Why isn't this a counterexample of Banach-Steinhaus theorem?

The theorem from Wikipedia is as follows

Let $X$ be Banach space, $Y$ be a normed space and $F$ be family of linear bounded operators $f:X \to Y$ such that $\forall x \in X \sup_{f \in F} \|f(x)\|_Y < \infty$. Then $$ \sup_{f \in F ,\ \|x\| = 1} \| f(x)\|_Y = \sup_{f \in F} \| f\| < \infty $$

I thought the following "counterexample"

Let $X = Y$ be a Banach space. It normalised linear (Hamel) basis $A = \{e_\alpha\}_{\alpha\in J}$ is uncountble ($|A| \ge 2^{\aleph_0}$) and let $J = \{e_j\}_{j \in \mathbb N} \subset A$ be a (countable) sequence thereof. Now let $\{f_n\}_{n\in \mathbb N}$ the countable family of operators, such that $$ f_n(e_\alpha) = \begin{cases} e_\alpha & e_\alpha \in A\backslash J \\ \frac{1}{\frac{1}{j+1} + \frac{1}{n+1}}e_j & e_\alpha = e_j \in J. \end{cases} $$ i.e. all what an operator does here is that it scales a countably infinite basis vectors by bounded sequence, from this I believe that these operators are bounded with $\|f_n\| = n+1$ (which may turn out not to be the case, as the comments below suggest). Considering $$ \|f_n(e_\alpha)\| = \begin{cases} 1 & e_\alpha \in A\backslash J \\ \frac{1}{\frac{1}{j+1} + \frac{1}{n+1}} & e_\alpha = e_j \in J \end{cases} $$ we find $\sup_{n \in \mathbb N}\|f_n(e_\alpha)\| = 1$ or $j+1$ (depending on $\alpha$), I thought that is enough to conclude, (since $x = \sum x_k e_k$) that condition $$ \sup_{n \in \mathbb N}\|f_n(x)\| \le \sup_{n \in \mathbb N} \sum_k |x_k| \|f_n(e_k)\| < \infty $$ for every $x \in X$ is statisfied. Now if we take the sequence $(\|f_n\|)_{n \in \mathbb N}$ it is not bounded.

What is the mistake ?

Added: to disprove this is a counerexample, I think, it is sufficient to show that in any norm on $X$ which has the Hamel basis $\{e_\alpha\}_{\alpha \in A}$ normalised we have (infinitely many elements in) $\{f_n\}_{n \in \mathbb N}$ not bounded. Otherwise there is a norm such that (most of) these maps are bounded and the sequence $(\|f_n\|)_{n \in \mathbb N}$ is not bounded.

Solution 1:

EDIT: turns out the "wrong counterexample" I present below is wrong for different reasons than that in the OP. I'll still leave it here, because in the future someone might ask the same question as the OP but based on a different "wrong counterexample".

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Your idea works (or doesn't) the same way in Hilbert spaces, where you don't have to worry about Hamel bases and choices of topology. To be precise, if you believe that your example is a counterexample, you probably also will accept this as a counterexample:

For a Hilbert space with countable orthonormal Hilbert basis $(x_n)_{n>0}$, define the diagonal operators $$ f_m(x_n)=\delta_{m, n}nx_n $$ which are all bounded, with $\|f_m\|=m$. For all $n$ we have $\sup_m \|f_m(x_n)\|=n<\infty$. Yet $\sup \|f_m\|=\infty$.

What went wrong? This sentence:

For all $n$ we have $\sup_m \|f_m(x_n)\|=n<\infty$.

is not the same as what you really need to apply Banach-Steinhaus, namely

For all $x$ we have $\sup_m\|f_m(x)\|<\infty$.

The reason they are not the same is that an arbitrary, non basis element, $x$ can combine the basis elements such as to get the worst out of each $f_m$. To be precise, you can define $x:=\sum_{n} a_n x_n$ such that $\|x\|=1$, yet $\|f_m(x)\|$ is unbounded. To be even more precise, for $$ a_n=\begin{cases} 1/2^j & n=4^j\\ 0 &\text{else} \end{cases} $$ you have $\|f_{4^j}(x)\|= \|4^j a_{4^j}x_{4^j}\|=2^j$ and therefore $\sup_m \|f_m(x)\|=\infty$.

PS: the principle I used is called condensation of singularities and is a popular way to prove Banach-Steinhaus (without using the Baire category theorem, and therefore getting slightly weaker statements): Assume the sequence of operators is not bounded. Then construct an $x$ using near-orthogonality (Riez lemma) that gets the almost-worst of each operator, and show that this $x$ violates the assumption $\sup_{m}\|f_m(x)\|<\infty$. Normally, the proof is done here because you derived a contradiction, but in your case, it's not a contradiction but rather points out the assumption that's not satisfied, thus making Banach-Steinhaus not hold in your example.

Solution 2:

Your question is:

What is the mistake ?

