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1d
revised Showing $f \equiv 0$ if $f$ is real analytic on $(a,b)$ and $f(x_k) = 0$ for a distinct sequence where $\lim_{k \to \infty} x_k = x_0$
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1d
comment Holomorphy on open unit disk and continuity to the closure implies absolutely convergence of coefficients?
@DavidC.Ullrich Sorry, I wasn't expressing myself clearly. I knew exactly what I meant, so it was clear to me, but … I should have said "For non-open $S$, it's common to define 'holomorphic on $S$ …'."
1d
comment Holomorphy on open unit disk and continuity to the closure implies absolutely convergence of coefficients?
@DavidC.Ullrich Here, however, we're looking at $A(\mathbb{D}) = C(\overline{\mathbb{D}}) \cap \mathscr{O}(\mathbb{D})$, not at the space of functions holomorphic (or analytic) on $\overline{\mathbb{D}}$.
1d
comment Holomorphy on open unit disk and continuity to the closure implies absolutely convergence of coefficients?
@DavidC.Ullrich One does not define "holomorphic" that way. As usual, if $U$ is open, a function $f\colon U \to \mathbb{C}$ is holomorphic if it is complex differentiable on $U$. Sometimes, one is interested in the family functions defined on a set $S$ that isn't open, and that are the restriction of holomorphic functions defined on some neighbourhood of $S$ - where the neighbourhood is allowed to depend on the function. Then it's convenient to call such functions "holomorphic on $S$".
1d
revised Given two metrics $\rho,\sigma$ on $\chi$ show that the following is a metric on $\chi$
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1d
revised For a normal operator is it true that $\|T^*T^2\| = \|T^3\|$?
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1d
comment For a normal operator is it true that $\|T^*T^2\| = \|T^3\|$?
The question originally had the tag functional-analysis, which I've just re-added. So likely, the OP considers infinite-dimensional spaces.
1d
revised For a normal operator is it true that $\|T^*T^2\| = \|T^3\|$?
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1d
revised Prove that $\|u - v\| \ge \|u\| - \|v\|$ for any $u, v \in \mathbb R^n$
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1d
reviewed Approve A fierce differential-delay equation: df/dx = f(f(x))
1d
revised Generalization of Cauchy-Schwarz to positive operators
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1d
answered Does the series converge
1d
comment Holomorphy on open unit disk and continuity to the closure implies absolutely convergence of coefficients?
You're having trouble proving it because the assertion is false. The map $\ell^1(\mathbb{N}) \to A(\mathbb{D})$ given by $(a_n) \mapsto \sum a_n z^n$ is clearly linear, injective and continuous (with norm $1$). If it were surjective, it would be a topological isomorphism. Can you think of an interesting family of functions in $A(\mathbb{D})$ of norm $1$?
1d
comment Holomorphy on open unit disk and continuity to the closure implies absolutely convergence of coefficients?
@DavidC.Ullrich It's common to define "holomorphic on $S$" as "holomorphic on some (unspecified) open set containing $S$".
1d
revised Use contraction mapping theorem to show that the integral equation has a unique continuous solution on $t \in [0,3]$
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1d
comment Finding the image of the annulus $\{1 < |z| < e\}$ under the principal logarithm
You can't define a (continuous) branch of the logarithm on that annulus.
1d
revised Simple inequality with a,b,c
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1d
comment Averge of sequence converges, sequence bounded?
Look for a counterexample. Let $a_n = 0$ for most $n$.