Bessel-Type Operators and a Refinement of Hardy’s Inequality

Abstract

The principal aim of this paper is to employ Bessel-type operators in proving the inequality

$$\displaystyle \begin{aligned} \int _0^\pi dx \, |f'(x)|{ }^2 \geq \dfrac {1}{4}\int _0^\pi dx \, \dfrac {|f(x)|{ }^2}{\sin ^2 (x)}+\dfrac {1}{4}\int _0^\pi dx \, |f(x)|{ }^2,\quad f\in H_0^1 ((0,\pi )), \end{aligned} $$

where both constants 1∕4 appearing in the above inequality are optimal. In addition, this inequality is strict in the sense that equality holds if and only if f ≡ 0. This inequality is derived with the help of the exactly solvable, strongly singular, Dirichlet-type Schrödinger operator associated with the differential expression

$$\displaystyle \begin{aligned} \tau _s=-\dfrac {d^2}{dx^2}+\dfrac {s^2-(1/4)}{\sin ^2 (x)}, \quad s \in [0,\infty ), \; x \in (0,\pi ). \end{aligned} $$

The new inequality represents a refinement of Hardy’s classical inequality

$$\displaystyle \begin{aligned} \int _0^\pi dx \, |f'(x)|{ }^2 \geq \dfrac {1}{4}\int _0^\pi dx \, \dfrac {|f(x)|{ }^2}{x^2}, \quad f\in H_0^1 ((0,\pi )), \end{aligned} $$

and it also improves upon one of its well-known extensions in the form

$$\displaystyle \begin{aligned} \int _0^\pi dx \, |f'(x)|{ }^2 \geq \dfrac {1}{4}\int _0^\pi dx \, \dfrac {|f(x)|{ }^2}{d_{(0,\pi )}(x)^2}, \quad f\in H_0^1 ((0,\pi )), \end{aligned} $$

where d _(0,π)(x) represents the distance from x ∈ (0, π) to the boundary {0, π} of (0, π).

Dedicated with great pleasure to Lance Littlejohn on the occasion of his 70th birthday.

Appendices

A.1 The Weyl–Titchmarsh–Kodaira m-Function Associated with T _s,F

We start by introducing a normalized fundamental system of solutions ϕ _0,s(z, ⋅) and θ _0,s(z, ⋅) of τ _s u = zu, s ∈ [0, 1), $z \in {\mathbb {C}}$, satisfying (cf. the generalized boundary values introduced in (2.14), (2.15))

$$\displaystyle \begin{aligned} \widetilde \theta_{0,s}(z,0)=1,\quad \widetilde \theta^{\, \prime}_{0,s}(z,0)=0, \quad \widetilde \phi_{0,s}(z,0)=0,\quad \widetilde \phi^{\, \prime}_{0,s}(z,0)=1, \end{aligned} $$

(A.1)

with ϕ _0,s(⋅ , x) and θ _0,s(⋅ , x) entire for fixed x ∈ (0, π). To this end, we introduce the two linearly independent solutions to τ _s y = zy (entire w.r.t. z for fixed x ∈ (0, π)) given by

(A.2)

Using the connection formula found in [1, Eq. 15.3.6] yields the behavior near x = 0, π,

(A.3)

Remark A.1.1

Before we turn to the case s = 0, we recall Gauss’s identity (cf. [1, no. 15.1.20])

$$\displaystyle \begin{aligned} F(a,b;c;1) = \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)}, \quad c \in{\mathbb{C}} \backslash \{ - {\mathbb{N}}_0\}, \; \text{Re}(c-a-b) > 0, {} \end{aligned} $$

(A.4)

and the differentiation formula (cf. [1, no. 15.2.1])

$$\displaystyle \begin{aligned} \frac{d}{dz} F(a,b;c;z) = \frac{ab}{c} F(a+1,b+1;c+1;z), \quad a, b, c \in {\mathbb{C}}, \; z \in \{\zeta \in {\mathbb{C}} \,|\, |\zeta| < 1\}, {} \end{aligned} $$

