MATHEMATICS

M. I. KADEC

ON WEAK AND STRONG CONVERGENCE

(Presented by Academician S. N. Bernstein on 5 V 1958)

The following proposition is known, connecting the notions of weak and strong convergence:

Theorem 1. If a sequence \(x_n\) of elements of a uniformly convex space converges weakly to an element \(x\) and \(\lim\limits_{n\to\infty}\|x_n\|=\|x\|\), then the sequence \(x_n\) converges to \(x\) strongly \((^{1})\).

This theorem was extended by Vyborny \((^{2})\) to locally uniformly convex spaces \((^{3})\).

In the present note the following will be proved:

Theorem 2. In any separable Banach space one can introduce a new norm, equivalent to the old one, such that for arbitrary elements \(x_n\) and \(x\) from

\[ x_n \xrightarrow{\mathrm{sl}} x,\qquad \lim_{n\to\infty}\|x_n\|=\|x\| \tag{1} \]

it follows that

\[ \lim_{n\to\infty}\|x_n-x\|=0. \tag{2} \]

We shall prove this theorem for the space \(C\) of continuous functions defined on \([0,1]\), and then use the universality of \(C\) in the class of separable Banach spaces.

On the set of continuous functions on \([0,1]\), define the norm

\[ \|f(t)\|=\max_{0\leq t\leq 1}|f(t)|+\sum_{k=1}^{\infty}\frac{1}{2^k}\,\omega\!\left(f,\frac{1}{k}\right), \tag{3} \]

where \(\omega(f,\delta)\) is the modulus of continuity of the function \(f(t)\):

\[ \omega(f,\delta)=\max_{|t'-t''|\leq \delta}|f(t')-f(t'')|. \]

Since \(\omega(f,\delta)\leq 2\max\limits_{0\leq t\leq 1}|f|\), the norm (3) is equivalent to the norm of the space \(C\):

\[ \max_{0\leq t\leq 1}|f(t)|\leq \|f(t)\|\leq 3\max_{0\leq t\leq 1}|f(t)|. \]

The space of continuous functions with the norm (3) will be denoted by \(C^*\). Let us prove some auxiliary propositions.

Lemma 1. If a sequence \(\varphi_n(t)\) converges to zero at every point of the interval \([0,1]\) and \(\|\varphi_n\|=1\), then for any \(\delta>0\)

\[ \lim_{n\to\infty}\omega(\varphi_n,\delta)\geq \frac{1}{3}. \tag{4} \]

Take \(n_0\) so large that for all \(n>n_0\) and arbitrary \(\varepsilon>0\) the inequality \(|\varphi_n(t)|<\varepsilon\) holds on a set \(E\), whose measure ...

greater than \(1-\delta\). If \(\varphi_n(t'_n)=\max_{0\leq t\leq 1}|\varphi_n(t)|\), \(t''_n\in E\), and \(|t'_n-t''_n|<\delta\), then

\[ |\varphi_n(t'_n)-\varphi_n(t''_n)|>\max_{0\leq t\leq 1}|\varphi_n(t)|-\varepsilon, \]

whence (4) follows.

Lemma 2. If the sequence \(\varphi_n(t)\) converges to zero at every point of the interval \([0,1]\), then

\[ \lim_{n\to\infty}\omega(f+\varphi_n,\delta)\geq \omega(f,\delta) \tag{5} \]

for every continuous function \(f(t)\) and \(\delta>0\).

Let \(|t'-t''|\leq\delta\) and \(|f(t')-f(t'')|=\omega(f,\delta)\). Choose \(n_0\) so large that, for all \(n>n_0\) and arbitrary \(\varepsilon>0\), the inequalities \(|\varphi_n(t')|<\varepsilon/2\) and \(|\varphi_n(t'')|<\varepsilon/2\) hold; then

\[ |f(t')+\varphi_n(t')-f(t'')-\varphi_n(t'')|\geq |f(t')-f(t'')|-\varepsilon, \]

whence (5) follows.

Now let \(\varphi_n\in C^*\) be a sequence weakly converging to zero such that \(\|\varphi_n\|>3\varepsilon>0\), and let \(f\) be an arbitrary element of \(C^*\). Define the index \(q=q(\varepsilon)\) so that

\[ \omega\!\left(f,\frac{1}{k}\right)<\frac{\varepsilon}{8}\quad\text{for } k>q. \tag{6} \]

Choose \(n\) so as to satisfy the conditions

\[ \sum_{k=q+1}^{\infty}\frac{1}{2^k}\,\omega\!\left(\varphi_n,\frac{1}{k}\right)>\frac{\varepsilon}{2^{q+1}}, \tag{7} \]

which can be done on the basis of (4), and

\[ \omega\!\left(f+\varphi_n,\frac{1}{k}\right)>\omega\!\left(f,\frac{1}{k}\right)-\frac{\varepsilon}{2^{q+3}} \quad\text{for } k\leq q. \tag{8} \]

