Application of the uniform boundedness Principle.

functional-analysismeasure-theorysolution-verification

Here is the question:

Let $A = [a_{ij}]_{i,j = 1}^{\infty}$ be an infinite matrix of real numbers and suppose that, for any $x \in \ell^2,$ the sequence $Ax$ belongs to $\ell^2.$ Prove that the operator $T,$ defined by $T(x) = Ax,$ is a bounded operator on $\ell^2.$

**Here is my trial: **

We will use the uniform bounded theorem Principle, since we showed that $(\ell^2,\| x\|_{2})$ is a normed space. and it is easy to show that its complete by showing that every Cauchy sequence converges using $\| .\|_{2}$, then $(\ell^2 , \| x\|_{2})$ is a Banach space and since by the given we have that for any $x \in \ell^2,$ the sequence $Ax$ belongs to $\ell^2.$ Which can be formulated mathematically as follows:

According to the givens we can define $A : \ell^2 \rightarrow \ell^2 $ by $$A x = A (\xi_{1}, \xi_{2}, …) = (\beta_{1}, \beta_{2}, …), $$
Where

$$
[a_{ij}]\begin{bmatrix}
\xi_{1} \\
\xi_{2} \\
\vdots
\end{bmatrix}=\begin{bmatrix}
\beta_{1} \\
\beta_{2} \\
\vdots
\end{bmatrix}
$$

i.e., $\beta_{i} = \sum_{j=1}^{\infty} a_{ij} \xi_{j}$ and $ \|Ax\|^{2} = \sum_{i=1}^{\infty} |\beta_{i}|^2 < \infty $. Hence $\sup \{\|A x \|\} < \infty $ and hence $\sup \{\|T \|\} = \sup \{\|A \|\} < \infty $ as required.

Is my solution correct?

Best Answer

Hint: Define $T_Nx= (\sum\limits_{k=1}^{N} a_{ij}x_j)$. Verify that $T_N$ is a bounded opeartor and $(\|T_N(x)\|)$ is a bounded sequence for each fixed $x$. Apply UBP to finish the proof.

Related Solutions

Functional Analysis – Uniform Boundedness Principle for Norm Convergence

First of all, in order to have $(iii)$ be equivalent to $(i)$ and $(ii)$, we need that $Y$ is complete (a Banach space) too.

As a counterexample when $Y$ is not complete, consider $X = \ell^p$, and $Y$ the subspace of sequences with only finitely many nonzero terms. Then let

$$(T_nx)_k = \begin{cases}x_k &, k \leqslant n\\ 0 &, k > n. \end{cases}$$

$T_n$ is a bounded sequence of linear operators $X\to Y$, and $T_n(x)$ converges for all $x$ in the dense subspace $Y$ of $X$. But $T_n(x)$ does converge only for $x\in Y$, so neither $(i)$ nor $(ii)$ hold in this example.

Your attempt to prove $(i) \Rightarrow (iii)$ does not work, there are two problems. First, from $\lVert T_n(x)\rVert < N$, you cannot deduce $\lVert T_n\rVert_{op}\lVert x\rVert < N$, since you only have the inequality $\lVert T_n(x)\rVert \leqslant \lVert T_n\rVert_{op}\lVert x\rVert$, not equality.

Second, the bound on the sequence $T_n(x)$ depends on $x$, you only have $\lVert T_n(x)\rVert \leqslant N(x)$.

To show the uniform boundedness, you consider the closed sets $$A_K = \{x \in X : \lVert T_n(x)\rVert \leqslant K \text{ for all } n\}.$$

Since $T_n(x)$ is bounded for every $x\in X$, you have

$$X = \bigcup_{k=1}^\infty A_k.$$

From Baire's theorem, deduce that some $A_k$ is a neighbourhood of $0$ in $X$.

For the implication $(iii) \Rightarrow (ii)$ (under the assumption that $Y$ is complete!), note that

the set $C = \left\{ x \in X : \lim\limits_{n\to\infty} T_n(x)\text{ exists}\right\}$ is a linear subspace of $X$,
the map $T\colon C \to Y$ defined by $T(x) = \lim\limits_{n\to\infty} T_n(x)$ is linear and continuous.

Then the general theory asserts the existence of a continuous (linear) extension $\tilde{T}\colon \overline{C}\to Y$. Since $C$ is by assumption dense, $\overline{C} = X$. Now use the boundedness of the norms - the equicontinuity of the family $(T_n)$ - to deduce that actually $C = X$.

[Math] Converse of uniform boundedness principle

Note: Here are two examples of partial converses which could be convenient.

Partial converse in section 27 of A Hilbert Space Problem Book by Paul R. Halmos:

Every bounded subset of a Hilbert Space is weakly bounded.

We can read in section $27$: Uniform boundedness

The celebrated principle of uniform boundedness (true for all Banach spaces) is the assertion that a pointwise bounded collection of bounded linear functionals is bounded. The assumption and the conclusion can be expressed in the terminology appropriate to a Hilbert space $\mathbf{H}$ as follows.