(A.5)

which imply that for s ∈ (0, 1), the two F(⋅ ,  ⋅ ;  ⋅ ;1) exist in (A.2) (indeed, for j = 1, 2 one obtains from (A.2) with s ∈ (0, 1) that c − a − b = s > 0) and hence the asymptotic behavior of y _j,s(z, x), j = 1, 2, as x ↓ 0 and x ↑ π is dominated by x ^(1−2s)∕2 and (π − x)^(1−2s)∕2, respectively. However, the analogous statement fails for $y_{j,s}^{\prime }(z,x)$, j = 1, 2, as, taking into account (A.5), the analog of the 2nd condition in (A.4), namely, Re[c + 1 − (a + 1) − (b + 1)] > 0, is not fulfilled (in this case (A.2) with s ∈ (0, 1) yields [c + 1 − (a + 1) − (b + 1)] = s − 1 < 0). The situation is similar for the first two F(⋅ ,  ⋅ ;  ⋅ ;x) for y _1,s(z, x) in (A.3) as x → π∕2 as in this case the two F(⋅ ,  ⋅ ;  ⋅ ;1) exist. Even though for y _2,s(z, x) in (A.3) the two F(⋅ ,  ⋅ ;  ⋅ ;1) do not exist individually, the limit of each term does exist due to the multiplication by the factor $\cos {}(x)$. To see this, one can instead consider the limit (cf. [51, no. 15.4.23])

$$\displaystyle \begin{aligned} \lim_{z\to1^-}\dfrac{F(a,b;c;z)}{(1-z)^{c-a-b}}=\dfrac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)},\quad c \in{\mathbb{C}} \backslash \{ - {\mathbb{N}}_0\}, \; \text{Re}(c-a-b) < 0, \end{aligned} $$

(A.6)

which through the appropriate change of variable reveals that the connection formula for y _2,s(z, x) as x → π∕2 approaches 0 as expected from evaluating y _2,s(z, π∕2) in (A.2). But once again, the analog of the 2nd condition in (A.4), namely, Re[c + 1 − (a + 1) − (b + 1)] > 0, fails for the four F′(⋅ ,  ⋅ ;  ⋅ ;x) in (A.3) as x → π∕2. ◇

Similarly, by Abramowitz and Stegun [1, Eq. 15.3.10] one obtains for the remaining case s = 0,

$$\displaystyle \begin{aligned} &y_{1,0}(z,x)=\dfrac{\pi^{1/2}[\sin (x)]^{1/2}}{\Gamma\big(\big[(1/2) + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(1/2) - z^{1/2}\big]\big/2\big)} \notag \\ & \hspace{1.7cm} \times \sum_{n=0}^\infty \dfrac{\big(\big[(1/2) + z^{1/2}\big]\big/2\big)_n\big(\big[(1/2) - z^{1/2}\big]\big/2\big)_n}{(n!)^2} \big[2\psi(n+1) \notag \\ &\hspace{2cm} -\psi\big(n+\big[(1/2) + z^{1/2}\big]\big/2\big)-\psi\big(n+\big[(1/2) - z^{1/2}\big]\big/2\big) \notag \\ & \hspace{2cm} -\text{ln}(\sin^2 (x))\big] [\sin (x)]^{2n}, {} \\ {} &y_{2,0}(z,x)=\dfrac{\pi^{1/2}\cos (x) [\sin (x)]^{1/2}}{2\Gamma\big(\big[(3/2) + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(3/2) - z^{1/2}\big]\big/2\big)} \notag \\ & \hspace{1.7cm} \times \sum_{n=0}^\infty \dfrac{\big(\big[(3/2) + z^{1/2}\big]\big/2\big)_n\big(\big[(3/2) - z^{1/2}\big]\big/2\big)_n}{(n!)^2} \big[2\psi(n+1) \notag \\ & \hspace{2cm} -\psi\big(n+\big[(3/2) + z^{1/2}\big]\big/2\big)-\psi\big(n+\big[(3/2) - z^{1/2}\big]\big/2\big) \notag \\ & \hspace{2cm} - \text{ln}\big(\sin^2 (x)\big)\big][\sin (x)]^{2n}, \notag \\ {} & \hspace{2.15cm} s=0,\; z\in{\mathbb{C}},\; x\in(0,\pi). \notag \end{aligned} $$

