Mathematics

G. M. Khenkin

On the embedding of the space of \(s\)-smooth functions of \(n\) variables into a space of sufficiently smooth functions of a smaller number of variables

(Presented by Academician A. N. Kolmogorov on 14 V 1963)

In the present note we consider the question of the existence of an isomorphic embedding of the space of continuous functions of \(n\) variables into a space of continuous functions of a smaller number of variables, under which functions of fixed smoothness from one space are mapped to sufficiently smooth functions from the other space. (By an isomorphism of a Banach space \(E_1\) into a Banach space \(E_2\) we mean a one-to-one continuous linear mapping of the space \(E_1\) onto some closed linear subspace of \(E_2\).)

Let \(I^n\) be the \(n\)-dimensional cube in the \(n\)-dimensional Euclidean space \(R^n\), defined by the inequalities \(|x_i|\leq 1\) \((i=1,2,\ldots,n)\). Denote by \(C(I^n)\) the space of all continuous real-valued (or complex-valued) functions defined on the cube \(I^n\), with norm
\[ \|f(x)\|=\sup_{x\in I^n}|f(x)|. \]
By \(C^{(s)}(I^n)\) we denote the space of all \(s\)-times continuously differentiable real-valued (or complex-valued) functions defined on the cube \(I^n\), with norm
\[ \|f(x)\|_s= \sum_{k_1+k_2+\cdots+k_n\leq s} \sup_{x\in I^n} \left| \frac{\partial^{k_1+k_2+\cdots+k_n} f(x)} {\partial x_1^{k_1}\partial x_2^{k_2}\cdots \partial x_n^{k_n}} \right|. \]

Let \(n>m\). Our main theorem is the following.

Theorem 1. If
\[ s>\left\langle \frac{n}{m}\right\rangle \left(1+\frac12\left\langle \frac{n}{m}\right\rangle\right)p, \]
then there exists an isomorphism
\[ T:C(I^n)\to C(I^m) \]
having the property
\[ T\bigl[C^{(s)}(I^n)\bigr]\subset C^{(p)}(I^m) \]
(by \(\left\langle n/m\right\rangle\) is denoted the integer nearest to \(n/m\) from the right).

For convenience of notation we shall assume that the functions take complex values. We shall need one assertion based on Whitney’s results. Denote by \(C_0^{(s)}(I_\pi^n)\) the space of all \(s\)-times continuously differentiable functions in \(R^n\) that vanish outside the cube \(I^n\), defined by the inequalities \(|x_i|\leq \pi\) \((i=1,2,\ldots,n)\).

Lemma 1. There exists a linear continuous operator
\[ M:C(I^n)\to C(I_\pi^n) \]
having the following properties: a)
\[ (Mf)(x)\equiv f(x),\qquad x\in I^n \]
for all \(f\in C(I^n)\); b)
\[ M\bigl[C^{(s)}(I^n)\bigr]\subset C_0^{(s)}(I_\pi^n). \]

The proof of Lemma 1, formulated in another form, can be found in \((^2)\).

Let \(f(x)\in C(I^n)\) and let \(M\) be the extension operator indicated in Lemma 1. Expand the function \((Mf)(x)\) in a Fourier series:
\[ (Mf)(x)\sim \sum_{\nu_1,\ldots,\nu_n=-\infty}^{\infty} c_{\nu_1,\ldots,\nu_n}(f)e^{i[\nu_1x_1+\cdots+\nu_nx_n]}, \qquad \text{where } x=(x_1,x_2,\ldots,x_n)\in I_\pi^n. \]

Lemma 2. There exists a constant \(\lambda_1\) such that

\[ \left(\sum_{\nu_1,\ldots,\nu_n=-\infty}^{\infty} c_{\nu_1,\ldots,\nu_n}^{\,2}(f)\right)^{1/2} \leq \lambda_1 \|f(x)\|, \quad \text{for all } f(x)\in C(I^n). \]

To prove Lemma 2 one must use the continuity of the operator \(M\).