The assumption of pointwise boundedness for a subset $\mathbf{T}$ of $\mathbf{H}$ could also be called weak boundedness; it means that foreach $f$ in $\mathbf{H}$ there exists a positive constant $\alpha(f)$ such that $|(f,g)|\leq \alpha(f)$ for all $g$ in $\mathbf{T}$.

The desired conclusion means that there exists a positive constant $\beta$ such that $|(f,g)|\leq \beta\|f\|$ for all $f$ in $\mathbf{H}$ and all $g$ in $\mathbf{T}$; this conclusion is equivalent to $\|g\|\leq \beta$ for all $g$ in $\mathbf{T}$.

It is clear that every bounded subset of a Hilbert space is weakly bounded. The principle of uniform boundedness (for vectors in a Hilbert space) is the converse: every weakly bounded set is bounded.

And later on in section $51$: Uniform boundedness of linear transformations

... the generalization of the principle of uniform boundedness from linear functionals to linear transformations is somewhat subtler. The generalization can be formulated almost exactly the same way as the special case: a pointwise bounded collection of bounded linear transformations is uniformly bounded. The assumption of pointwise boundedness can be formulated in a weak manner and a strong one.

A set $\mathbf{Q}$ of linear transformations (from $\mathbf{H}$ into $\mathbf{K}$) is weakly bounded if for each $f$ in $\mathbf{H}$ and each $g$ in $\mathbf{K}$ there exists a positive constant $\alpha(f,g)$ such that $|(Af,g)|\leq \alpha(f,g)$ for all $A$ in $\mathbf{Q}$. The set $\mathbf{Q}$ is strongly bounded if for each $f$ in $\mathbf{H}$ there exists a positive constant $\beta(f)$ such that $\|Af\|\leq \beta(f)$ for all $A$ in $\mathbf{Q}$.

It is clear that every bounded set is strongly bounded and every strongly bounded set is weakly bounded. The principle of uniform boundedness for linear transformations is the best possible converse.

Partial converse in section $18.2$ of Mathematical Methods in Physics: Distributions, Hilbert Space Operators, and Variational Methods by Philipp Blanchard and Erwin Brüning

Every uniformly bounded family of continuous linear functionals is pointwise bounded

We can read in section $18.2$: The weak Topology:

$\mathbf{Definition\ 18.2.3}$ Let $X$ be a Banach space with norm $\|\cdot\|$ and $\{T_\alpha:\alpha\in A\}$ a family of continuous linear functionals on $X$ ($A$ an arbitrary index set). One says that this family is

$\mathbf{pointwise\ bounded}$ if, and only if, for every $x\in X$ there is a real constant $C_x<\infty$ such that $$\sup_{\alpha\in A}|T_\alpha(x)|\leq C_x$$

$\mathbf{uniformly\ bounded}$ or $\mathbf{norm\ bounded}$ if, and only if $$\sup_{\alpha \in A}\sup\{|T_\alpha(x)|:x\in X,\|x\|\leq 1\}=C< \infty.$$ Clearly, every uniformly bounded family of continuous linear functionals is pointwise bounded.

For a certain class of spaces the converse is also true and is called the principle of uniform boundedness or the uniform boundedness principle.

Maybe also interesting is the following generalisation to Hausdorff locally convex topological vector spaces stated in Appendix C: The Uniform Boundedness Principle

$\mathbf{Definition\ C.1.1}$ Suppose $(X,\mathcal{P})$ and $(Y,\mathcal{Q})$ are two Hausdorff locally convex topological vector spaces over the field $\mathbb{K}$. Denote the set of linear functions $T:X\rightarrow Y$ with $L(X,Y)$. A subset $\Lambda\subset L(X,Y)$ is called

a) $\mathbf{pointwise\ bounded}$ if, and only if, for every $x\in X$ the set $\{Tx:T\in\Lambda\}$ is bounded in $(Y,\mathcal{Q})$, i.e., for every semi-norm $q\in \mathcal{Q}$, $$\sup\{q(Tx):T\in\Lambda\}=C_{x,q}<\infty;$$ b) $\mathbf{equi}$-$\mathbf{continuous}$ if, and only if, for every semi-norm $q\in\mathcal{Q}$ there is a semi-norm $p\in\mathcal{P}$ and a constant $C\geq 0$ such that $$q(Tx)\leq Cp(x)\qquad\qquad\forall x\in X, \forall T\in\Lambda.$$ Obviously, the elements of an equi-continuous family of linear mappings are continuous and such a family is pointwise bounded. For an important class of spaces $(X,\mathcal{P})$ the converse holds too, i.e., a pointwise bounded family of continuous linear mappings $\Lambda\subset L(X,Y)$ is equi-continuous.

Note: The conclusion of uniform boundedness on bounded sets in the classical uniform boundedness principle for normed spaces is equivalent to the condition that the family of operators is equicontinuous with respect to the original topology of the domain spaces. See e.g. this paper by Ronglu Li and Charles Swartz.

Best Answer

Related Solutions

Functional Analysis – Uniform Boundedness Principle for Norm Convergence

[Math] Converse of uniform boundedness principle

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