Abstract
Full Text
MATHEMATICS
B. S. MITIAGIN
ON THE SECOND MIXED DERIVATIVE
(Presented by Academician S. L. Sobolev on 1 VII 1958)
Let \(E\) be some Banach space of functions on the two-dimensional torus\(^*\), with \(D_\infty\)\(^ {**}\) dense in \(E\), and let it be known for a continuous function \(f(t,s)\) that \(\partial^2 f/\partial t^2\) and \(\partial^2 f/\partial s^2\) belong to \(E\). Can one assert that \(\partial^2 f/\partial t\,\partial s\) also belongs to \(E\)?
S. N. Bernstein showed\(^1\) that when \(E\) is \(L^2\) or\(^ {***}\) \(W\), the answer is affirmative. For the case \(E=C\), the question was not considered. In the present note it is shown that in this case the answer is negative.
For the proof, an example is given of a function \(F(t,s)\) such that \(\partial^2 F/\partial t^2\) and \(\partial^2 F/\partial s^2 \in C\), but \(\partial^2 F/\partial t\,\partial s \notin C\). In constructing the example the following theorem is essentially used:
Petrini’s Theorem\(^2\). Let \(\mu(\xi,\eta)\) be a continuous function;
\[ H(t,s)=\int_{C^{2a}}\mu\log\frac{1}{r}\,d\xi\,d\eta \]
(\(C^a\) is the disk of radius \(a\), \(4a<1\), with center at the point \((0,0)\); \(r=\sqrt{(\xi-t)^2+(\eta-s)^2}\))\(^ {****}\).
Then inside \(C^{2a}\), \(H(t,s)\) is continuously differentiable. For the existence at a point \((t,s)\) of the derivative \(\partial^2H/\partial l_1\partial l_2\) (\(l_1,l_2\) are directions making with the \(t\)-direction respectively the angles \(\psi_1,\psi_2\)) it is necessary and sufficient that the limit exist
\[ K_{l_1l_2}(t,s)=\lim_{h\to0}K^h_{l_1l_2}(t,s);\qquad K^h_{l_1l_2}(t,s)= \int_{C^{2a}\setminus C^h_{ts}}\cos(2\nu-\psi_1-\psi_2)\,\mu\,\frac{d\nu\,dr}{r} \tag{1} \]
(\(C^h_{ts}\) is the disk of radius \(h\) with center at the point \((t,s)\); \(r,\nu\) are polar coordinates with pole \((t,s)\)); moreover
\[ \frac{\partial^2H(t,s)}{\partial l_1\partial l_2} = K_{l_1l_2}(t,s)-\pi\mu(t,s)\cos\widehat{(l_1,l_2)}. \tag{2} \]
Lemma. Let \(\mu=\dfrac{\cos2\varphi}{\log(1/\rho)}\) (\(\rho,\varphi\) are polar coordinates with pole \((0,0)\)).
Then
\[ H(t,s)=\int_{C^{2a}}\mu\log\frac{1}{r}\,d\xi\,d\eta \]
has, for \(\rho\le a\), continuous \(\Delta H\) and \(\partial^2H/\partial t\,\partial s\), but \(\partial^2H/\partial t^2\) and \(\partial^2H/\partial s^2\) do not exist at the point \((0,0)\).
\(^*\) For simplicity, doubly periodic functions are considered.
\(^ {**}\) \(D_\infty\) is the space of all infinitely differentiable functions.
\(^ {***}\) \(W\) is the space of functions with an absolutely convergent Fourier series
\[
f(t,s)=\sum_{m,n=-\infty}^{\infty} a_{mn}e^{i(mt+ns)}
\]
and norm
\[
\|f\|_W=\sum_{m,n=-\infty}^{\infty}|a_{mn}|.
\]
\(^ {****}\) On the torus a metric is introduced in the natural way; distances less than one are thereby determined uniquely.