Number the totality of all sets \(\{\nu_1,\nu_2,\ldots,\nu_n\}\). The set \(\{0,0,\ldots,0\}\) receives number 1. Further, if all sets satisfying the condition \(\max_{1\leq i\leq n}|\nu_i|<\mu\), where \(\mu\) is a natural number, have been numbered, we number the sets satisfying the condition \(\max_{1\leq i\leq n}|\nu_i|=\mu\). The set that has received number \(k\) will be denoted by \(\{\nu_1(k),\ldots,\nu_n(k)\}\). Put
\[ c_k(f)=c_{\nu_1(k),\ldots,\nu_n(k)}(f). \]

Lemma 3. If \(f(x)\in C^{(s)}(I^n)\), then the sequence \(\{k^{s/n}c_k(f)\}=\{\gamma_k\}\in l_2\), i.e.
\[ \sum_{k=1}^{\infty}\gamma_k^2<\infty . \]

Proof. By Lemma 1, \((Mf)(x)\in C_0^{(s)}(I_\pi^n)\). Hence, as is known, it follows that the sequence
\[ \left\{\left[\sum_{i=1}^{n}\nu_i^s(k)\right]c_k(f)\right\}\in l_2. \]
By the numbering, for any number \(k\) one of the numbers \(\nu_i(k)\) \((i=1,2,\ldots,n)\) is not less than \(\tfrac12(k^{1/n}-1)\). Therefore the sequence \(\{k^{s/n}c_k(f)\}\in l_2\).

The following assertion is known:

Lemma 4. There exists a continuous mapping of the segment \(I\) onto the cube \(I^n\),
\[ x_i=x_i(t)\quad (i=1,2,\ldots,n), \]
such that the functions \(x_i(t)\) satisfy the Hölder condition with exponent \(1/n\) and constant \(\lambda\):
\[ |x_i(t+\delta)-x_i(\delta)|\leq \lambda \delta^{1/n} \quad (i=1,2,\ldots,n). \]

Lemma 5. Put
\[ \varphi_k(t)=e^{\,i[\nu_1(k)x_1(t)+\cdots+\nu_n(k)x_n(t)]}, \]
where \(x_i(t)\) \((i=1,2,\ldots,n)\) is the mapping indicated in Lemma 4. Then
\[ |\varphi_k(t+\delta)-\varphi_k(t)| \leq n\lambda (k\delta)^{1/n}, \quad t\in I \quad (k=1,2,\ldots). \]

Proof.
\[ |\varphi_k(t+\delta)-\varphi_k(t)| \leq [\nu_1(k)+\cdots+\nu_n(k)]\lambda\delta^{1/n} \leq nk^{1/n}\lambda\delta^{1/n}, \]
where the first estimate is obtained by using Lemma 4, and the second estimate follows from the fact that, by the numbering, for any number \(k\) we have
\[ |\nu_i(k)|^n\leq k \quad (i=1,2,\ldots,n). \]

We smooth the functions \(\varphi_k(t)\) by averaging:
\[ \widetilde{\varphi}_k(t)= \frac{1}{(2\delta_k)^p} \int_{|t-t_p|\leq \delta_k} dt_p \int_{|t_p-t_{p-1}|\leq \delta_k} dt_{p-1}\cdots \int_{|t_2-t_1|\leq \delta_k}\varphi_k(t_1)\,dt_1 . \tag{1} \]

Lemma 6. The inequality holds
\[ \|\varphi_k(t)-\widetilde{\varphi}_k(t)\| \leq n\lambda (pk\delta_k)^{1/n} \quad (k=1,2,\ldots). \]

Lemma 7. The inequality is valid
\[ \|\varphi_k^{(p)}(t)\| \leq \frac{n\lambda}{2}\, \frac{(2k\delta_k)^{1/n}}{\delta_k^p} \quad (k=1,2,\ldots). \]

Proof. Lemma 6 follows directly from Lemma 5. To prove Lemma 7 one must differentiate expression (1) \(p\) times and use Lemma 5.