Split the sum in (3) into two:

\[ \sum_{k=1}^{\infty}\frac{1}{2^k}\,\omega\!\left(f+\varphi_n,\frac{1}{k}\right) = \sum_{k=1}^{q}\frac{1}{2^k}\,\omega\!\left(f+\varphi_n,\frac{1}{k}\right) + \sum_{k=q+1}^{\infty}\frac{1}{2^k}\,\omega\!\left(f+\varphi_n,\frac{1}{k}\right). \]

Estimate from below each of the sums obtained:

\[ \sum_{k=1}^{q}\frac{1}{2^k}\,\omega\!\left(f+\varphi_n,\frac{1}{k}\right) > \sum_{k=1}^{q}\frac{1}{2^k} \left[ \omega\!\left(f,\frac{1}{k}\right)-\frac{\varepsilon}{2^{q+3}} \right] > \sum_{k=1}^{q}\frac{1}{2^k}\omega\!\left(f,\frac{1}{k}\right) -\frac{\varepsilon}{2^{q+3}} > \]

\[ > \sum_{k=1}^{\infty}\frac{1}{2^k}\omega\!\left(f,\frac{1}{k}\right) - \sum_{k=q+1}^{\infty}\frac{1}{2^k}\frac{\varepsilon}{8} -\frac{\varepsilon}{2^{q+3}} = \sum_{k=1}^{\infty}\frac{1}{2^k}\omega\!\left(f,\frac{1}{k}\right) -\frac{\varepsilon}{2^{q+2}}; \tag{9} \]

in this estimate we have used inequalities (8) and (6).

We estimate the second sum with the aid of inequalities (7) and (6):

\[ \sum_{k=q+1}^{\infty}\frac{1}{2^k}\,\omega\!\left(f+\varphi_n,\frac{1}{k}\right) \geq \sum_{k=q+1}^{\infty}\frac{1}{2^k}\,\omega\!\left(\varphi_n,\frac{1}{k}\right) - \sum_{k=q+1}^{\infty}\frac{1}{2^k}\,\omega\!\left(f,\frac{1}{k}\right) > \]

\[ > \frac{\varepsilon}{2^{q+1}}-\frac{\varepsilon}{2^{q+3}} = \frac{3\varepsilon}{2^{q+3}}. \tag{10} \]

Adding (9) and (10), we obtain

\[ \sum_{k=1}^{\infty}\frac{1}{2^k}\,\omega\!\left(f+\varphi_n,\frac{1}{k}\right) > \sum_{k=1}^{\infty}\frac{1}{2^k}\,\omega\!\left(f,\frac{1}{k}\right) + \frac{\varepsilon}{2^{q+3}}. \]

Since, moreover,

\[ \lim_{n\to\infty}\max_{0\le t\le 1}|f+\varphi_n|\ge \max_{0\le t\le 1} f, \]

we have

\[ \lim_{n\to\infty}\|f+\varphi_n\|>\|f\|, \]

whence the validity of Theorem 2 follows.

In what follows we shall consider spaces satisfying conditions (1), (2) and, in addition, strictly normed. Such a norm can be obtained by modifying (3) in the following way:

\[ \|f\|=\max_{0\le t\le 1}|f(t)|+ \left[\int_0^1 f^2(t)\,dt\right]^{1/2} +\sum_{k=1}^{\infty}\frac{1}{2^k}\,\omega\left(f,\frac1k\right). \tag{3a} \]

Let us apply Theorem 2 to the proof of the following assertion.

Theorem 3. Separable reflexive Banach spaces are monotonically equivalent.

Let \(X\) be a separable reflexive space satisfying (1), (2) and strictly normed. Consider a system of linear subspaces
\(P_0 \supset P_1 \supset P_2 \supset \cdots\) such that \(\operatorname{def} P_n=n\), and the intersection of all \(P_n\) contains only the zero element of the space; such a system exists in every separable Banach space. For each element \(x\in X\) introduce the sequence of deviations \(H_n(x)\):

\[ H_n(x)=\min_{y\in P_n}\|x-y\|=\|x-x^{(n)}\|\qquad (n=1,2,\ldots). \]

The existence and uniqueness of the element of best approximation \(x^{(n)}\) are ensured respectively by the reflexivity and the strict normedness of the space \(X\). For every \(n\), evidently,

\[ H_n(x)\le H_{n+1}(x)\le \|x\|. \]

We shall use the fact that each \(P_n\) divides \(P_{n-1}\) into three parts

\[ P_{n-1}=P_{n-1}^{+}+P_n+P_{n-1}^{-}, \]

and construct a sequence of functionals \(\varepsilon_n(x)\):

\[ \begin{aligned} \varepsilon_n(x)&=+1, &&\text{if } x^{(n-1)}\in P_{n-1}^{+};\\ \varepsilon_n(x)&=\varepsilon_{n-1}(x), &&\text{if } x^{(n-1)}\in P_n\quad(\text{i.e. } H_{n-1}(x)=H_n(x));\\ \varepsilon_n(x)&=-1, &&\text{if } x^{(n-1)}\in P_{n-1}^{-}. \end{aligned} \]

In addition, we shall put \(\varepsilon_0(x)=0\) for every \(x\).

