Problems to Solve

for a sequence of continuous linear functionals that converges weakly-*;

for a sequence of elements of an NLS that converges weakly.

Prove that the sequence $\left\{ x^{(n)} = \left(x_{1}^{(n)}, x_{2}^{(n)}, \ldots \right): n \geq 1\right\} \subset c_{0}$ converges weakly to the element $x = \left(x_{1}, x_{2}, \ldots \right) \in c_{0}$ if and only if the following conditions are fulfilled:

(a) $\exists C > 0 \forall n \geq 1:\left\| x^{(n)}\right\| \leq C$;

(b) $\forall k \geq 1: x_{k}^{(n)} \rightarrow x_{k}, n \rightarrow \infty$.

Formulate and prove the criterion of weak convergence in the space $c$.

$X = l_{p}, 1 < p < +\infty , x^{(n)} = (\underbrace{0, \ldots , 0}_{n}, 1, \frac{1}{2}, \frac{1}{3}, \ldots )$;

$X = l_{p}, 1 < p < +\infty , x^{(n)} = (\underbrace{0, \ldots , 0}_{n}, \frac{1}{n+1}, \frac{1}{n+2}, \ldots )$;

$X = l_{p}, 1 < p < +\infty , x^{(n)} = (\underbrace{1, \ldots , 1}_{n-1}, \frac{1}{n}, \frac{1}{n+1}, \ldots )$;

$X = l_{p}, 1 < p < +\infty , x^{(n)} = \left(\frac{1}{n}, \frac{1}{n+1}, \ldots \right)$;

$X = l_{p}, 1 < p < +\infty , x^{(n)} = \left(\frac{1+n}{1+n}, \frac{1+n}{1+2 n}, \ldots , \frac{1+n}{1+k n}, \ldots \right)$;

$X = l_{p}, 1 \leq p < +\infty , x^{(n)} = \left(\frac{1}{2}, \frac{2}{2^{2}}, \ldots , \frac{n}{2^{n}}, \frac{1}{(n+1)^{2}}, \frac{1}{(n+2)^{2}}, \ldots \right)$;

$X = l_{p}, 1 \leq p < +\infty , x^{(n)} = \left(1,0, \frac{1}{3^{2}}, 0, \frac{1}{5^{2}}, \ldots , 0, \frac{1}{(2 n-1)^{2}}, \frac{1}{(2 n)^{2}}, \frac{1}{(2 n+1)^{2}}, \ldots \right)$;

$X = l_{\infty }, x^{(n)} = (\underbrace{0, \ldots , 0}_{n-1}, \ln n, 0,0, \ldots )$;

$X = L_{p}(\left[0,1\right]), 1 \leq p < +\infty , x_{n}(t) = t^{n}$;

$X = L_{p}(\left[0,1\right]), 1 \leq p \leq +\infty , x_{n}(t) = \begin{cases} \sqrt{n}, & t \in \left[0, \frac{1}{n}\right), \\ 0, & t \in \left[\frac{1}{n}, 1\right] ; \end{cases}$

$X = L_{p}(\left[0,1\right]), 1 < p < +\infty , x_{n}(t) = e^{i n t}$;

$X = L_{p}(\left[0,1\right]), 1 < p < +\infty , x_{n}(t) = \sin (t \cdot \ln n)$;

$X = L_{p}(\left[0,1\right]), 1 < p < +\infty , x_{n}(t) = \cos \left(n^{2} t\right)$;

$X = L_{p}(\left[0,1\right]), 1 < p < +\infty , x_{n}(t) = \sin \left(2^{n} t\right)$;

$X = L_{p}(\left[0,1\right]), 1 < p < +\infty , x_{n}(t) = \begin{cases} \sqrt{n}-n \sqrt{n} t, & t \in \left[0, \frac{1}{n}\right), \\ 0, & t \in \left[\frac{1}{n}, 1\right] ; \end{cases}$

$X = L_{p}(\left[0,1\right]), 1 < p \leq +\infty , x_{n}(t) = \begin{cases} 2 n(1-n t), & t \in \left[0, \frac{1}{n}\right), \\ 0, & t \in \left[\frac{1}{n}, 1\right] ; \end{cases}$

$X = L_{p}(\left[0,1\right]), 1 \leq p < +\infty , x_{n}(t) = \sin \left(t^{n}\right)$;

$X = L_{p}(\mathbb {R}), 1 \leq p \leq +\infty , x_{n}(t) = \sqrt[4]{n} \chi_{\left[n, n+\frac{1}{n}\right]}(t)$;

$X = L_{p}(\mathbb {R}), 1 < p < +\infty , x_{n}(t) = \chi_{\left[n, n+1\right]}(t)$;

$X = L_{p}(\mathbb {R}), 1 \leq p \leq +\infty , x_{n}(t) = \frac{1}{\sqrt{n}} \chi_{\left[n, 2 n\right]}(t)$.

