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.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Pedro
    Nov 21, 2021 at 14:41

4 Answers 4


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".


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.


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.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Pedro
    Nov 22, 2021 at 14:12

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.


Final answer. I think I finally have a definitive answer to your question, at least I hope so. In what follows I prove that all but a finite number of the operators $f_n$ are unbounded without directly using the Banach-Steinhaus theorem and without making extra-assumptions on the space or on the Hamel basis. I guess this was the question that you had in mind from the beginning reading your comments. I think that in any case what I wrote could be interesting to read (it would be interesting for my past self of some hours ago).

Just a small note on the notation: in my answer, $A$ and $J$ refer to the sets of indexes, while in your question they denote the corresponding sets of elements of the Hamel basis.

Consider the space $C=\operatorname{span}\{e_\alpha\}_{\alpha\in A\setminus J}$. The closure of this space $\overline C$ is a Banach space. Since $C$ has countable codimention, $\overline C$ must necessarily have finite codimension. In fact, the quotient space $X/\overline C$ is a Banach space of at most countable dimension, therefore its dimension has to be finite. I’ll split the proof in two cases just for the sake of a clear exposition.

Case 1.

If $\overline C=X$, then the operators you have defined are all unbounded. In fact, any continuous linear operator $T$ such that $T(e_\alpha)=e_\alpha,\;\forall \alpha\in A\setminus J$ must be the identity on $C$. By continuity, it has to be the identity operator on the whole $\overline C=X$.

Case 2.

Assume instead that $\overline C$ has nonzero codimension. Then, $X=Z\oplus\overline C$, with $\operatorname{dim}(Z)=k<\infty$. Since $\{e_\alpha\}_{\alpha\in A} $ is a Hamel basis, we can assume that $k$ elements in the Hamel basis are a basis of $Z$ (we are constructing $Z$ in this way, e.g., completing a Hamel basis of the vector space $\overline C$ with element from $\{e_\alpha\}_{\alpha\in J}$). Without loss of generality, assume that $Z=\operatorname{span}\{e_1,…,e_k\}$ (you should see in what follows that this is not compromising your example).

Now, assume we have a continuous linear operator $T$ such that $T(e_\alpha)=e_\alpha,\;\forall \alpha\in A\setminus J$. As before, $T$ must be the identity on $\overline C$. Therefore, by the direct sum decomposition of $X$, the operator $T$ is uniquely determined by the values $T(e_1),…,T(e_k)$. Just this should already suggest that in general, the operators you wrote are not continuous, as you are choosing more elements than you should in the definition of the operators than those you need to define the linear continuous operator (the operator is overdetermined).

Now, call $M_T:=\sup\{\|T(x)\|\;|\;x\in Z,\;\|x\|=1\}$, which is finite by the boundedness of $T$. I claim that $$ \|T\|\leq K,$$ where the constant $K$ depends only on $M_T$, $X$, the choice of the Hamel basis $\{e_\alpha\}_{\alpha\in A} $, the choice of $J$ and the choice of the elements $e_1,…,e_k$ among those in $J$. This follows from the fact that, if |x|=1, $$\|T(x)\|\leq\|T(z)\|+\|T(c)\|\leq DM_T+\|c\|\leq DM_T+D,$$ where $x=z+c$, $z\in Z$, $c\in \overline C$ is the unique decomposition of x in the direct summands, and where the constant $D$ (depending on $X$, $\overline C$, and $Z$) is such that $$\|c\|+\|z\|\leq D\|x\|$$ (in fact, the norm of the Banach space $X$ is equivalent to the sum of the norms of the spaces involved in the direct sum; in other words, the direct sum $X=Z\oplus \overline C$ is actually a direct sum of Banach spaces).

Now, this is not the case for your sequence of operators. Taken $k$ indexes among the set $J$, the corresponding number $M_{f_n}$ is bounded by a constant which is independent of $n$ (this is an easy consequence of the pointwise boundedness of the sequence $f_n$). It follows from what I wrote above that your sequence of operators, if made by continuous operators, must be uniformly bounded by a constant, which is not true by construction. It follows that all the operators starting from a certain index are all unbounded.

Comment 1. The existence of the constant $D$ above is a consequence of the open mapping theorem. More precisely (very briefly), if you have two norms on a vector space $X$ such that both of them make the space a Banach space, and you can bound one with the other (up to a multiplicative constant), then the two norms are equivalent. As a consequence, if a Banach space is the direct sum (as vector spaces) of two closed subspaces, then the sum is actually a direct sum of Banach spaces, i.e., with equivalence of norms. See this other post for more details.

Comment 2. Another way of proving that the operators $f_n$ are (almost) all unbounded in the second case is to make an interpolation with the identity operator. In fact, consider $T$ as above, with the restriction $$T(e_\alpha)=\lambda_\alpha e_\alpha\;\forall\alpha\in A$$ (as in your example). Let’s consider the linear combinations $$T_\theta=(1-\theta)\operatorname{Id}+\theta T.$$ The coefficients $\lambda_\alpha$ of the new operator $T_\theta$ will then be the corresponding combinations of the coefficients of $T$ and of $\operatorname{Id}$. It should then follow that the coefficients $\lambda_j$, $j>k$, which as I said above have to depend on the first ones (i.e., they are uniquely determined from $\lambda_j$, $1\leq j\leq k$), actually depend in an affine way on the coefficients $\lambda_j$, $1\leq j\leq k$ (here I mean that the dependence is linear up to an additive constant). Then, the uniform boundedness of the coefficients $\lambda_j$, $j>k$ should follow from the pointwise boundedness of all the indexes before and after the $k$-th one (maybe I should check better this last argument, but I think it works). Then again we would have a contradiction assuming an infinite number of the $f_n$ are bounded, as uniform boundedness does not hold by construction.

Comment 3. I think that in general it is not possible to say that a single one of them, e.g. $f_1$, is necessarily unbounded for all choices of $X$, the Hamel basis, etc… It very much depends on how you choose the Hamel basis. That would be a nice thing to prove if I am wrong, though.


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