(A.7)

Here ψ(⋅) =  Γ′(⋅)∕ Γ(⋅) denotes the Digamma function, γ _E = −ψ(1) = 0.57721… represents Euler’s constant, and

$$\displaystyle \begin{aligned} (\zeta)_0 =1, \quad (\zeta)_n = \Gamma(\zeta + n)/\Gamma(\zeta), \; n \in {\mathbb{N}}, \quad \zeta \in {\mathbb{C}} \backslash (-{\mathbb{N}}_0), {} \end{aligned} $$

(A.8)

abbreviates Pochhammer’s symbol (see, e.g., [1, Ch. 6]). Direct computation now yields

$$\displaystyle \begin{aligned} \widetilde y_{1,s}(z,0)&=-\widetilde y_{1,s}(z,\pi) \notag \\ &=\dfrac{2\pi^{1/2}\Gamma(1+s)}{\Gamma\big(\big[(1/2) + s + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(1/2) + s - z^{1/2}\big]\big/2\big)}, \notag \\ {} \widetilde y^{\, \prime}_{1,s}(z,0)&=\widetilde y^{\, \prime}_{1,s}(z,\pi)=\dfrac{\pi^{1/2}\Gamma(-s)}{\Gamma\big(\big[(1/2) - s + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(1/2) - s - z^{1/2}\big]\big/2\big)}, \notag \\ {} \widetilde y_{2,s}(z,0)&=\widetilde y_{2,s}(z,\pi)=\dfrac{\pi^{1/2}\Gamma(1+s)}{\Gamma\big(\big[(3/2) + s + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(3/2) + s - z^{1/2}\big]\big/2\big)}, \notag \\ {} \widetilde y^{\, \prime}_{2,s}(z,0)&=-\widetilde y^{\, \prime}_{2,s}(z,\pi) \notag \\ &=\dfrac{\pi^{1/2}\Gamma(-s)}{2\Gamma\big(\big[(3/2) - s + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(3/2) - s - z^{1/2}\big]\big/2\big)}, \notag \\ {} &\hspace{5.7cm} s\in(0,1),\, z\in{\mathbb{C}}, {} \end{aligned} $$

(A.9)

$$\displaystyle \begin{aligned} \widetilde y_{1,0}(z,0)&=-\widetilde y_{1,0}(z,\pi)=\dfrac{2\pi^{1/2}}{\Gamma\big(\big[(1/2) + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(1/2) - z^{1/2}\big]\big/2\big)}, \notag \\ {} \widetilde y^{\, \prime}_{1,0}(z,0)&=\widetilde y^{\, \prime}_{1,0}(z,\pi) \notag \\ & =\dfrac{-\pi^{1/2}\big[2\gamma_E+\psi\big(\big[(1/2) + z^{1/2}\big]\big/2\big)+\psi\big(\big[(1/2) - z^{1/2}\big]\big/2\big)\big]}{\Gamma\big(\big[(1/2) + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(1/2) - z^{1/2}\big]\big/2\big)}, \notag \\ {} \widetilde y_{2,0}(z,0)&=\widetilde y_{2,0}(z,\pi)=\dfrac{\pi^{1/2}}{\Gamma\big(\big[(3/2) + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(3/2) - z^{1/2}\big]\big/2\big)}, \notag \\ {} \widetilde y^{\, \prime}_{2,0}(z,0)&=-\widetilde y^{\, \prime}_{2,0}(z,\pi) \notag \\ & =\dfrac{-\pi^{1/2}\big[2\gamma_E+\psi\big(\big[(3/2) + z^{1/2}\big]\big/2\big)+\psi\big(\big[(3/2) - z^{1/2}\big]\big/2\big)\big]}{2\Gamma\big(\big[(3/2) + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(3/2) - z^{1/2}\big]\big/2\big)},\notag \\ {} &\hspace{6.7cm} s=0,\, z\in{\mathbb{C}}. {} \end{aligned} $$