$x_{n}(t) = t^{n}-t^{3 n}$;

$x_{n}(t) = e^{-\frac{t}{n}}$;

$x_{n}(t) = e^{-n t}$;

$x_{n}(t) = t e^{-n t}$;

$x_{n}(t) = n t e^{-n t}$;

$x_{n}(t) = \frac{n^{2} t}{1+n^{4} t^{2}}$;

$x_{n}(t) = \frac{n^{3} t^{2}}{1+4 n^{4} t^{4}}$.

$X = L_{2}(\left[0,1\right]), x_{n}(t) = \sin \left(n t^{2}\right)$;

$X = L_{2}(\left[0,1\right]), x_{n}(t) = \sin \left(n e^{t}\right)$;

$X = L_{p}(\left[0,1\right]), 1 < p < +\infty , x_{n}(t) = x_{0}\left(2^{n} t\right), t \in [0,1]$, where $x_{0}(t) = (-1)^{k}$ for $t \in [k, k+1), k \in \mathbb {N} \cup \left\{ 0\right\}$;

$X = L_{1}(\mathbb {R}), x_{n}(t) = \chi_{\left[n, n+1\right]}(t)$;

$X = l_{1}, x_{n} = e_{n}$.

$X = L_{2}(\left[0,1\right]), f_{n}(x) = \int_{0}^{1} \cos 2 \pi n t x(t) d t$;

$X = L_{2}(\left[0,1\right]), f_{n}(x) = \int_{0}^{1} e^{2 \pi i n t} x(t) d t$;

$X = C(\left[0,1\right]), f_{n}(x) = n \int_{0}^{\frac{1}{n}}(1-2 n t) x(t) d t ;$

$X = C(\left[0,1\right]), f_{n}(x) = n \int_{0}^{1} x(t) t^{n} d t ;$

$X = C^{1}(\left[-1,1\right]), f_{n}(x) = \frac{n}{2}\left(x\left(\frac{1}{n}\right)-x\left(-\frac{1}{n}\right)\right)$.

Prove that for each $x \in C^{1}(\left[-1,1\right])$ the sequence $\left\{ f_{n}(x): n \geq 1\right\}$ is convergent, and find its limit.

Prove that the sequence $\left\{ f_{n}: n \geq 1\right\}$ does not converge weakly-*.

Prove that the sequence $\left\{ \left\| f_{n}\right\| : n \geq 1\right\}$ is not bounded.

Do these statements contradict the Uniform boundedness principle?

Let the functionals $\left\{ f_{n}: n \geq 1\right\} \subset c_{0}^{*}$ be defined as follows: $f_{n}(x) = x_{n}$, $x \in c_{0}, n \in \mathbb {N}$. Prove that $f_{n} \xrightarrow {w-*} 0$, but $f_{n} \not\xrightarrow {w} 0, n \rightarrow \infty$.

Let the functionals $\left\{ f_{n}: n \geq 1\right\} \subset l_{1}^{*}$ be defined as follows: $f_{n}(x) = \sum_{k = n+1}^{\infty } x_{k}$, $x \in l_{1}, n \in \mathbb {N}$. Prove that $f_{n} \xrightarrow {w-*} 0$, but $f_{n} \not\xrightarrow {w} 0, n \rightarrow \infty$.

Let the functionals $\left\{ f_{n}: n \geq 1\right\} \subset (C(\left[0,1\right]))^{*}, f_{0} \in (C(\left[0,1\right]))^{*}$ be defined as follows: $f_{n}(x) = x\left(\frac{1}{n}\right), n \in \mathbb {N}, f_{0}(x) = x(0), x \in C(\left[0,1\right])$. Prove that $f_{n} \xrightarrow {w-*} f_{0}$, but $f_{n} \not\xrightarrow {w} f_{0}, n \rightarrow \infty$.

$x_{n} \xrightarrow {w} x$ if and only if $\left(x_{n}, a\right) \rightarrow (x, a), n \rightarrow \infty$, for all $a \in H$;

if $x_{n} \xrightarrow {w} x$ and $y_{n} \rightarrow y$ in norm, then $\left(x_{n}, y_{n}\right) \rightarrow (x, y), n \rightarrow \infty$;

every orthonormal sequence converges weakly to zero;

if $x_{n} \xrightarrow {w} x$ and $\left\| x_{n}\right\| \rightarrow \left\| x\right\|$ (or $\limsup_{n \rightarrow \infty }\left\| x_{n}\right\| \leq \left\| x\right\|$), then $x_{n} \rightarrow x$ in norm.

