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Let $\{e_{\alpha}\}_{\alpha \in \mathcal{A}}$ be a Hamel basis of an infinite dimensional Banach space $(X, \| \cdot \|_X)$. Without loss of generality, we assume $\|e_{\alpha}\|_X = 1$. It is known that one can define an unbounded functional on $X$ with Hamel basis: choose countable subset $\{e_i\}_{i \in \mathbb{N}}$ from $\{e_{\alpha}\}_{\alpha \in \mathcal{A}}$, set $\varphi(e_i) = i$ and $\varphi(e_{\alpha}) = 0$ for any other element of Hamel basis. By linearity, since every element of $X$ can be uniquely expressed as a finite linear combination of elements from Hamel basis, $\varphi$ is a linear functional on $X$ but not bounded.

One can define $\varphi_n$ as a truncation of $\varphi$, i.e. for $1 \leq i \leq n$ we put $\varphi(e_i) = i$ and $\varphi(e_{\alpha}) = 0$ for all other indices. Then, $\varphi_n \in X^*$ with dual norm $\|\varphi_n \| = n$. Fix $x \in X$ and write $x = \sum_{i=1}^{n_x} a_i^x e_{j_i^x}$ where I used index $x$ to stress dependence of this decomposition on $x$. For any $k \in \mathbb{N}$, $$ \left|\varphi_k (x)\right| = \left|\sum_{i=1}^{n_x} a_i^x \varphi_k(e_{j_i^x}) \right| \leq \sum_{i=1}^{n_x} \left|a_i^x\right| \left|\varphi(e_{j_i^x})\right| < \infty $$ for fixed $x \in X$ independently of $k \in \mathbb{N}$. On the other hand, $\sup_{n \in \mathbb{N}} \| \varphi_n \| = \infty$. This would contradict Uniform Boundedness Principle.

Question: I would like to understand where is the mistake in this reasoning.

qbini
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    One problem could be the assumption that $\varphi_n$ is bounded. Just because a $\varphi_n$ map a Hamel basis contained in the unit sphere to a bounded set, doesn't mean necessarily that $\varphi_n$ is bounded. You can construct an example by finding a codimension $1$ dense subspace of an infinite-dimensional Banach space, creating a (normalised) Hamel basis for it, then extending it with one other norm-$1$ vector. If $\varphi$ maps every basis vector to $0$, except the one extension vector, then $\varphi$ has a kernel that is not closed, and hence $\varphi$ is not continuous. – Theo Bendit Oct 15 '19 at 07:23
  • I see! So the point is that even projection functionals on elements of Hamel basis can be unbounded like in this problem. Thanks! – qbini Oct 15 '19 at 14:04

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