# Tagged Questions

A Banach space is a complete normed vector space: A vector space equipped with a norm such that every Cauchy sequence converges.

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### Does the open mapping theorem imply the Baire category theorem?

A nice observation by C.E. Blair1, 2, 3 shows that the Baire category theorem for complete metric spaces is equivalent to the axiom of (countable) dependent choice. On the other hand, the three ...
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### Is it possible for a function to be in $L^p$ for only one $p$?

I'm wondering if it's possible for a function to be an $L^p$ space for only one value of $p \in [1,\infty)$ (on either a bounded domain or an unbounded domain). One can use interpolation to show that ...
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### Norm for pointwise convergence

Does there exist a norm on the space of all real-valued functions on the real line (or on an open set? a compact set?) such that convergence in this norm is equivalent to pointwise convergence?
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### Norms on C[0, 1] inducing the same topology as the sup norm

This is an old homework problem of mine that I was never able to solve. The solution may or may not involve the Baire category theorem, which I am terrible at applying. Let $C[0, 1]$ denote the ...
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### Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.

Let $X$ be an infinite dimensional Banach space. Prove that every basis of $X$ is uncountable. Can anyone help how can I solve the above problem?
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### Strong and weak convergence in $\ell^1$

Let $\ell^1$ be the space of absolutely summable real or complex sequences. Let us say that a sequence $(x_1, x_2, \ldots)$ of vectors in $\ell^1$ converges weakly to $x \in \ell^1$ if for every ...
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### If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic

If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic unless $p=2$. Maybe I would have to use the Rademacher's functions.
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### Space of Complex Measures is Banach (proof?)

How can we prove that the space of Complex Measures is Complete? with the norm of Total Variation. I have stuck on the last part of the proof where I have to prove that the limit function of a Cauchy ...
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### Nested sequences of balls in a Banach space

This seems to be a fairly easy question but I'm looking for new points of view on it and was wondering if anyone might be able to help (by the way- this question does come from home-work, but I've ...
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### If weak topology and weak* topology on $X^*$ agree, must $X$ be reflexive?

Let $X$ be a Banach space and suppose that the weak topology on $X^*$ agrees with the weak* topology on $X^*$. Must $X$ be reflexive? To prove the contrapositive, it will suffice to assume that $X$ ...
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### Banach spaces isomorphic to square

This is another exercise from Allan's book "Introduction to Banach Spaces and Algebras". Exercise 2.9: A Banach space $E$ is said to be homeomorphic to its square if $E\oplus E$ is linearly ...
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### A Hamel basis for $l^{\,p}$?

I am looking for an explicit example for a Hamel basis for $l^{\,p}$?. As we know that for a Banach space a Hamel basis has either finite or uncountably infinite cardinality and for such a basis one ...
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### Isomorphic embedding of $L^{p}(\Omega)$ into $L^{p}(\Omega \times \Omega)$?

Let $(\Omega,\mu)$ be a finite measure space such that $\mu(\Omega)=1$. Suppose $1\leq p \leq \infty$. Let $\psi \colon L^p(\Omega) \to L^p(\Omega \times \Omega)$ be the map which maps $f$ onto the ...
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### $\ell_\infty$ a Grothendieck space

The problem I am considering stated formally is this: Show that if a sequence in $\ell_\infty^*$ is weak*-convergent, then it is also weakly convergent. We may reduce this to the case where the ...
### Ideals of the algebra of all bounded linear operators on $\ell_p \oplus \ell_q$
Let $\mathcal{L}(X)$ be the algebra of all bounded linear operators from $X$ to $X$ for Banach space $X$. I need to show that $\mathcal{L}(\ell_p \oplus \ell_q)$ for $p \neq q$ contains at least two ...
Let $B$ be a (complex) Banach space. A function $f : \mathbb{C} \to B$ is holomorphic if $\lim_{w \to z} \frac{f(w) - f(z)}{w - z}$ exists for all $z$, just as in the ordinary case where \$B = \mathbb{...