(A.10)

In particular, one obtains

(A.11)

since

$$\displaystyle \begin{aligned} W(y_{1,s}(z,\,\cdot), y_{2,s}(z,\,\cdot)) = \widetilde y_{1,s}(z,0)\widetilde y^{\, \prime}_{2,s}(z,0) - \widetilde y^{\, \prime}_{1,s}(z,0)\widetilde y_{2,s}(z,0) = - 1, {} \end{aligned} $$

(A.12)

with the generalized boundary values given by (A.9), (A.10). To prove (A.12) one recalls Euler’s reflection formula (cf. [1, no. 6.1.17])

$$\displaystyle \begin{aligned} \Gamma(z)\Gamma(1-z)=\dfrac{\pi}{\sin{}(\pi z)}, \quad z \in {\mathbb{C}} \backslash {\mathbb{Z}}, \end{aligned} $$

(A.13)

and hence concludes that

$$\displaystyle \begin{aligned} & \Gamma\big(\big[(1/2) + \varepsilon s \pm z^{1/2}\big]\big/2\big)\Gamma\big(\big[(3/2) - \varepsilon s \mp z^{1/2}\big]\big/2\big) \\ & \quad =\dfrac{\pi}{\sin\big(\pi\big[(1/2) + \varepsilon \pm z^{1/2}\big]\big/2\big)}, \quad \varepsilon \in \{-1,1\}. \end{aligned} $$

(A.14)

Thus one computes for s ∈ (0, 1),

$$\displaystyle \begin{aligned} & W(y_{1,s}(z,\,\cdot), y_{2,s}(z,\,\cdot)) \notag \\ & \quad =-[\sin{}(\pi s)]^{-1}\big\{\sin\big(\pi\big[(1/2)+s + z^{1/2}\big]\big/2\big)\sin\big(\pi\big[(1/2)+s - z^{1/2}\big]\big/2\big) \notag \\ & \quad \hspace{2.05cm}-\sin\big(\pi\big[(1/2)-s + z^{1/2}\big]\big/2\big)\sin\big(\pi\big[(1/2)-s - z^{1/2}\big]\big/2\big) \big\} \notag\\ & \quad =-[2\sin{}(\pi s)]^{-1}\{-\cos{}(\pi[(1/2)+s ])+\cos{}(\pi[(1/2)-s ]) \} = -1. \end{aligned} $$

(A.15)

For the case s = 0, one recalls the reflection formula for the Digamma function (cf. [1, no. 6.3.7])

$$\displaystyle \begin{aligned} \psi(1-z)-\psi(z)=\pi\cot{}(\pi z), \quad z \in {\mathbb{C}} \backslash {\mathbb{Z}}, \end{aligned} $$

(A.16)

and applies trigonometric identities to obtain W(y _1,0(z, ⋅), y _2,0(z, ⋅)) = −1. The singular Weyl–Titchmarsh–Kodaira function m _0,0,s(z) is then uniquely determined (cf. [30, Eq. (3.18)] and [29] for background on m-functions) to be²

$$\displaystyle \begin{aligned} m_{0,0,s}(z)= - \frac{\widetilde\theta_{0,s} (z,\pi)}{\widetilde\phi_{0,s} (z,\pi)}, \quad s\in[0,1),\; z\in\rho(T_{s,F}). \end{aligned} $$

(A.17)

