MATHEMATICS

Academician B. S. STECHKIN and S. B. STECHKIN

QUADRATIC MEAN AND ARITHMETIC MEAN

Let \(y(x)\in L^2[0,1]\). Put

\[ \varphi_0(x)=y(x), \qquad y_0=\int_0^1 |\varphi_0(x)|\,dx, \]

\[ \varphi_k(x)=|\varphi_{k-1}(x)|-y_{k-1}, \qquad y_k=\int_0^1 |\varphi_k(x)|\,dx \qquad (k=1,2,\ldots). \]

Then the formula

\[ \int_0^1 y^2(x)\,dx = \sum_{k=0}^{\infty} y_k^2 = \sum_{k=0}^{\infty} \left\{\int_0^1 |\varphi_k(x)|\,dx\right\}^2 . \tag{1} \]

is valid.

This formula relates the quadratic mean of the function \(y(x)\) to the arithmetic means of the functions \(|\varphi_k(x)|\), which depend very simply on \(y(x)\).

Applying the identity

\[ \int_0^1 \varphi^2(x)\,dx = \left\{\int_0^1 \varphi(x)\,dx\right\}^2 + \int_0^1 \left\{\varphi(x)-\int_0^1 \varphi(t)\,dt\right\}^2 dx \]

to the functions \(|\varphi_k(x)|\) \((k=0,1,\ldots,n-1)\), we infer that

\[ \int_0^1 \varphi_k^2(x)\,dx = y_k^2+\int_0^1 \varphi_{k+1}^2(x)\,dx \qquad (k=0,1,\ldots,n-1), \]

whence

\[ \int_0^1 y^2(x)\,dx = \sum_{k=0}^{n-1} y_k^2 + \int_0^1 \varphi_n^2(x)\,dx; \tag{2} \]

\[ \sum_{k=0}^{\infty} y_k^2 \leq \int_0^1 y^2(x)\,dx. \tag{3} \]

It remains to show that for every function \(y(x)\) this inequality becomes an equality, i.e., that

\[ \int_0^1 \varphi_n^2(x)\,dx \to 0 \qquad (n\to\infty). \tag{4} \]

For this we shall need two lemmas.

Lemma 1. Let \(p\geq 0\), and on a set \(E\), \(\operatorname{mes} E=\delta>0\), the inequality \(|\varphi_p(x)|\geq M\) be satisfied. Then

\[ K_p=\sum_{k=p}^{\infty} y_k \geq M. \tag{5} \]

Proof. Suppose that \(K_p < M\), and let us derive a contradiction. On the set \(E\) we have

\[ \varphi_{p+1}(x)=|\varphi_p(x)|-y_p \geq M-y_p \geq M-K_p>0, \]

\[ \varphi_{p+2}(x)=|\varphi_{p+1}(x)|-y_{p+1}\geq M-y_p-y_{p+1}\geq M-K_p>0, \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

\[ \varphi_n(x)\geq M-K_p>0 \qquad (n=p+1,\ p+2,\ldots). \]

Hence

\[ y_n=\int_0^1 |\varphi_n(x)|\,dx \geq \int_E |\varphi_n(x)|\,dx \geq \delta(M-K_p) \qquad (n=p+1,\ p+2,\ldots), \]

which contradicts the condition \(y_n\to 0\) \((n\to\infty)\) (see (3)). Inequality (5) is proved.

Lemma 2. Let \(0\leq p\leq n\). Put \(\varepsilon_p=\max_{m>p} y_m\).

1) If at the point \(x_0\in[0,1]\)

\[ |\varphi_p(x_0)|\geq \sum_{k=p}^{n} y_k, \]

then

\[ |\varphi_{n+1}(x_0)|=|\varphi_p(x_0)|-\sum_{k=p}^{n} y_k. \]

2) If at the point \(x_0\in[0,1]\)

\[ |\varphi_p(x_0)|<\sum_{k=p}^{n} y_k, \]

then

\[ |\varphi_{n+1}(x_0)|\leq \varepsilon_p. \]

Proof. 1) We have

\[ \varphi_{p+1}(x_0)=|\varphi_p(x_0)|-y_p \geq \sum_{k=p+1}^{n} y_k \geq 0, \]

\[ \varphi_{p+2}(x_0)=|\varphi_{p+1}(x_0)|-y_{p+1} =|\varphi_p(x_0)|-(y_p+y_{p+1})\geq \sum_{k=p+1}^{n} y_k, \]

\[ \cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots \]

\[ \varphi_{n+1}(x_0)=|\varphi_n(x_0)|-y_n=\varphi_n(x_0)-y_n =|\varphi_p(x_0)|-\sum_{k=p}^{n} y_k\geq 0, \]

i.e.

\[ |\varphi_{n+1}(x_0)|=|\varphi_p(x_0)|-\sum_{k=p}^{n} y_k. \]

