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E. M. SEMENOV
ON A SCALE OF SPACES WITH THE INTERPOLATION PROPERTY
(Presented by Academician S. L. Sobolev on 13 VIII 1962)
In (¹) a maximal scale of spaces connecting the spaces \(L_1\) and \(L_\infty\) was constructed. These spaces were first introduced by G. G. Lorentz (²). In his notation the Banach space \(\Lambda(\alpha)\) \((0<\alpha<1)\) consists of all functions measurable on \([0,1]\) for which
\[ \|x\|_{\Lambda(\alpha)}=\alpha\int_0^1 x^*(t)t^{\alpha-1}\,dt<\infty, \tag{1} \]
where \(x^*(t)\) denotes the nonincreasing function equimeasurable with \(|x(t)|\). The spaces conjugate to the spaces \(\Lambda(\alpha)\) are the spaces \(M(\alpha)\), consisting of all functions measurable on \([0,1]\) for which
\[ \|x\|_{M(\alpha)}=\sup_{0<h\le1}\frac{\int_0^h x^*(t)\,dt}{h^\alpha}<\infty. \tag{2} \]
In (²) it is shown that the spaces \(\Lambda(\alpha)\) are nonreflexive. In the present note a scale of spaces \(M_0(\alpha)\) will be constructed, to which \(\Lambda(\alpha)\) are conjugate, and interpolation theorems for this scale will be proved.
- Let \(M_0(\alpha)\) denote the set of functions in \(M(\alpha)\) for which
\[ \lim_{h\to0}\frac{\int_0^h x^*(t)\,dt}{h^\alpha}=0. \tag{3} \]
It is easy to verify that every bounded function belongs to \(M_0(\alpha)\) and, moreover,
\[ M_0(\alpha)\subset L_{\frac{1}{1-\alpha}}. \tag{4} \]
Lemma 1. \(M_0(\alpha)\) coincides with the closure of the set of bounded functions in the norm of the space \(M(\alpha)\).
Proof. Let \(x_N(t)\) denote the truncation of the function \(x(t)\in M(\alpha)\). Then
\[ \|x-x_N\|_{M(\alpha)} = \sup_{0<h\le1} \frac{\int_0^h (x-x_N)^*\,dt}{h^\alpha} = \]
\[ = \sup_{0<h\le mE(|x|>N)} \frac{\int_0^h (x-N\operatorname{sign}x)^*\,dt}{h^\alpha} < \sup_{0<h\le mE(|x|>N)} \frac{\int_0^h x^*\,dt}{h^\alpha}. \]
Since \(\lim_{N\to\infty} mE(|x|>N)=0\), it follows by virtue of (3) that
\(\lim_{N\to\infty}\|x-x_N\|_{M(\alpha)}=0\).
On the other hand, if \(x(t)\in M(\alpha)\) and \(x(t)\notin M_0(\alpha)\), then there exists a sequence \(h_k\downarrow0\) such that
\[ \lim_{k\to\infty}\int_0^{h_k} x^*(t)\,dt \,/\, h_k^\alpha=\lambda>0. \]
Then for any \(y(t)\in M_0(\alpha)\)
\[ \|x-y\|_{M(\alpha)} = \sup_{0<h\le1} \frac{\int_0^h (x-y)^*\,dt}{h^\alpha} \ge \lim_{k\to\infty} \frac{\int_0^{h_k}(x-y)^*\,dt}{h_k^\alpha} \ge \]
\[ \geq \lim_{k\to\infty}\left( \frac{\displaystyle\int_0^{h_k} x^*(t)\,dt}{h_k^\alpha} - \frac{\displaystyle\int_0^{h_k} y^*(t)\,dt}{h_k^\alpha} \right) = \lim_{k\to\infty} \frac{\displaystyle\int_0^{h_k} x^*(t)\,dt}{h_k^\alpha} =\lambda . \]
The lemma is proved.