Proof. \(\Delta H=-2\pi\mu\), and therefore is discontinuous. \(\partial^2 H/\partial t^2\) (as also \(\partial^2 H/\partial s^2\)) does not exist at the point \((0,0)\), since the condition of Petrini’s theorem is not satisfied:
\[ \int_0^{2\pi}\cos 2\nu\,d\nu \int_h^{2a}\frac{\cos 2\nu}{\log(1/r)}\,\frac{dr}{r} \longrightarrow \infty \quad \text{as } h\to 0 . \]
\(\partial^2 H/\partial t\partial s\) for \((t,s)\ne(0,0)\) exists and is continuous, since \(\mu\) is continuously differentiable everywhere except at the point \((0,0)\); at the point \((0,0)\)
\[ K_{ts}^{h} = -\int_0^{2\pi}\sin 2\nu\,d\nu \int_h^{2a}\frac{\cos 2\nu}{\log(1/r)}\,\frac{dr}{r} =0 \]
for all \(h\), so that \(\partial^2 H/\partial t\partial s\) exists at the point \((0,0)\) and is equal to zero. The continuity of \(\partial^2 H/\partial t\partial s\) at the point \((0,0)\) is verified directly. For this purpose, by Petrini’s theorem, it is enough to show that \(K_{ts}(\varepsilon,\varphi_0)\to 0\) as \(\varepsilon\to 0\), uniformly with respect to \(\varphi_0\). We examine
\[ -K_{ts}(\varepsilon,\varphi_0) = \lim_{h\to 0} \int_0^{2\pi}\int_0^{2a}{}' \sin 2\nu\, \frac{\cos 2\varphi}{\log(1/\rho)}\, \frac{\rho\,d\rho\,d\varphi}{r^2}, \]
where \(\rho\cos\varphi=r\cos\nu+\varepsilon\cos\varphi_0\), \(\rho\sin\varphi=r\sin\nu+\varepsilon\sin\varphi_0\), and the prime sign means that on rays intersecting the circle \(C_{\varepsilon\varphi_0}^{h}\), integration over the interval common with this circle is not performed.
We integrate by parts with respect to \(\rho\); for this we put
\[ \frac{\sin 2\nu}{r^2}\,d\rho=d_\rho v, \qquad \frac{\cos 2\varphi}{\log(1/\rho)}\,\rho=u. \]
Then
\[ v=\frac{\varepsilon\sin(\varphi-\varphi_0)-\rho\sin 2\varphi} {\rho^2-2\varepsilon\rho\cos(\varphi-\varphi_0)+\varepsilon^2}; \]
\[ \begin{aligned} -K_{t,s}^{h}(\varepsilon,\varphi_0) &= \int_0^{2\pi} \left\{ uv\big|_0^{\rho_h(\varphi)} + uv\big|_{\rho^h(\varphi)}^{2a} - \int_0^{2a}vu_\rho'\,d\rho \right\}\,d\varphi \\ &= \int_0^{2\pi}uv\big|_0^{2a}\,d\varphi - \int_\Delta uv\big|_{\rho_h(\varphi)}^{\rho^h(\varphi)}\,d\varphi - \int_0^{2\pi}\int_0^{2a}vu_\rho'\,d\rho\,d\varphi, \end{aligned} \tag{3} \]
since \(\rho_h(\varphi)=\rho^h(\varphi)\) for \(\varphi\notin\Delta=\left(\varphi_0-\arcsin\frac{h}{\varepsilon},\,\varphi_0+\arcsin\frac{h}{\varepsilon}\right)\); in the interval \(\Delta\), \(\rho_h(\varphi)\) and \(\rho^h(\varphi)\) are determined as the roots of the equation \(r^2=h^2\).
\(u(0,0)=0\), and therefore the first term in (3) is
\[ g_1(\varepsilon,\varphi_0) = \int_0^{2\pi} \frac{\varepsilon\sin(\varphi+\varphi_0)-2a\sin 2\varphi} {4a^2-4a\varepsilon\cos(\varphi-\varphi_0)+\varepsilon^2} \, \frac{2a\cos 2\varphi}{\log(1/2a)} \,d\varphi; \]
it is easy to see that, for small \(\varepsilon\),
\[ |g_1(\varepsilon,\varphi_0)| < \frac{3\varepsilon}{a\log(1/2a)}. \tag{4} \]
The second term in formula (3) is
\[ \begin{aligned} I_h(\varepsilon,\varphi_0) &= -\int_\Delta \left( \frac{\varepsilon\sin(\varphi+\varphi_0)-\rho\sin 2\varphi}{r^2} \, \frac{\rho}{\log(1/\rho)} \cos 2\varphi \right)\Bigg|_{\rho_h(\varphi)}^{\rho^h(\varphi)} d\varphi \\ &= -\frac{1}{h^2} \int_\Delta \Bigg\{ \varepsilon\sin(\varphi+\varphi_0) \left[ \frac{\rho^h}{\log(1/\rho^h)} - \frac{\rho_h}{\log(1/\rho_h)} \right] \\ &\qquad -\sin 2\varphi \left[ \frac{(\rho^h)^2}{\log(1/\rho^h)} - \frac{\rho_h^2}{\log(1/\rho_h)} \right] \Bigg\} \cos 2\varphi\,d\varphi . \end{aligned} \tag{5} \]