Lemma 3. Let \(h_1,h_2,\ldots,h_n\) be a sequence of real numbers satisfying the conditions

\[ |h_j|\le |h_{j+1}|;\qquad \text{if } |h_j|=|h_{j+1}|,\text{ then } h_j=h_{j+1}. \tag{11} \]

The set \(Q_n\) of elements \(x\) such that

\[ \varepsilon_k(x)H_k(x)=h_k\quad (k\le n), \tag{12} \]

is nonempty and is obtained from \(P_n\) if to each \(y\in P_n\) one adds one and the same element \(x^*\), satisfying (12).

The proof of this lemma is given in \((^4)\).

Lemma 4. Whatever bounded sequence of real numbers \(h_1,h_2,h_3,\ldots\) satisfying (11) may be, there exists a unique element \(x\) for which

\[ \varepsilon_n(x)H_n(x)=h_n\quad (n=1,2,\ldots); \tag{13} \]

moreover \(\|x\|=\lim_{n\to\infty}H_n(x)\).

The set of elements satisfying (13) is the intersection \(\prod_1^\infty Q^n\) of the sets satisfying (12). It is therefore sufficient to show,

that this intersection contains exactly one point. Suppose that

\[ \prod_{1}^{\infty} Q_n \]

contains the element \(x\); then, according to Lemma 4, each \(Q_n = x + P_n\), and the whole family \(\{Q_n\}_1^\infty\) is congruent to \(\{P_n\}_1^\infty\). Since

\[ \prod_{1}^{\infty} P_n \]

by definition contains a single element, it follows that

\[ \prod_{1}^{\infty} Q_n \]

also contains a single element. It remains to show that

\[ \prod_{1}^{\infty} Q_n \]

is nonempty. Consider the intersections of the sets \(Q_n\) and of the ball \(S\) with arbitrary radius
\[ h > \lim_{n \to \infty} |h_n|. \]
The resulting nonempty closed convex sets form a decreasing sequence. By a theorem of V. L. Shmul’yan \({}^{5}\), the intersection of these sets is nonempty. Thus, the existence and uniqueness of the element \(x\) are proved. Since \(h\) may be taken arbitrarily close to \(\lim_{n\to\infty}|h_n|\), the equality
\[ \|x\| = \lim_{n\to\infty} H_n(x) \]
has thereby also been proved.

Before passing to the proof of Theorem 3, let us note that if the sequence \(x_n\) converges weakly to \(x\), then the element \(x_n\), as \(n\) increases, approaches without bound each \(Q_k\) containing \(x\), and therefore

\[ \begin{aligned} H_k(x) &= \lim_{n\to\infty} H_n(x_n) \quad (k=1,2,\ldots);\\ \varepsilon_k(x) &= \lim_{n\to\infty} \varepsilon_k(x_n), \quad \text{if } H_{k-1}(x) < H_k(x);\\ \varepsilon_k(x) &= \varepsilon_{k-1}(x), \quad \text{if } H_{k-1}(x) = H_k(x). \end{aligned} \tag{14} \]

Now let \(X\) and \(Y\) be separable reflexive spaces satisfying (1), (2) and strictly normed; suppose that for the elements of each of them the deviations \(H_k(x)\) and \(H_k(y)\) and the functionals \(\varepsilon_k(x)\) and \(\varepsilon_k(y)\) are defined. To each element \(x \in X\) assign an element \(y \in Y\) so that

\[ \varepsilon_k(x) H_k(x) = \varepsilon_k(y) H_k(y) \quad (k=1,2,\ldots). \tag{16} \]

This correspondence is one-to-one. We shall show that it is continuous. Let \(x_n\) and \(x\) be arbitrary elements of \(X\), and

\[ \lim_{n\to\infty} x_n = x. \tag{16} \]

Since \(Y\) is reflexive, the corresponding sequence \(y_n\) is weakly compact. According to (14), no subsequence of it can converge weakly to an element different from the element \(y\) corresponding to \(x\), and therefore

\[ y_n \xrightarrow{\mathrm{w}} y. \tag{17} \]

According to Lemma 4, \(\|x_n\|=\|y_n\|\), \(\|x\|=\|y\|\), whence

\[ \lim_{n\to\infty} \|y_n\| = \|y\|. \tag{18} \]

From (17) and (18) it follows that

\[ \lim_{n\to\infty} y_n = y. \]

The continuity of the inverse correspondence is proved in exactly the same way. Thus, the spaces \(X\) and \(Y\) are homeomorphic, which proves Theorem 3.

Kharkov Automobile and Highway Institute

Received
4 IV 1958

References

\({}^{1}\) V. L. Shmul’yan, DAN, 24, No. 7, 647 (1939).
\({}^{2}\) R. Výborný, Casop. Pest. Mat., 81, No. 3, 352 (1956).
\({}^{3}\) A. R. Lovaglia, Trans. Am. Math. Soc., 78, No. 1, 225 (1955).
\({}^{4}\) M. I. Kadets, Uspekhi Mat. Nauk, 10, No. 4, 137 (1955).
\({}^{5}\) V. L. Shmul’yan, Matem. sborn., 5 (47), No. 2, 317 (1939).