The mistake is precisely in this sentence:

I believe that these operators are bounded with $\|f_n\| = n + 1$.

In order to have a counter example, you need to prove this statement above. But this statement is false. You cannot conclude that the $f_n$ are bounded. I will show you an example where your construction gives an unbounded operator.

Of course, it will always be unbounded simply because the theorem you want to disprove is true. Unless you break something else.

I think that the fact that you made $\|f_n(e_\alpha)\| \leq n + 1$ is giving you the impression that $\|f_n\| \leq n + 1$.

I will show to you that your construction gives a discontinuous (and therefore unbounded) $f := f_1$ in many cases.

Consider the vector $$\vec{v} = \sum_{j = 1}^{\infty} \frac{1}{2^j} e_j.$$

Now, since $A$ is a Hamel basis, there are $a_1, \dotsc, a_p \in A \setminus J$ and $b_1, \dotsc, b_q \in J$ such that $$ \vec{v} = \sum_{s=1}^{p} \alpha_s a_s + \sum_{t=1}^{q} \beta_t b_s $$ for appropriate $\alpha_s$ and $\beta_t$. In other words, $$ \vec{w} := \vec{v} - \sum_{t=1}^{q} \beta_t b_s = \sum_{s=1}^{p} \alpha_s a_s $$ is in the span of $A \setminus J$, and can also be written as a series in $J$.

Linearity of $f$ implies that $f(\vec{w}) = \vec{w}$, because the $a_s$ are fixed points.

In many cases, continuity would imply that $f(\vec{w}) \neq \vec{w}$. Notice, for example, that $$f(\vec{w}) - \vec{w} = \sum_{j=1}^\infty k_j e_j,$$ where only finitely many $k_j$ can be nonzero. So, for example, if we are on a Hilbert space, and $e_j$ are orthogonal, this cannot be zero, because $k_j = (f(\vec{w}) - \vec{w}) \cdot e_j$. Since there are nonzero $k_j$, $f(\vec{w}) - \vec{w} \neq 0$.

Edit: I have completely changed the example because my first attempt was wrong. Thanks to @LorenzoPompili.

Solution 3:

Edit: this is still somewhat a partial answer, as I am not checking explicitly that the operators in the question are unbounded. Check my other answer in this page for a more complete one.

What I want to do here is try to give a slightly different example to give an idea on why defining an operator to be “bounded on all elements of a Hamel basis” doesn’t necessarily mean that the operator is bounded. I am trying to exhibit an example with the same spirit of the one in the question, but easier do deal with (check my other answer for a more general approach that works in your case).

Assume the Hamel basis is all made by unit vectors. Define a linear operator $T_n$ to be $$T_n(e_\alpha)=\left\{\begin{aligned}&e_\alpha&&\text{if }\alpha\in A\setminus J\\& (1+n\delta_{n,j}) e_j&&\text{if }\alpha=j\in J\end{aligned}\right.$$

This sequence has essentially the same properties of your sequence (namely, it is pointwise bounded and clearly $\|T_n\|\geq n+1$ if the norm is finite). We can very trivially write the above operator as $$ T_n(x)=x+n\varphi_n(x)e_n,$$ where $\varphi_\alpha(x)$ is defined as the $\alpha$-th coordinate of $x$ in the Hamel basis (namely, $\varphi_\alpha(x)=x_\alpha$ if $x=x_\alpha e_\alpha+\sum_{\beta\neq\alpha} x_\beta e_\beta$, where the sum is finite). For every $\alpha$, $\varphi_\alpha\colon X\to \mathbb R$ is well-defined and linear.

Since the identity operator is continuous on $X$, and since $e_n$ is a fixed vector in $X$, $T_n$ is continuous iff $\varphi_n$ is continuous. Now it all reduces to the fact that among the $\varphi_\alpha$, i.e., the coordinate functions with respect to a Hamel basis, only a finite number of them is continuous.

Proposition. Let $X$ be an infinite dimensional Banach space and $\{e_\alpha\}_{\alpha\in A}$ a Hamel basis of $X$. Then, there exists a finite set $B\subset A$ such that $\varphi_\alpha$ is not continuous for all $\alpha\in A\setminus B$.

Proof. Assume if possible to have $e_i$, $i\in \mathbb N$ a sequence contained in the Hamel basis such that $\varphi_i$ is continuous. Consider $v_n=\sum_{i=1}^n \frac{1}{n^2}e_n.$ Since this is a banach space, $v_n\to v\in X$. But $\varphi_i(v_n)\to \frac{1}{i^2}$ for fixed n. Therefore, by continuity, $\varphi_i(v)\neq 0$ for all $i\in \mathbb N$, which is a contradiction.

(See also Are the coordinate functions of a Hamel basis for an infinite dimensional Banach space discontinuous?).

This means that only a finite number of the $T_n$ can be continuous. So, as expected, the Banach Steinhaus theorem is safe and we haven’t broken math.