Must Linear operators on entire Banach space $l^1$ be bounded?

Question

Assume that it is possible to construct an unbounded linear functional $f$ defined on entire Banach space $l^1$. Then let us consider in unit ball the standard linearly independent sequence $\{e_i\}_{i=1}^\infty$, $e_i=(\underbrace{0,0,\ldots,0}_{n-1},1,0,\ldots)$. We choose $k$ such that for subsequence $\{e_{i_k}\}$, $f(e_{i_k})\geq k^2$. Partial sums of the series $\sum_{k=1}^\infty {k^{-2}}e_{i_k}$ form a Cauchy sequence in $l^1$, converging to $x_0\in l^1$ with coordinates $\frac1{k^2}$ in the $i_k$ places and zeros otherwise. Then $\forall\ n,\ f(x_0)\geq \sum_{k=1}^n k^{-2} f(e_{i_k})\geq \sum_{k=1}^n 1=n$. Thus, functional $f$ is undefined on $x_0\in l^1$. Contradiction.
However, I found the claims that the axiom of choice allows to prove existence of unbounded functional defined on some Banach spaces.

Can I use this counterexample to conclude that space $l^1$ does not satisfy this property and there is no unbounded functionals defined on entire $l^1$?

Answer 1

Of course it's well known and not hard to show that the conclusion is false. Raising the question of where the error is. The following is at least one error that seems important:

You define $x_0$ to be a certain infinite sum; say the partial sums are $s_n$. You say that $f(s_n)$ has a certain property and then deduce something about $f(x_0)$. No! No, no, no. Since we don't know that $f$ is continuous, saying something about $f(s_n)$ says nothing about $f(x_0)$.

Hmm, the post does contain a correct proof that there cannot be a continuous unbounded functional, but of course we knew that...

Edit: The OP insists that he or she is not talking about convergence of infinite sums. But there's no justification given for the key inequality $f(x_0)\geq \sum_{k=1}^n k^{-2} f(e_{i_k})$, and in fact it need not hold.

The Truth of the Matter

Oops, that word "truth" is a little tricky here. I'm going to assume the axiom of choice and then show that AC implies there exists a discontinuous linear functional $f$ on $\ell^1$, in particular one such that $f(x_0)< \sum_{k=1}^n k^{-2} f(e_{i_k})$.

Does that prove it's true? Maybe not; it's known that we can't prove AC (speaking loosely), so you're free to disbelieve in AC, hence in the validity of the argument. But NOTE it's also known that we can't disprove AC. So, while a proof that AC implies the existence of an unbounded functional may not show that an unbounded functional exists, it does show that any proof that every functional is bounded must be wrong! (Because that would prove AC was false, known to be impossible.)

So, as of now we as.assume AC.

Exercise. Suppose $x=(x_j)\in\ell^1$ and $x_j\ne0$ for infinitely many $j$. Suppose $\alpha, \alpha_1,\alpha_2,\dots\in\Bbb R$. There exists a linear functional $f$ on $\ell^1$ such that $fx=\alpha$ and $fe_j=\alpha_j$.

Proof: The set $S=\{x\}\cup\{e_1,e_2,\dots\}$ is linearly independent (in the straight linear algebra finite linear combination sense), so this follows from elementary linear algebra (AC gives us a basis containing $S$, etc.)

Elementary Linear Algebra Suppose $X$ and $Y$ are vector spaces (over the same field), $S\subset X$ is independent, and $\phi:S\to Y$. There exists a linear map $f:X\to Y$ such that $f|_S=\phi$.

Answer 2

As Kavi Rama Murthy says in the comments, there is nothing that ensures that $f(e_i)\ge0$ for all $i$. Also, your argument cannot possibly be correct because of the following reason:

Let $X$ be an infinite dimensional Banach space and assume that every linear functional $X\to\mathbb{C}$ is bounded. Let $\{e_i\}_{i\in I}$ be a Hamel basis for $X$ consisting of vectors each of which has norm $1$. Since $\{e_i\}_{i\in I}$ is an infinite set, pick a countable subset, say $\{e_{i_k}\}_{k=1}^\infty$, so this is a linearly independent set.

Define $\phi:X\to\mathbb{C}$ by declaring $\phi(e_{i_k})=k^2$, $\phi(e_i)=0$ for $i\in I\setminus\{i_k:k\in\mathbb{N}\}$ and extend this linearly to the entire space $X$. By assumption, $\phi$ must be bounded, i.e. continuous. But, if $x=\sum_{k=1}^\infty\frac{1}{k^2}e_{i_k}$ (the series converges in $X$ because $X$ is Banach and the series is absolutely convergent), then $$\phi(x)=\sum_{k=1}^\infty\frac{1}{k^2}\phi(e_{i_k})=\sum_{k=1}^\infty1=\infty$$ a contradiction.

One can come up with a slightly more involved argument for infinite dimensional normed spaces in general, without assuming completeness, I think you need the Hahn-Banach theorem for this.