2) Let us find a number \(m\) \((p\leq m\leq n)\) such that

\[ \sum_{k=p}^{m-1} y_k \leq |\varphi_p(x_0)| < \sum_{k=p}^{m} y_k. \]

Then, according to part 1),

\[ |\varphi_m(x_0)|=|\varphi_p(x_0)|-\sum_{k=p}^{m-1} y_k \]

Next we have

\[ \varphi_{m+1}(x_0)=|\varphi_m(x_0)|-y_m=|\varphi_p(x_0)|-\sum_{k=p}^{m} y_k<0, \]

\[ |\varphi_{m+1}(x_0)|\leqslant y_m\leqslant \varepsilon_p, \]

\[ \varphi_{m+2}(x_0)=|\varphi_{m+1}(x_0)|-y_{m+1}\leqslant |\varphi_{m+1}(x_0)|\leqslant \varepsilon_p, \]

\[ \varphi_{m+2}(x_0)\geqslant -y_{m+1}\geqslant -\varepsilon_p, \]

i.e.

\[ |\varphi_{m+2}(x_0)|\leqslant \varepsilon_p,\ldots,|\varphi_{n+1}(x_0)|\leqslant \varepsilon_p, \]

and the lemma is proved.

We pass to the proof of relation (4). Consider two cases.

1st case. \(\sup_x |\varphi_0(x)|=M_0<\infty\)*. Then, obviously, for any \(p\geqslant 0\)

\[ \sup_x |\varphi_p(x)|=M_p<\infty. \]

Fix \(\varepsilon>0\), and let \(p\) be so large that

\[ \varepsilon_p=\max_{m\geqslant p} y_m\leqslant \varepsilon. \]

According to Lemma 1 we can find a number \(n=n(\varepsilon)\) for which

\[ \sum_{k=p}^{n} y_k\geqslant M_p-\varepsilon_p. \]

Then, by Lemma 2, for almost all \(x\) for which

\[ |\varphi_p(x)|\geqslant \sum_{k=p}^{n} y_k, \]

the inequality

\[ |\varphi_{n+1}(x)|=|\varphi_p(x)|-\sum_{k=p}^{n} y_k\leqslant M_p-(M_p-\varepsilon_p)=\varepsilon_p \]

holds, and for all \(x\) for which

\[ |\varphi_p(x)|<\sum_{k=p}^{n} y_k, \]

the inequality

\[ |\varphi_{n+1}(x)|\leqslant \varepsilon_p \]

holds. Hence

\[ \sup_x |\varphi_{n+1}(x)|\leqslant \varepsilon_p\leqslant \varepsilon, \]

\[ r_{n+1}^2=\int_0^1 \varphi_{n+1}^2(x)\,dx\leqslant \varepsilon^2. \]

Since, moreover,

\[ r_{n+1}^2=y_{n+1}^2+r_{n+2}^2\geqslant r_{n+2}^2, \]

it follows that \(r_n^2\to 0\) \((n\to\infty)\). We note that in this case \(M_n\to 0\) \((n\to\infty)\), and for any \(n\geqslant 0\)

\[ \int_0^1 y^2(x)\,dx=\sum_{k=0}^{n} y_k^2+\theta_n M_n^2,\qquad (0\leqslant \theta_n\leqslant 1). \]

* Here and below \(\sup_x |\varphi(x)|\) denotes the essential supremum of the function \(|\varphi(x)|\) on the interval \([0,1]\).

2nd case. \(\sup_x |\varphi_0(x)|=\infty\). Then for any \(p \geqslant 0\)

\[ \sup_x |\varphi_p(x)|=\infty . \]

Fix \(\varepsilon>0\) and find \(p \geqslant 0\) and \(N>0\) such that

\[ \varepsilon_p \leqslant \varepsilon,\qquad \int_{|\varphi_p|\geqslant N}\varphi_p^2(x)\,dx \leqslant \varepsilon^2 . \]

By Lemma 1, for all sufficiently large \(n\) we have

\[ K_{p,n}=\sum_{k=p}^{n} y_k > N . \]

Hence, as above,

\[ \begin{aligned} r_{n+1}^2 &=\int_0^1 \varphi_{n+1}^2(x)\,dx \\ &=\int_{|\varphi_p|\geqslant K_{p,n}}\varphi_{n+1}^2(x)\,dx +\int_{|\varphi_p|<K_{p,n}}\varphi_{n+1}^2(x)\,dx \\ &\leqslant \int_{|\varphi_p|\geqslant N}\varphi_p^2(x)\,dx+\varepsilon_p^2 \leqslant 2\varepsilon^2, \end{aligned} \]

and, consequently, \(r_n^2 \to 0\) \((n\to\infty)\).

Thus, in both cases

\[ \int_0^1 \varphi_n^2(x)\,dx \to 0 \qquad (n\to\infty), \]

and formula (1) is proved.

Received
14 XII 1960