Direct calculation once again yields

$$\displaystyle \begin{aligned} m_{0,0}(z)&= -\dfrac{\widetilde y^{\, \prime}_{2,s}(z,0)\widetilde y_{1,s}(z,\pi)-\widetilde y^{\, \prime}_{1,s}(z,0)\widetilde y_{2,s}(z,\pi)}{2\widetilde y_{1,s}(z,0)\widetilde y_{2,s}(z,0)} \notag \\ {} &=\begin{cases} \dfrac{\pi\Gamma(-s)}{4\Gamma(1+s)}\bigg[\dfrac{\Gamma\big(\big[(3/2) + s + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(3/2) + s - z^{1/2}\big]\big/2\big)}{\Gamma\big(\big[(3/2) - s + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(3/2) - s - z^{1/2}\big]\big/2\big)} \\ {} \qquad \quad \ \ +\dfrac{\Gamma\big(\big[(1/2) + s + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(1/2) + s - z^{1/2}\big]\big/2\big)}{\Gamma\big(\big[(1/2) - s + z^{1/2}\big]\big/2\big)\Gamma\big(\big[(1/2) - s - z^{1/2}\big]\big/2\big)}\bigg], \\ \hspace{7.55cm} s\in(0,1),\\ {} -\big[4\gamma_E+\psi\big(\big[(1/2) + z^{1/2}\big]\big/2\big)+\psi\big(\big[(1/2) - z^{1/2}\big]\big/2\big) \\ {} \quad \, +\psi\big(\big[(3/2) + z^{1/2}\big]\big/2\big)+\psi\big(\big[(3/2) - z^{1/2}\big]\big/2\big)\big]/4, \quad s=0, \end{cases} \notag \\ &\hspace{7.3cm} z\in\rho(T_{s,F}), \end{aligned} $$

(A.18)

which has simple poles precisely at the simple eigenvalues of T _s,F given by

$$\displaystyle \begin{aligned} \sigma(T_{s,F})=\big\{[(1/2)+s+n]^2\big\}_{n\in{\mathbb{N}}_0},\quad s\in[0,1). \end{aligned} $$

(A.19)

Remark A.1.2

For the limit point case at both endpoints, that is, for s ∈ [1, ∞), the solutions y _j,s(z, ⋅) in (A.2) remain linearly independent and also the connection formulas (A.3) remain valid for $s \in [1,\infty ) \backslash {\mathbb {N}}$. Moreover, employing once again (A.4) and (A.5) one verifies that the two F(⋅ ,  ⋅ ;  ⋅ ;1) as well as F′(⋅ ,  ⋅ ;  ⋅ ;1) are well defined in (A.2) and hence for s ∈ [1, ∞), the asymptotic behavior of y _j,s(z, x) and $y_{j,s}^{\prime }(z,x)$, j = 1, 2, as x ↓ 0 and x ↑ π is dominated by x ^(1−2s)∕2 and x ^{−(1+2s)∕2} and (π − x)^(1−2s)∕2 and (π − x)^{−(1+2s)∕2}, respectively. Since in connection with (A.3) one has c − a − b = ±1∕2, independently of the value of s ∈ (0, ∞), the situation described in Remark A.1.1 for (A.3) and s ∈ (0, 1) applies without change to the current case s ∈ [1, ∞).

Actually, some of these failures (as x → π∕2 in $y_{j,s}^{\prime }(z,x)$, j = 1, 2) are crucial for the following elementary reason: The function

(A.20)

(i.e., the analog of the second part of y _1,s(z, ⋅) on the right-hand side in (A.3)) generates an L ²((0, π);dx)-element near x = 0, π, and hence if this function and its x-derivative were locally absolutely continuous in a neighborhood of x = π∕2 (the only possibly nontrivial point in the interval (0, π)), the self-adjoint maximal operator T _s,max, s ∈ [1, ∞), would have eigenvalues for all $z \in {\mathbb {C}}$, an obvious contradiction. ◇

Because of the subtlety pointed out in Remark A.1.2 we omit further details on the limit point case s ∈ [1, ∞) and refer to [23, Sect. 4], instead. In particular, [23, Theorem 4.1 b)] extends (A.19) to s ∈ [1, ∞) and hence one actually has

$$\displaystyle \begin{aligned} \sigma(T_{s,F})=\big\{[(1/2)+s+n]^2\big\}_{n\in{\mathbb{N}}_0}, \quad s\in[0,\infty). \end{aligned} $$

(A.21)

B.1 Remarks on Hardy-Type Inequalities

In this appendix we recall a Hardy-type inequality useful in Sect. 2.