$x_{n} \xrightarrow {w} x, y_{n} \xrightarrow {w} y$, where $x, y \in H$, but $\left(x_{n}, y_{n}\right) \nrightarrow (x, y)$;

$x_{n} \xrightarrow {w} x, y_{n} \xrightarrow {w} y$, where $x, y \in H, x_{n} \nrightarrow x, y_{n} \nrightarrow y$ in norm, but $\left(x_{n}, y_{n}\right) \rightarrow (x, y)$.

the series $\sum_{n = 1}^{\infty } x_{n}$ converges strongly;

the series $\sum_{n = 1}^{\infty } x_{n}$ converges weakly;

the series $\sum_{n = 1}^{\infty }\left\| x_{n}\right\|^{2}$ converges.

Let $X$ be a Banach space, and the sequence $\left\{ f_{n}: n \geq 1\right\} \subset X^{*}$ is such that for all $x \in X$ there exists $\lim_{n \rightarrow \infty } f_{n}(x) = f(x)$. Prove that $f \in X^{*}$ and $\left\| f\right\| \leq \liminf_{n \rightarrow \infty }\left\| f_{n}\right\|$.

Let $X$ be an NLS, and the sequence $\left\{ x_{n}: n \geq 1\right\} \subset X$ converges weakly to $x \in X$. Prove that $\left\| x\right\| \leq \liminf_{n \rightarrow \infty }\left\| x_{n}\right\|$.

$C(\left[0,1\right])$;

$c_{0}$;

$c$.

$x_{n} \rightarrow x, f_{n} \rightarrow f$ in norm;

$x_{n} \xrightarrow {w} x, f_{n} \rightarrow f$ in norm;

$x_{n} \rightarrow x$ in norm, $f_{n} \xrightarrow {w-*} f$.

every weakly closed set in $X$ is strongly closed;

the converse statement to statement (1) is false;

every subspace in $X$ is a weakly closed set;

every closed ball in $X$ is a weakly closed set.

Assume that a sequence $a = \left\{ a_{n}: n \geq 1\right\}$ is such that for every $x \in l_{p}$ the series $\sum_{n = 1}^{\infty } a_{n} x_{n}$ converges. Prove that $a \in l_{q}$.

Assume that a measurable function $h$ is such that for every $x \in L_{p}(\left[a, b\right])$ the Lebesgue integral $\int_{a}^{b} h(t) x(t) d t$ is finite. Prove that $h \in L_{q}(\left[a, b\right])$.

Let $p_{n}(t) \leq p_{n+1}(t)$ almost everywhere on $\left[0,1\right]$ with respect to the Lebesgue measure for all $n \in \mathbb {N}$, and also for each $x \in C(\left[0,1\right])$ the sequence $\left\{ f_{n}(x): n \geq 1\right\}$ is bounded. Show that the sequence $\left\{ f_{n}: n \geq 1\right\}$ converges weakly-*.

Let $f_{n} \xrightarrow {w-*} 0$. Is it necessarily true that there exists a subsequence of the sequence $\left\{ p_{n}: n \geq 1\right\}$, which converges almost everywhere on $\left[0,1\right]$?

The same question as in item (2), if $f_{n} \rightarrow 0$ in norm.

weakly-*?

in norm?

for each $x \in C(\left[a, b\right])$ and each $\varepsilon > 0$ the following inequality holds:

where the number $\delta = \delta (x, \varepsilon ) > 0$ is such that $\left|x(t)-x(c)\right| < \varepsilon$ for all $t \in [a, b]$, for which $\left|t-c\right| < \delta$.

(Korovkin) $f_{n} \xrightarrow {w-*} f$, if one of the following conditions is fulfilled:

(a) $f_{n}\left(e_{0}\right) \rightarrow f\left(e_{0}\right)$ and $f_{n}(y) \rightarrow f(y), n \rightarrow \infty$;

(b) $f_{n}\left(e_{k}\right) \rightarrow f\left(e_{k}\right), n \rightarrow \infty$, for $k = 0,1,2$.

$\forall k \geq 1: \lim_{n \rightarrow \infty } a_{n k} = 0$;

$\lim_{n \rightarrow \infty } \sum_{k = 1}^{\infty } a_{n k} = 1$;

$\exists C > 0 \forall n \geq 1: \sum_{k = 1}^{\infty }\left|a_{n k}\right| \leq C$.

The mapping $\sum_{n = 1}^{\infty } a_{n} \mapsto \sum_{n = 1}^{\infty } a_{n} b_{n}$, defined by the sequence $\left\{ b_{n}: n \geq 1\right\} \subset \mathbb {C}$, transforms convergent series into convergent ones if and only if $\sum_{n = 1}^{\infty } \mid b_{n+1}- b_{n} \mid < +\infty ;$

The series $\sum_{n = 1}^{\infty } a_{n} b_{n}$ is convergent for all sequences $\left\{ a_{n}: n \geq 1\right\} \subset \mathbb {C}$, for which the partial sums $\sum_{k = 1}^{n} a_{k}, n \geq 1$, are uniformly bounded, if and only if $b_{n} \rightarrow 0$, $n \rightarrow \infty$, and $\sum_{n = 1}^{\infty }\left|b_{n+1}-b_{n}\right| < +\infty$.