\(M_0(\alpha)\), being a closed subspace of \(M(\alpha)\), is a complete normed space. If \(\chi_E(t)\) is the characteristic function of a measurable set \(E\subset[0,1]\), then
\[ \|\chi_E(t)\|_{M(\alpha)} = \sup_{0<h\leq 1} \frac{\displaystyle\int_0^h \chi_E^*(t)\,dt}{h^\alpha} = \sup_{0<h\leq mE} \frac{\displaystyle\int_0^h 1\cdot dt}{h^\alpha} = mE^{1-\alpha}. \tag{5} \]
Theorem 1. The formula
\[ f(x)=\int_0^1 x(t)y(t)\,dt,\qquad \text{where } y(t)\in \Lambda(\alpha), \tag{6} \]
gives the general form of a linear functional on \(M_0(\alpha)\), and moreover \(\|f\|=\|y\|_{\Lambda(\alpha)}\).
Proof. As shown in \((^2)\),
\[ |f(x)|\leq \|x\|_{M(\alpha)}\|y\|_{\Lambda(\alpha)}, \]
therefore \(f(x)\) is defined on the whole space \(M_0(\alpha)\) and \(\|f\|\leq \|y\|_{\Lambda(\alpha)}\).
Let us prove that equality holds. It is clear that for every function \(y(t)\in\Lambda(\alpha)\) there exists a function \(\theta(t)\) such that \(\theta^*(t)=\alpha t^{\alpha-1}\),
\[ \int_0^1 \theta(t)y(t)\,dt = \int_0^1 \alpha t^{\alpha-1}y^*(t)\,dt. \]
Denote by \(\theta_N(t)\) the truncation of the function \(\theta(t)\) and
\[ a_N=(\alpha/N)^{1/(1-\alpha)}. \]
Then \(\|\theta_N\|_{M(\alpha)}\leq \|\theta\|_{M(\alpha)}=1\), \(\theta_N(t)\in M_0(\alpha)\), and
\[ f(\theta_N) = \int_0^1 \theta_N(t)y(t)\,dt = N\int_0^{a_N} y^*(t)\,dt + \int_{a_N}^1 \alpha t^{\alpha-1}y^*(t)\,dt \geq \]
\[ \geq \alpha\int_{a_N}^1 y^*(t)t^{\alpha-1}\,dt = \alpha\int_0^1 y^*(t)t^{\alpha-1}\,dt - \alpha\int_0^{a_N} y^*(t)t^{\alpha-1}\,dt. \]
Since the function \(y^*(t)t^{\alpha-1}\) is summable on \([0,1]\), it follows that
\[ \sup_N f(\theta_N)\geq \lim_{N\to\infty} \left( \alpha\int_0^1 y^*(t)t^{\alpha-1}\,dt - \alpha\int_0^{a_N} y^*(t)t^{\alpha-1}\,dt \right) = \|y\|_{\Lambda(\alpha)}. \]
We shall show that every linear functional on \(M_0(\alpha)\) is representable in the form (6). Let \(f(x)\) be an arbitrary linear functional on \(M_0(\alpha)\). Then, by virtue of (5),
\[ |f(\chi_E)|\leq \|f\|\,\|\chi_E\|_{M(\alpha)} = \|f\|\,mE^{1-\alpha}. \]
Therefore \(f(\chi_E)\), considered as a function of subsets of the interval \([0,1]\), is absolutely continuous and, by the Radon–Nikodym theorem, is representable in the form
\[ f(\chi_E)=\int_E y(t)\,dt, \tag{7} \]
where \(y(t)\) is a summable function on \([0,1]\). It follows from (7) that for every step function \(x(t)\) the equality (6) is valid.
Now let \(x(t)\in M(\alpha)\). One can construct a sequence of step functions \(z_N(t)\) such that \(z_N(t)\) converge to \(x(t)\) almost everywhere and \(|z_N(t)|\leq |x(t)|\). Then \(|z_N(t)y(t)|\) will converge almost everywhere to the function \(|x(t)y(t)|\), and the inequality
\[ \|z_N\|_{M(\alpha)}\leq \|x\|_{M(\alpha)}. \tag{8} \]
will obviously be satisfied.