We shall present the proof of Theorem 1 in the case where \(m=1\) (the case of arbitrary \(m\) causes no additional difficulties). The desired isomor-

the morphism \(T: C(I^n)\to C(I)\) is constructed by the formula

\[ T[f(x_1,x_2,\ldots,x_n)] = \]

\[ = f(x_1(t),x_2(t),\ldots,x_n(t))+ \sum_{k=1}^{\infty} c_k(f)\,[\widetilde{\varphi}_k(t)-\varphi_k(t)]. \tag{2} \]

In order that the series on the right-hand side of (2) converge uniformly, it is sufficient that the inequality

\[ \sum_{k=1}^{\infty}\|\widetilde{\varphi}_k(t)-\varphi_k(t)\|^2<\infty \]

hold. To this end we fix \(0<\varepsilon<1\) and put \(\delta_k=\lambda_2 k^{-s/np}\), where the constant \(\lambda_2\) is chosen from the condition

\[ \sum_{k=1}^{\infty}\|\widetilde{\varphi}_k(t)-\varphi_k(t)\|^2< \left(\frac{\varepsilon}{\lambda_1}\right)^2 . \tag{3} \]

This can be done, since, by Lemma 6,

\[ \sum_{k=1}^{\infty}\|\widetilde{\varphi}_k(t)-\varphi_k(t)\|^2 \le p^{2/n}(n\lambda)^2\sum_{k=1}^{\infty}(k\delta_k)^{2/n} = (n\lambda)^2(\lambda_2 p)^{2/n} \sum_{k=1}^{\infty} k^{2/n-2s/(n^2p)}<\infty . \tag{4} \]

The last inequality is obvious if one recalls that \(s>(n+n^2/2)p\). Using the Cauchy–Bunyakovsky inequality, Lemma 2, and inequality (3), we have

\[ \left\|\sum_{k=1}^{\infty} c_k(f)(\widetilde{\varphi}_k-\varphi_k)\right\| \le \left(\sum_{k=1}^{\infty} c_k^2(f)\right)^{1/2} \left(\sum_{k=1}^{\infty}\|\widetilde{\varphi}_k-\varphi_k\|^2\right)^{1/2} \le \varepsilon\|f\|. \]

From the last inequality and formula (2) we obtain

\[ (1-\varepsilon)\|f(x_1,\ldots,x_n)\| \le \|T[f(x_1,x_2,\ldots,x_n)]\| \le \]

\[ \le (1+\varepsilon)\|f(x_1,x_2,\ldots,x_n)\|. \tag{5} \]

The operator \(T\) is linear by virtue of formula (2) and the linearity of the operator \(M\). The operator \(T\) is continuous and one-to-one by virtue of (5). Consequently, \(T\) is an isomorphism of \(M(I^n)\) into \(C(I)\). It remains to verify that \(T[C^{(s)}(I^n)]\subset C^{(p)}(I)\). If \(f(x)\in C^{(s)}(I^n)\), then the Fourier series of the function \((Mf)(x)\) converges to it uniformly. Consequently,

\[ f(x_1(t),x_2(t),\ldots,x_n(t))= \sum_{k=1}^{\infty} c_k(f)\varphi_k(t), \]

where convergence is understood in the sense of the norm of the space \(C(I)\). Therefore formula (2) for \(f(x)\in C^{(s)}(I^n)\) takes the form

\[ T[f(x_1,x_2,\ldots,x_n)] = \sum_{k=1}^{\infty} c_k(f)\widetilde{\varphi}_k(t) = \widetilde{f}(t). \]

The function \(\widetilde{f}(t)\), evidently, will have \(p\) continuous derivatives if we prove that the series

\[ \sum_{k=1}^{\infty} c_k(f)\widetilde{\varphi}_k^{(p)}(t) \]

converges uniformly. Using Lemma 3, Lemma 7, the Cauchy–Bunyakovsky inequality, and inequality 4, we have

\[ \sum_{k=1}^{\infty}|c_k(f)|\cdot \|\widetilde{\varphi}_k^{(p)}(t)\| \le \frac{n\lambda}{2} \sum_{k=1}^{\infty} \frac{\gamma_k(2k\delta_k)^{1/n}}{k^{s/n}\delta_k^p} = \]

\[ = \frac{n\lambda}{2\lambda_2^p} \sum_{k=1}^{\infty}\gamma_k(2k\delta_k)^{1/n} \le \frac{n\lambda}{2\lambda_2^p} \left(\sum_{k=1}^{\infty}\gamma_k^2\right)^{1/2} \left(\sum_{k=1}^{\infty}(2k\delta_k)^{2/n}\right)^{1/2} <\infty . \]

Consequently, \(\widetilde{f}(t)\in C^{(p)}(I)\). The theorem is proved.

