ ∗ f of f with a Schwartz function is defined by letting ϕ ∗ f (s) = +∞ −∞ ϕ(s − t)f (t)dt Proposition (Young inequality for Banach spaces) Let f ∈ BC(R → X), ϕ ∈ S (R) Then ϕ ∗ f ∈ BC(R → X) and ϕ∗f ∞ ≤ f ∞ ϕ The Beurling spectrum Spec(f ) of a function f ∈ BC(R → X) is defined by Spec(f ) = {ξ ∈ R : ∀ > 0, ∃ϕ ∈ S (R) : supp ϕ ⊂ (ξ − , ξ + )), ϕ ∗ f ≡ 0} The Sequence of Norm of Derivatives (or Primitives) of Functions The Beurling spectral radius ρ(f ) of f is defined by ρ(f ) = sup{|ξ | : ξ ∈ Spec(f )} Note that Spec(f ) is always a closed subset of R Moreover, Proposition We have the following properties of Beurling spectrum: – – – if ϕ = on √ Spec(f ) then ϕ ∗ f = 0; if ϕ = 1/ 2π on Spec(f ) then ϕ ∗ f = f ; Spec((λ − D)−1 f ) = Spec(f ) for every f ∈ BC(R → X) and λ ∈ C \ iR See [21] for more details Now, we define D = −iD and the differential operator P (D) is obtained from P (x) by substituting x → −iD (for P (x) = nk=0 ak x k then P (D)f = n k k=0 ak D f ) Theorem Let P (x) be a polynomial and f ∈ BC(R → X), f ≡ Assume that Spec(f ) is a compact set Then lim m→∞ P m (D)f 1/m ∞ = P m (D)f 1/m ∞ ≥ sup |P (ξ )| (1) sup |P (ξ )| (2) ξ ∈Spec(f ) Proof Let us first show that lim m→∞ ξ ∈Spec(f ) Indeed, let σ ∈ Spec(f ) satisfy P (σ ) = For a small enough positive number , we have P (x) = for all x ∈ B(σ, ) From the definition of Beurling spectrum, there exists a function ϕ ∈ C ∞ (R), supp ϕ ⊂ B(σ, ) such that ϕq∗ f ≡ We put ϕm = F −1 (ϕ(x)/P m (x)) Then ϕm is well defined and it follows from Dk ϕm = F −1 (x k ϕm (x)) = F −1 (x k ϕ(x)/P m (x)) ∀k ∈ Z+ that P m (D)ϕm = F −1 (P m (x)ϕ(x)/P m (x)) = ϕq Since ϕm ∗ (D k f ) = (D k ϕm ) ∗ f (3) for all k ∈ Z+ , we have ϕm ∗ (P m (D)f ) = (P m (D)ϕm ) ∗ f Thus, by (3), we get < ϕq ∗ f ∞ = ϕm ∗ P m (D)f ∞ Applying Young inequality for Banach spaces, we obtain < ϕq ∗ f ∞ = ϕm ∗ P m (D)f ∞ ≤ P m (D)f and then lim m→∞ P m (D)f 1/m ∞ ≥ lim m→∞ ϕm 1/m −1 According to [14] and ϕ(x) = ∀x ∈ / B(σ, ), we conclude that lim m→∞ ϕm 1/m ≤ sup |1/P (ξ )| ξ ∈B(σ, ) ∞ ϕm (4) H H Bang, V N Huy Therefore, by (4), we obtain Letting 1/m ∞ P m (D)f lim m→∞ ≥ inf ξ ∈B(σ, ) |P (ξ )| → 0, we get 1/m ∞ P m (D)f lim m→∞ ≥ |P (σ )| (5) Because (5) holds for any σ ∈ Spec(f ), P (σ ) = 0, we arrive at (2) Finally, we show 1/m lim P m (D)f ∞ ≤ sup |P (ξ )|, m→∞ ξ ∈K (6) ∞ where K = Spec(f ) Indeed, √ for > 0, we can choose a function h ∈ C (R) satisfying the conditions h(ξ ) = 1/ 2π for ξ ∈ K /2 and h(ξ ) = for ξ ∈ K Using Proposition 2, ˇ ∗ f for all k ∈ Z+ That gives we have f = hˇ ∗ f Clearly, D k f = D k (hˇ ∗ f ) = (D k h) ˇ ∗ f P m (D)f = (P m (D)h) Then, since P m (D)hˇ = F −1 (h(x)P m (x)), we get P m (D)f = F −1 (h(x)P m (x)) ∗ f Applying Young inequality, we obtain P m (D)f ∞ F −1 (h(ξ )P m (ξ )) ≤ f ∞ 1/m ∞ m→∞ Hence, lim m→∞ P m (D)f ≤ lim F (h(ξ )P m (ξ )) 1/m (7) We have known in [11] that lim m→∞ F (h(ξ )P m (ξ )) 1/m = lim m→∞ gm 1/m ≤ sup |P (ξ )| ξ ∈K (8) Combining (7) and (8), we get lim m→∞ P m (D)f 1/m ∞ ≤ sup |P (ξ )| ξ ∈K and then (6) by letting → Combing (2) and (6), we arrive at (1) The proof is complete We define Iλ = (λ − D)−1 , where λ ∈ C \ iR The integral operator P (Iλ ) is obtained from P (x) by substituting x → Iλ We have the following theorem: Theorem Let P (x) be a polynomial, f ∈ BC(R → X), f ≡ and λ ∈ C \ iR Assume that Spec(f ) is a compact set Then there always exists the following limit lim m→∞ P m (Iλ )f 1/m ∞ and lim m→∞ P m (Iλ )f 1/m ∞ = P m (Iλ )f 1/m ∞ ≥ sup |P (1/(λ − iξ ))| sup |P (1/(λ − iξ ))| ξ ∈Spec(f ) Proof Now we show that lim m→∞ ξ ∈Spec(f ) (9) The Sequence of Norm of Derivatives (or Primitives) of Functions Indeed, let σ be an arbitrary element in Spec(f ) satisfying P (1/(λ − iσ )) = For a small enough positive number , we have P m (1/(λ − iξ )) = for all ξ ∈ B(σ, ) From the definition of Beurling spectrum, there exists ϕ ∈ C0∞ (R), supp ϕ ⊂ B(σ, ) such that ϕq ∗ f ≡ Put ϕm = F −1 ϕ(ξ )/P m (1/(λ − iξ )) Then, ϕm is well defined, ϕm ∈ S (R) and P m (Iλ )ϕm = ϕq Clearly, ϕm ∗(Iλk f ) = (Iλk ϕm )∗f for all k ∈ Z+ That gives ϕm ∗ (P m (Iλ )f ) = (P m (Iλ )ϕm ) ∗ f and then ϕm ∗ P m (Iλ )f = ϕq ∗ f So, applying Young inequality, we obtain < ϕq ∗ f ∞ = ϕm ∗ P m (Iλ )f ∞ ≤ P m (Iλ )f ∞ (10) ϕm Clearly, sup |(1 + x )ϕm (x)| x∈R ≤ √ 2π D ϕ(ξ )/P m (1/(λ − iξ )) + ϕ(ξ )/P m (1/(λ − iξ )) ξ ∈B(σ, ) dξ So sup (1 + x )ϕm (x) ≤ C1 m2 sup ξ ∈B(σ, ) x∈R 1/P m+2 (1/(λ − iξ )) , (11) where C1 := √ 2π |(D ϕ)(ξ )P (1/(λ − iξ ))| + 2|(Dϕ)(ξ )D(P (1/(λ − iξ ))) ξ ∈K ×P (1/(λ − iξ ))| + |ϕ(ξ )D (P (1/(λ − iξ )))P (1/(λ − iξ )) +|ϕ(ξ )(D(P (1/(λ − iξ ))))2 | + |ϕ(ξ )P (1/(λ − iξ ))| dξ Further, we have R |ϕm (x)|dx ≤ π sup |(1 + x )ϕm (x)| (12) x∈R Combining (11)–(12), we obtain lim m→∞ ϕm 1/m ≤ sup |1/P (1/(λ − iξ ))| ξ ∈B(σ, ) (13) Relations (10) and (13) imply P m (Iλ )f lim m→∞ Letting 1/m ∞ ≥ inf ξ ∈B(σ, ) |P (1/(λ − iξ ))| → 0, we get P m (Iλ )f lim m→∞ 1/m ∞ ≥ |P (1/(λ − iσ ))| (14) Because (14) holds for any σ ∈ Spec(f ), P (1/(λ − iσ )) = 0, we confirm (9) Put K = Spec(f ) Now we claim that lim m→∞ P m (Iλ )f 1/m ∞ ≤ sup |P (1/(λ − iξ ))| ξ ∈K (15) √ Indeed, for any > 0, there exists h ∈ C ∞ (R) such that h(x) = 1/ 2π if x ∈ K /2 and h(x) = if x ∈ / K Note that λ − iξ = for all ξ ∈ K , so the following function is well defined: Hm = F −1 (h(ξ )P m (1/(λ − iξ ))) H H Bang, V N Huy ˇ ∗ f ∀k ∈ Z+ ˇ Using Proposition 2, we have f = h∗f and observe from Iλk f = (Iλk h) that ˇ ∗ f (16) P m (Iλ )f = (P m (Iλ )h) k −1 k m −1 m ˇ ˇ From Iλ h = F (h(ξ )/(λ − iξ ) ), we conclude P (Iλ )h = F (h(ξ )P (1/(λ − iξ ))) = Hm Therefore, by using (16), we have P m (Iλ )f = Hm ∗ f Hence, it follows from Proposition that P m (Iλ )f ∞ ≤ f ∞ Hm So lim m→∞ P m (Iλ )f 1/m ∞ ≤ lim m→∞ Hm 1/m (17) Clearly, √ 2π sup |(1 + x )Hm (x)| x∈R ≤ ξ ∈K ≤ ξ ∈K |D (h(ξ )P m (1/(λ − iξ )))| + |h(ξ )P m (1/(λ − iξ ))| dξ |D h(ξ )P m (1/(λ−iξ ))|+2m|Dh(ξ )D(P (1/(λ − iξ )))P m−1 (1/(λ − iξ ))| +m(m − 1)|h(ξ )(D(P (1/(λ − iξ ))))2 P m−2 (1/(λ − iξ ))| +m|h(ξ )D (P (1/(λ − iξ )))P m−1 (1/(λ − iξ ))| + |h(ξ )P m (1/(λ − iξ ))| dξ So sup |(1 + x )Hm (x)| ≤ C2 m2 sup |P m−2 (1/(λ − iξ ))|, ξ ∈K x∈R (18) where C2 does not depend on m Further, we have R |Hm (x)|dx ≤ π sup |(1 + x )Hm (x)| (19) x∈R Combining (18)–(19), we obtain lim m→∞ Hm 1/m ≤ sup |P (1/(λ − iξ ))| ξ ∈K (20) We combine inequality (17) with inequality (20) to conclude that lim m→∞ Letting P m (Iλ )f 1/m ∞ ≤ sup |P (1/(λ − iξ ))| ξ ∈K → 0, we confirm (15) Combining (15) and (9), we have lim m→∞ P m (Iλ )f 1/m ∞ = sup ξ ∈Spec(f ) |P (1/(λ − iξ ))| The proof is complete Based on the results about Bernstein inequality in Lp (R)-spaces (see [23–25]), we obtain Bernstein inequality for Banach spaces in the following theorem: Theorem Let σ > 0, f ∈ BC(R → X) and Spec(f ) ⊂ [−σ, σ ] Then there exists a constant C not depending on f, m, σ such that Dm f ∞ ≤ Cσ m f ∞, m = 1, 2, (21) The Sequence of Norm of Derivatives (or Primitives) of Functions Proof Let us first prove (21) for the case σ = Indeed, put K := [−1, 1] and the function (ξ ) is defined as follows C1 e1/(ξ (ξ ) = where C1 is chosen such that via the formula −1) if |ξ | < 1, if |ξ | ≥ 1, (ξ )dξ = We define the sequence of functions φm (ξ )m≥1 R φm (ξ ) = (1K3/(4m) ∗ 1/(4m) )(ξ ), where = 4m (4mξ ) 1/(4m) (ξ ) Then 1/(4m) (ξ ) = for all ξ ∈ [−1/(4m), 1/(4m)] and R 1/(4m) (ξ )dξ = Hence, for any m ≥ 1, we have φm (ξ ) ∈ C0∞ (R), and φm (ξ ) = ∀ξ ∈ K1/(2m) , φm (ξ ) = ∀ξ ∈ / K1/m So, it follows from Spec(f ) ⊂ K that f = (2π )−1/2 F (φm ) ∗ f Hence, Dm f ∞ = (2π )−1/2 F (φm (ξ )ξ m ) ∗ f ∞ Therefore, applying Young inequality (Proposition 1), we obtain the estimate Dm f ∞ = (2π )−1/2 F (φm (ξ )ξ m ) ∗ f ∞ ≤ (2π )−1/2 f F (φm (ξ )ξ m ) ∞ (22) Define km := + , m ξ km gm (ξ ) = φm = φm (ξ ) − gm (ξ ) m (ξ ) , Then (F (gm (ξ )ξ m ))(x) = (km )m F φm ξ km m ξ km (x) = (km )m+1 (F (φm (ξ )ξ m ))(km x) So F (gm (ξ )ξ m ) Then, it follows from (km )m = (1 + F( m (ξ )ξ m (ξ ) m ) 1 ≥ that m m) m F (gm (ξ )ξ ) Therefore, since = (km )m F (φm (ξ )ξ m ) ≥ F (φm (ξ )ξ m ) 1 = φm (ξ ) − gm (ξ ) we get ≥ F (gm (ξ )ξ m ) − F (φm (ξ )ξ m ) ≥ F (φm (ξ )ξ m ) (23) From (22)–(23), we obtain Dm f ∞ ≤ (2π )−1/2 f ∞ F( m (ξ )ξ m (24) ) Now, we show the existence of a constant C such that F( m (ξ )ξ m ) ≤ (2π )1/2 C for all m ≥ Indeed, we put C2 = max{ (j ) , j ≤ 3} Since (j ) we obtain that 1/(4m) (x) = (4m)j +1 (j ) (4mx) and then (j ) 1/(4m) = (4m)j (j ) ≤ C2 (4m)j 1/(4m) (x) = 4m (4mx), ∀j ≤ Therefore, (j ) φm ∞ = (1K3/(4m) ∗ (j ) 1/(4m) ) ∞ ≤ (j ) 1/(4m) ≤ (4m)j C2 ∀j ≤ (25) H H Bang, V N Huy Note that φm (ξ ) = ∀ξ ∈ (−1 − (1/2m), + (1/2m)), and φm (ξ ) = ∀ξ ∈ (−∞, −1 − (1/m))∪(1+(1/m), +∞) So, if |ξ | < then |ξ/km | < |ξ | < and φm (ξ ) = φm (ξ/km ) = 1, i.e., m (ξ ) = Further, if |ξ | > + (3/m) then |ξ | > |ξ/km | > + (1/m) and then φm (ξ ) = φm (ξ/km ) = 0, i.e., m (ξ ) = So we have supp m ⊂ [1, + (3/m)] ∪ [−1 − (3/m), −1] (26) Now, if ξ ∈ [1, + (3/m)] ∪ [−1 − (3/m), −1] then ξ− (km − 1)ξ ξ ξ = = ≤ km km mkm m (27) From (25) and (27), we get the following estimate for ξ ∈ [1, 1+(3/m)]∪[−1−(3/m), −1] | m (ξ )| = |φm (ξ ) − gm (ξ )| = φm (ξ ) − φ ≤ ξ− m (ξ ) ξ · φm km ∞ ≤ ξ km 4mC2 = 16C2 , m = φm (ξ ) − gm (ξ ) = φm (ξ ) − φm = φm (ξ ) − ≤ ξ− ≤ φ km m ξ · φm km ξ km ∞ ξ km ≤ φm (ξ ) − φm + 1− · φm km (28) ξ km 1− + km φm ξ km ∞ (4m)2 C2 + − 4mC2 ≤ 68mC2 , m km (29) and m (ξ ) = φm (ξ ) − gm (ξ ) = φm (ξ ) − φm = φm (ξ ) − ≤ ξ− ≤ φ m km ξ · φm km ξ km ∞ ξ km ≤ φm (ξ ) − φm + 1− · φm km ξ km 1− + km φm ∞ (4m)3 C2 + − (4m)2 C2 ≤ 528m2 C2 m km Put Hm (x) = (F ( m (ξ )ξ m ))(x) ξ km (30) Then Hm (x) = √ e−ixξ m (ξ )ξ m dξ 2π R Therefore, from (26), we obtain 1 m sup |Hm (x)| ≤ √ dξ = √ m (ξ )ξ 2π 2π 1≤|ξ |≤1+ m3 R x∈R m (ξ )ξ m dξ and it follows from (25) that sup |Hm (x)| ≤ x∈R sup | √ m 2π ξ ∈R m (ξ )| 1+ m m ≤ 96e3 C2 √ m 2π (31) The Sequence of Norm of Derivatives (or Primitives) of Functions We also have x Hm (x) = √ 2π ≤ √ 2π and then R R e−ixξ ( m (ξ )m(m − 1)ξ m−2 + m (ξ )2mξ m−1 + m (ξ )ξ e−ixξ ( m (ξ )m(m − 1)ξ m−2 + m (ξ )2mξ m−1 + m ξ m)dξ m )dξ dξ sup x Hm (x) x∈R ≤ √ 2π ≤ 1≤|ξ |≤1+ m | m (ξ )m(m − 1)ξ √ m 2π ξ ∈R + sup | (ξ )|2m + sup | (ξ )|m(m − 1) + ξ ∈R m−2 + (ξ )2mξ m−1 + m−2 m m−1 m (ξ )ξ m |dξ + sup | (ξ )| + ξ ∈R m m So, from (28)–(30), we get sup x Hm (x) ≤ x∈R ≤ 16C2 m(m − 1)e3 + 68mC2 2me3 + 528m2 C2 e3 √ m 2π 5280e3 C2 m √ 2π (32) We have Hm = |x|≤m |Hm (x)|dx + ≤ sup |Hm (x)| |x|≤m x∈R |x|≥m |Hm (x)|dx 1dx + sup |x Hm (x)| x∈R |x|≥m = 2m sup |Hm (x)| + sup |x Hm (x)| m x∈R x∈R dx x2 (33) From (31)–(33), we obtain Hm ≤ 2m 96e3 C2 √ m 2π + 5280e3 C2 m √ m 2π Therefore, from (24), we can choose a constant C such that Dm f ∞ ≤C f ∞ So, (21) has been proved with σ = Next, we prove (21) for any σ > Indeed, put x g(x) = f σ Since Spec(f ) ⊂ [−σ, σ ], Spec(g) ⊂ [−1, 1] Therefore Dm g Since g(x) = f ( σx ), ∞ ≤C g (34) ∞ we have g ∞ = f ∞, Dm g ∞ = σ −m D m f ∞ H H Bang, V N Huy Hence, it follows from (34) that σ −m D m f ≤C f ∞ ∞ We conclude that Dm f ∞ ≤ Cσ m f ∞ The proof is complete The minimum of C satisfying the above inequalities is called the Bernstein constant Theorem Let f ∈ BC(R → X), σ ∈ R+ Assume that Spec(f ) ⊂ [−σ, σ ] Then lim σ −m D m f m→∞ ∞ ≤ C lim λ→1− f (·) − f (λ·) (35) ∞, where C is the Bernstein constant Proof Fix λ ∈ (0, 1) Put fλ (x) = f (λx) Then Spec(fλ ) = λSpec(f ) Then it follows from Spec(f ) ⊂ [−σ, σ ] that Spec(fλ ) ⊂ [−λσ, λσ ] (36) Applying the Bernstein inequality for Beurling spectrum, we get D m fλ ∞ ≤ C(λσ )m fλ (37) ∞ for all m ∈ N Using (36) and Spec(f ) ⊂ [−σ, σ ], we obtain Spec(f − fλ ) ⊂ [−σ, σ ] Then it follows from the Bernstein inequality for Beurling spectrum that D m (f − fλ ) ∞ ≤ Cσ m f − fλ (38) ∞ Therefore, using (37) and (38), we have σ −m D m f ∞ ≤ σ −m D m (f − fλ ) ∞ m + Cλ + σ −m D m fλ ≤ C f − fλ ∞ = C f − fλ m f ∞ + Cλ fλ ∞ ∞ ∞ for all m ∈ N Hence, lim σ −m D m f m→∞ Letting λ → 1− , we get limm→∞ σ −m D m f is complete ∞ ≤ C f − fλ ∞ ∞ ≤ C limλ→1− f (·)−f (λ·) ∞ The proof Corollary Let f ∈ BC(R → X), σ ∈ R+ Assume that Spec(f ) ⊂ [−σ, σ ] Then the sequence {σ −m D m f ∞ } is bounded Moreover, if limλ→1− f (·) − f (λ·) ∞ = then lim σ −m D m f m→∞ ∞ = Acknowledgments This research is funded by Vietnam National Foundation for Science and Technology Development (NAFOSTED) under grant number 101.01-2011.32 The authors would like to thank the referees for the exact corrections References Abreu, L.D.: Real Paley–Wiener theorems for the Koornwinder–Swarttouw q-Hankel transform J Math Anal Appl 334, 223–231 (2007) The Sequence of Norm of Derivatives (or Primitives) of Functions Albrecht, E., Ricker, W.J.: Functional calculi and decomposability of unbounded multiplier operators in Lp (RN ) Proc Edinb Math Soc Ser 38, 151–166 (1995) Albrecht, E., Ricker, 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L.D.: Real Paley–Wiener theorems for the Koornwinder–Swarttouw q-Hankel transform J Math Anal Appl 334, 223–231 (2007) The Sequence of Norm of Derivatives (or Primitives) of Functions Albrecht,...H H Bang, V N Huy the Theory of Banach algebras: Let A be a unital Banach algebra and x ∈ A Then the limit limm→∞ x m 1/m always exists and satisfies xm lim m→∞ = rA (x), 1/m where rA (x) =