Taking (8) into account, we have
\[ \sup_N \int_0^1 |z_N(t)y(t)|\,dt = \sup_N f\bigl(|z_N(t)|\operatorname{sign} y(t)\bigr) \le \|f\|\sup_N \|z_N\|_{M(\alpha)} \le \|f\|\|x\|_{M(\alpha)}<\infty . \]
By Fatou’s theorem,
\[ \left|\int_0^1 x(t)y(t)\,dt\right|\le \|f\|\|x\|_{M(\alpha)}. \]
In particular, when \(x(t)=\theta(t)\), then
\[ \int_0^1 x(t)y(t)\,dt = \int_0^1 \theta(t)y(t)\,dt = \int_0^1 \alpha t^{\alpha-1}y^*(t)\,dt = \|y\|_{\Lambda(\alpha)}. \]
Thus, \(y(t)\in \Lambda(\alpha)\). Using Lemma 1 and the continuity of the functional \(f(x)\), it is easy to show by passage to the limit that formula (6) holds for all \(x\in M_0(\alpha)\). The theorem is proved. Thus, \(\Lambda(\alpha)\) is conjugate to \(M_0(\alpha)\), and \(M(\alpha)\) is the second conjugate of \(M_0(\alpha)\).
Theorem 2. If \(x(t)\in M(\alpha)\), then
\[ d(x,M_0(\alpha))=\lim_{N\to\infty}\|x-x_N\|_{M(\alpha)}, \]
where \(d(x,M_0(\alpha))\) denotes the distance from the function \(x(t)\) to \(M_0(\alpha)\), and \(x_N(t)\) are the cutoffs of the function \(x(t)\).
Proof. It is evidently enough to show
\[ d(x,M_0(\alpha))\ge \lim_{N\to\infty}\|x-x_N\|_{M(\alpha)}. \]
Let \(u_N(t)\in M_0(\alpha)\) and
\[ d(x,M_0(\alpha))=\lim_{N\to\infty}\|x-u_N\|_{M(\alpha)}. \]
Since the set of bounded functions is dense in \(M_0(\alpha)\), we may assume that \(u_N(t)\) are bounded functions. Repeating each function the required number of times, one can arrange that
\[ |u_N(t)|\le N \tag{9} \]
for sufficiently large \(N\). If (9) is satisfied, then
\[ \|x-u_N\|_{M(\alpha)}\ge \|x-x_N\|_{M(\alpha)},\qquad d(x,M_0(\alpha))\ge \lim_{N\to\infty}\|x-x_N\|_{M(\alpha)}. \]
The theorem is proved.
2. In the well-known interpolation theorem of Marcinkiewicz (see \((^3)\)), functions satisfying the condition
\[ \sup_{0<\tau<\infty}\tau n_x^{\,1-\alpha}(\tau)<\infty, \tag{10} \]
are studied, where \(n_x(\tau)=mE(|x|>\tau)\).
The functional \(\sup_{0<\tau<\infty}\tau n_x^{\,1-\alpha}(\tau)\), defined on the set of functions for which (10) is satisfied, does not have the properties of a norm (the triangle inequality is not satisfied). However, the following assertion is true.
Lemma 2. The collection of functions satisfying the Marcinkiewicz condition (10) coincides with the space \(M(\alpha)\), and moreover
\[ \sup_{0<\tau<\infty}\tau n_x^{\,1-\alpha}(\tau) \le \|x\|_{M(\alpha)} \le \frac1\alpha \sup_{0<\tau<\infty}\tau n_x^{\,1-\alpha}(\tau). \]
Proof. If we put \(n_x(\tau)=t\), then by the definition of the rearrangement of a function (see \((^4)\), p. 332), \(\tau=x^*(t)\). Therefore it is enough to prove the inequality
\[ \sup_{0<t\le 1}x^*(t)t^{1-\alpha} \le \|x\|_{M(\alpha)} \le \frac1\alpha \sup_{0<t\le 1}x^*(t)t^{1-\alpha}. \]
If
\[ \frac{1}{\alpha}\sup_{0<t\leqslant 1} x^*(t)t^{1-\alpha}=C, \]
then \(x^*(t)\leqslant C\alpha t^{\alpha-1}\), and
\[ \|x\|_{M(\alpha)} =\sup_{0<h\leqslant 1} \frac{\displaystyle\int_0^h x^*(t)\,dt}{h^\alpha} \leqslant \sup_{0<h\leqslant 1} \frac{\displaystyle\int_0^h C\alpha t^{\alpha-1}\,dt}{h^\alpha} =C. \]
On the other hand, since \(x^*(t)\) is nonincreasing,
\[ \|x\|_{M(\alpha)} =\sup_{0<h\leqslant 1} \frac{\displaystyle\int_0^h x^*(t)\,dt}{h^\alpha} \geqslant \sup_{0<h\leqslant 1} \frac{h x^*(h)}{h^\alpha} = \sup_{0<t\leqslant 1} x^*(t)t^{1-\alpha}. \]
The lemma is proved.