Introducing the differential expressions α _s, $\alpha ^+_s$ (cf. (3.28) for s = 0),

$$\displaystyle \begin{aligned} \alpha_s = \frac{d}{dx} - \frac{s + (1/2)}{x}, \quad \alpha_s^+ = - \frac{d}{dx} - \frac{s + (1/2)}{x}, \quad s \in [0,\infty), \; x \in (0,\pi), {} \end{aligned} $$

(B.1)

one confirms that

$$\displaystyle \begin{aligned} \alpha_s^+ \alpha_s = \omega_s = - \frac{d^2}{dx^2} + \frac{s^2 - (1/4)}{x^2}, \quad s \in [0,\infty), \; x \in (0,\pi). {} \end{aligned} $$

(B.2)

Following the Hardy inequality considerations in [24, 38, 40], one obtains the following basic facts.

Lemma B.1.1

Suppose f ∈ AC _loc((0, π)), α _s f ∈ L ²((0, π);dx) for some $s \in {\mathbb {R}}$ , and 0 < r ₀ < r ₁ < π < R < ∞. Then,

$$\displaystyle \begin{aligned} & \int_{r_0}^{r_1} dx \, |(\alpha_s f)(x)|{}^2 \geq s^2 \int_{r_0}^{r_1} dx \, \frac{|f(x)|{}^2}{x^2} + \frac{1}{4} \int_{r_0}^{r_1} dx \, \frac{|f(x)|{}^2}{x^2 [\mathit{\text{ln}}(R/x)]^2} {} \\ & \hspace{3.25cm} - s \frac{|f(x)|{}^2}{x}\bigg|{}_{x = r_0}^{r_1} - \frac{|f(x)|{}^2}{2 x [\mathit{\text{ln}}(R/x)]}\bigg|{}_{x = r_0}^{r_1}, \end{aligned} $$

(B.3)

(B.4)

$$\displaystyle \begin{aligned} & \int_{r_0}^{r_1} dx \, |(\alpha_s f)(x)|{}^2 = \int_{r_0}^{r_1} dx \, \bigg[|f'(x)|{}^2 + \big[s^2 - (1/4)\big] \frac{|f(x)|{}^2}{x^2}\bigg] {} \\ & \hspace{3.25cm} - [s + (1/2)] \frac{|f(x)|{}^2}{x}\bigg|{}_{x = r_0}^{r_1} \geq 0. \end{aligned} $$

(B.5)

If s = 0,

$$\displaystyle \begin{aligned} \int_0^{r_1} dx \, \frac{|f(x)|{}^2}{x^2 [\mathit{\text{ln}}(R/x)]^2} < \infty, \quad \lim_{x \downarrow 0} \frac{|f(x)|}{[x \mathit{\text{ln}}(R/x)]^{1/2}} = 0. {} \end{aligned} $$

(B.6)

If s ∈ (0, ∞), then

$$\displaystyle \begin{aligned} \int_0^{r_1} dx \, |f'(x)|{}^2 < \infty, \quad \int_0^{r_1} dx \, \frac{|f(x)|{}^2}{x^2} < \infty, \quad \lim_{x \downarrow 0} \frac{|f(x)|}{x^{1/2}} = 0, {} \end{aligned} $$

(B.7)

in particular,

$$\displaystyle \begin{aligned} f \widetilde \chi_{[0,r_1/2]} \in H^1_0((0,r_1)), \end{aligned} $$

(B.8)

where

$$\displaystyle \begin{aligned} \widetilde \chi_{[0,r/2]} (x) = \begin{cases} 1, & x \in [0,r/4], \\ 0, & x \in [3r/4,r], \end{cases} \quad \widetilde \chi_{[0,r/2]} \in C^{\infty}([0,r]), \; r \in (0,\infty). \end{aligned} $$

(B.9)

Proof

Relations (B.4) and (B.5) are straightforward (yet somewhat tedious) identities; together they yield (B.3). The 1st relation in (B.6) is an instant consequence of (B.3), so is the fact that lim_x↓0|f(x)|²∕[xln(R∕x)] exists. Moreover, since [xln(R∕x)]⁻¹ is not integrable at x = 0, the 1st relation in (B.6) yields liminf_x↓0|f(x)|²∕[xln(R∕x)] = 0, implying the 2nd relation in (B.6).