we estimate as follows: since \(\rho^h-\rho_h \leq 2h\) and
\[ \left(\frac{\rho}{\log(1/\rho)}\right)'= \frac{1}{\log(1/\rho)}+\frac{1}{\log^2(1/\rho)}, \]
then, by Lagrange’s theorem,
\[ \frac{\rho^h}{\log(1/\rho^h)}-\frac{\rho_h}{\log(1/\rho_h)} \leq 2h\left(\frac{1}{\log(1/\xi)}+\frac{1}{\log^2(1/\xi)}\right) \qquad(\rho_h<\xi<\rho^h), \]
and, consequently, does not exceed \(\dfrac{4h}{\log(1/\varepsilon)}\) [for sufficiently small \(\varepsilon\)]. Analogously, for the second square bracket in (5) we obtain the estimate
\[ \frac{(\rho^h)^2}{\log(1/\rho^h)}-\frac{\rho_h^2}{\log(1/\rho_h)} \leq \frac{6\varepsilon h}{\log(1/\varepsilon)}. \]
Then
\[ |I_h(\varepsilon,\varphi_0)| \leq 2\arcsin \frac{h}{\varepsilon}\cdot \frac{1}{h^2}\, \frac{10h}{\log(1/\varepsilon)} \leq \frac{10\pi}{\log(1/\varepsilon)} \]
and for \(I(\varepsilon,\varphi_0)=\lim I_h(\varepsilon,\varphi_0)\) we have
\[ |I(\varepsilon,\varphi_0)|\leq \frac{10\pi}{\log(1/\varepsilon)}. \tag{6} \]
The third term in (3) is
\[ J_h(\varepsilon,\varphi_0)= -\int_{0}^{2\pi}\int_{0}^{2a} \frac{\varepsilon\sin(\varphi+\varphi_0)-\rho\sin 2\varphi} {\rho^2-2\rho\varepsilon\cos(\varphi-\varphi_0)+\varepsilon^2} \left(\frac{1}{\log(1/\rho)}+\frac{1}{\log^2(1/\rho)}\right) \cos 2\varphi\,d\varphi\,d\rho. \]
We note that the integrand \(v u'_\rho\) has a single singularity of the first order at the point \((\varepsilon,\varphi_0)\) and is integrable over the whole circle \(C^{2a}\). Therefore
\[ J(\varepsilon,\varphi_0)=\lim J_h(\varepsilon,\varphi_0) = -\int_{\rho\leq 2a} vu'_\rho\,d\varphi\,d\rho. \]
We use the formula
\[ \frac{1}{1-2k\cos\alpha+k^2} = \frac{1}{1-k^2} + \frac{2}{1-k^2}\sum_{n=1}^{\infty} k^n\cos n\alpha \]
for \(|k|<1\), uniformly in \(\alpha\). Expanding \(v\) in the corresponding series for \(\rho<\varepsilon\) and \(\rho>\varepsilon\) and integrating them, we obtain
\[ J(\varepsilon,\varphi_0) = 4\pi\sin 4\varphi_0\cdot \frac{1}{\varepsilon^4} \int_{0}^{\varepsilon} \rho^3 \left( \frac{1}{\log(1/\rho)} + \frac{1}{\log^2(1/\rho)} \right)d\rho, \]
whence
\[ |J(\varepsilon,\varphi_0)| \leq \frac{8\pi}{\log(1/\varepsilon)}. \tag{7} \]
Inequalities (4), (6), and (7) give
\[ |K_{ts}(\varepsilon,\varphi_0)|\leq \frac{20\pi}{\log(1/\varepsilon)}. \]
This completes the proof of the lemma.
Let \(e(\rho)\) be an infinitely differentiable function equal to one for \(\rho\leq a/2\) and to zero for \(\rho\geq a\). Then \(\overline{H}(t,s)=H(t,s)e(\rho)\) is defined on the whole torus. It is not difficult to see that the function \(F(t,s)=\overline{H}(t+s,t-s)\) has continuous \(\partial^2 F/\partial t^2\) and \(\partial^2 F/\partial s^2\) on the whole torus, but \(\partial^2 F/\partial t\,\partial s\) does not exist at the point \((0,0)\).
Thus it has been proved:
Theorem. The space \(B\), obtained as the closure of \(D_\infty\) in the norm
\[ \|f\|=\max_{t,s}\left\{|f(t,s)|,\left|\frac{\partial f(t,s)}{\partial t}\right|,\left|\frac{\partial f(t,s)}{\partial s}\right|,\left|\frac{\partial^2 f(t,s)}{\partial t^2}\right|,\left|\frac{\partial^2 f(t,s)}{\partial s^2}\right|\right\} \]
is distinct from \(D_2\)—the space of twice continuously differentiable functions.
The author expresses his gratitude to Prof. G. E. Shilov for valuable advice and attention to the work.
Moscow State University
named after M. V. Lomonosov
Received
24 VI 1958
REFERENCES
- S. N. Bernstein, Collected Works, 1, 1952, p. 97.
- H. Petrini, J. Math. Pures et Appl., ser. 6, 5, 127 (1909).