The Marcinkiewicz theorem itself may be formulated as follows. Let \(1\leqslant p_k<q_k<\infty\) \((k=0,1)\), \(q_0\ne q_1\). If a linear operator \(T\) is a bounded operator from \(L_{p_k}\) to \(M(1-1/q_k)\) \((k=0,1)\), then \(T\) acts boundedly from \(L_{p_\tau}\) to \(L_{q_\tau}\), where
\[ \frac{1}{p_\tau}=\frac{1-\tau}{p_0}+\frac{\tau}{p_1},\qquad \frac{1}{q_\tau}=\frac{1-\tau}{q_0}+\frac{\tau}{q_1},\qquad 0<\tau<1. \]
Here \(M(0)\) is the set of functions \(x(t)\) for which \(\sup_{0<t\leqslant 1} x^*(t)t<\infty\).
Correspondingly, the interpolation theorems from \((^{1,5})\) can be reformulated.
Theorem 3. If \(T\) is a bounded operator from \(M(1-\alpha_k)\) to \(M(1-\beta_k)\) \((k=1,2)\), \(1>\alpha_k\geqslant\beta_k>0\), \(\alpha_1\ne\alpha_2\), \(\beta_1\ne\beta_2\), then \(T\) acts boundedly from \(M_0(1-\alpha_\tau)\) to \(M_0(1-\beta_\tau)\), where \(\alpha_\tau=(1-\tau)\alpha_1+\tau\alpha_2\), \(\beta_\tau=(1-\tau)\beta_1+\tau\beta_2\).
Proof. From interpolation theorem 1 in \((^1)\) it follows that \(T\) is a bounded operator from \(M(1-\alpha_\tau)\) to \(M(1-\beta_\tau)\), and
\[ \|Tx\|_{M(1-\beta_\tau)}\leqslant C_\tau\|x\|_{M(1-\alpha_\tau)}. \tag{11} \]
It remains to show that if \(x\in M_0(1-\alpha_\tau)\), then also \(Tx\in M_0(1-\beta_\tau)\). For any sufficiently small \(\varepsilon>0\) the inclusion
\[ M(1-\beta_\tau+\varepsilon)\subset L_{\frac{1}{1-\beta_\tau}} \]
holds. By virtue of (4), \(M(1-\beta_\tau+\varepsilon)\subset M_0(1-\beta_\tau)\). Let \(x(t)\) be bounded on \([0,1]\). If \(Tx\in M_0(1-\beta_\tau)\) for some \(0<\tau<1\), then, by the preceding, \(Tx\in M(1-\beta_\tau+\varepsilon)\), which contradicts (11). Thus \(Tx\in M_0(1-\beta_\tau)\) for all \(0<\tau<1\) and all \(x\in L_\infty\).
Every function \(x(t)\in M_0(1-\alpha_\tau)\) can be approximated by bounded functions \(u_N\) (in the norm of \(M(1-\alpha_\tau)\)). By virtue of (11), the sequence \(Tu_N\) is fundamental in \(M(1-\beta_\tau)\), and, as was shown, \(Tu_N\in M_0(1-\beta_\tau)\). From the closedness of \(M_0(1-\beta_\tau)\) in \(M(1-\beta_\tau)\) it follows that \(Tx\in M_0(1-\beta_\tau)\).
The theorem is proved.
Theorem 4. If \(T\) is a bounded operator from \(M_0(1-\alpha_k)\) to \(M_0(1-\beta_k)\) \((k=1,2)\), \(1>\alpha_k\geqslant\beta_k>0\), \(\alpha_1\ne\alpha_2\), \(\beta_1\ne\beta_2\), then \(T\) acts boundedly from \(M_0(1-\alpha_\tau)\) to \(M_0(1-\beta_\tau)\).
The proof follows from theorem 1 in \((^1)\) and theorem 3.
The author expresses his sincere gratitude to S. G. Krein for his guidance.
Received
7 VIII 1962
REFERENCES
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