Finally, if s ∈ (0, ∞), then inequality (B.3) implies the 2nd relation in (B.7); together with α _s f ∈ L ²((0, π);dx), this yields the 1st relation in (B.7). By inequality (B.3), lim_x↓0|f(x)|²∕x exists, but then the second relation in (B.7) yields liminf_x↓0|f(x)|²∕x = 0 and hence also lim_x↓0|f(x)|²∕x = 0. □

We also recall the following elementary fact.

Lemma B.1.2

Suppose f ∈ H ¹((0, r)) for some r ∈ (0, ∞). Then, for all x ∈ (0, r),

(B.10)

Thus, if f ∈ H ¹((0, r)), then $\int _0^r dx \, |f(x)|{ }^2/x^2 < \infty$ if and only if f(0) = 0, that is, if and only if $f \widetilde \chi _{[0,r/2]} \in H^1_0((0,r))$.

In particular, if f ∈ H ¹((0, r)) and f(0) = 0, then actually,

$$\displaystyle \begin{aligned} \lim_{x \downarrow 0} \frac{|f(x)|}{x^{1/2}} = 0. {} \end{aligned} $$

(B.11)

Proof

Since (B.10) is obvious, we briefly discuss the remaining assertions in Lemma B.2. If f ∈ H ¹((0, r)) and $\int _0^r dx \, |f(x)|{ }^2/x^2 < \infty$, then identity (B.5) for s < −1∕2, that is,

$$\displaystyle \begin{aligned} & \int_{r_0}^{r_1} dx \, |(\alpha_s f)(x)|{}^2 = \int_{r_0}^{r_1} dx \, \bigg[|f'(x)|{}^2 + \big[s^2 - (1/4)\big] \frac{|f(x)|{}^2}{x^2}\bigg] {} \\ & \hspace{3.25cm} - [s + (1/2)] \frac{|f(x)|{}^2}{x}\bigg|{}_{x = r_0}^{r_1} \geq 0, \quad s < - 1/2, \end{aligned} $$

(B.12)

yields the existence of lim_x↓0|f(x)|²∕x. Since $\int _0^r dx \, |f(x)|{ }^2/x^2 < \infty$ implies that liminf_x↓0|f(x)|²∕x = 0, one concludes that lim_x↓0|f(x)|²∕x = 0 and hence f behaves locally like an $H^1_0$-function in a right neighborhood of x = 0. Conversely, if f(0) = 0, then $\int _0^r dx \, |f(x)|{ }^2/x^2 < \infty$ by Hardy’s inequality as discussed in Remark 3.4. Relation (B.11) is clear from (B.10) with f(0) = 0. □

Remark B.1.3

(i) If f ∈ AC _loc((0, r)) and f′∈ L ^p((0, r);dx) for some p ∈ [1, ∞), the Hölder estimate analogous to (B.10),

$$\displaystyle \begin{aligned} |f(d) - f(c)| = \bigg|\int_c^d dt \, f'(t) \bigg| \leq |d-c|{}^{1/p'} \bigg(\int_c^d dt \, |f'(t)|{}^p\bigg)^{1/p}, \\ (c,d) \subset (0,r), \; \frac{1}{p} + \frac{1}{p'} = 1, \end{aligned} $$

(B.13)

implies the existence of lim_c↓0 f(c) = f(0) and lim_d↑r f(d) = f(r) and hence yields f ∈ AC([0, r]).

(ii) The fact that f ∈ H ¹((0, r)) and $\int _0^r dx \, |f(x)|{ }^2/x^2 < \infty$ implies $f \widetilde \chi _{[0,r/2]} \in H^1_0((0,r))$ is a special case of a multi-dimensional result recorded, for instance, in [19, Theorem 5.3.4].

(iii) When replacing x ⁻², x ∈ (0, r), by $[\sin {}(x)]^{-2}$, x ∈ (0, π), due to locality, the considerations in Lemmas 5.1 and 5.2 at the left endpoint x = 0 apply of course to the right interval endpoint π. ◇