Remark. Let us note that for any \(0<\varepsilon<1\) the desired isomorphism
\(T:C(I^n)\to C(I^m)\) can be constructed in the form of an \(\varepsilon\)-isometric isomorphism:

\[ (1-\varepsilon)\|f\|\leqslant \|Tf\|\leqslant (1+\varepsilon)\|f\|;\qquad f\in C(I^n);\quad Tf\in C(I^m). \]

It is not difficult to prove that there does not exist an isometric isomorphism
\(T:C(I^n)\to C(I^m)\) having the property
\(T[C^{(s)}(I^n)]\subset C^{(p)}(I^m)\) for any \(s,p\) and \(n>m\).

Theorem 2. If \(s<\dfrac{n}{m}p\), then there does not exist an isomorphism
\(T:C(I^n)\to C(I^m)\) having the property

\[ T[C^{(s)}(I^n)]\subset C^{(p)}(I^m). \]

This theorem, as we shall now see, is a simple consequence of a theorem of A. N. Kolmogorov \((^1)\). The scheme of the argument is borrowed from the works of A. A. Milyutin. Let \(K^n\) be the unit ball of the space \(C(I^n)\), and let \(K_s^n\) be the unit ball of the space \(C^{(s)}(I^n)\). Let \(N(K_s^n,\varepsilon K^n)\) be equal to the smallest number of translates of the set \(\varepsilon K^n\) by which the set \(K_s^n\) can be covered. A. N. Kolmogorov’s theorem asserts that there exist two constants \(A_{n,s}\) and \(B_{n,s}\), independent of \(\varepsilon\), such that

\[ B_{n,s}\left(\frac{1}{\varepsilon}\right)^{n/s} \leqslant \log N(K_s^n,\varepsilon K^n) \leqslant A_{n,s}\left(\frac{1}{\varepsilon}\right)^{n/s}. \]

Proof of Theorem 2. Suppose that under the hypotheses of the theorem there exists an isomorphism
\(T:C(I^n)\to C(I^m)\) having the property
\(T[C^{(s)}(I^n)]\subset C^{(p)}(I^m)\). Then the mapping
\(T:C^{(s)}(I^n)\to C^{(p)}(I^m)\) is closed. Indeed, if the sequence
\(\{g_k\}\in C^{(s)}(I^n)\) and \(\|g_k-g\|_s\to 0\), while the sequence
\(\{Tg_k\}\in C^{(p)}(I^m)\) and \(\|Tg_k-\varphi\|_p\to 0\), then
\(\|g_k-g\|\to 0\) and \(\|Tg_k-\varphi\|\to 0\) in \(C(I^m)\), since
\(\|g_k-g\|\leqslant \|g_k-g\|_s\), and
\(\|Tg_k-\varphi\|\leqslant \|Tg_k-\varphi\|_p\).

From the continuity of the mapping \(T:C(I^n)\to C(I^m)\) it follows that
\(\varphi=Tg\). By the closed graph theorem the mapping
\(T:C^{(s)}(I^n)\to C^{(p)}(I^m)\) is continuous. Consequently, there exists a constant \(R_1\) such that
\(T(K_s^n)\subset R_1K_p^m\). Since \(T:C(I^n)\to C(I^m)\) is an isomorphism, there exists a constant \(R_2\), independent of \(\varepsilon\), such that
\[ N(K_s^n,\varepsilon K^n)\leqslant N(R_2T(K_s^n),\varepsilon K^m) \leqslant N(R_2R_1K_p^m,\varepsilon K^m) \]
for every \(\varepsilon>0\). Hence, by A. N. Kolmogorov’s theorem,

\[ B_{n,s}\left(\frac{1}{\varepsilon}\right)^{n/s} \leqslant \log N(K_s^n,\varepsilon K^n) \leqslant \log N(R_2R_1K_p^m,\varepsilon K^m) \leqslant A_{m,p}\left(\frac{R_2R_1}{\varepsilon}\right)^{m/p} \]

for every \(\varepsilon>0\). The latter is impossible, since \(n/s>m/p\). The theorem is proved.

The author expresses his deep gratitude to A. G. Vitushkin for posing the problem and for supervising the work.

Received
22 IV 1963

REFERENCES

1. A. N. Kolmogorov, V. M. Tikhomirov, UMN, 14, no. 2 (1959).
2. G. M. Fichtenholz, A Course of Differential and Integral Calculus, 1, 1958, pp. 590–597.