# Tag Info

### Extreme point of the unit ball in H(U)

There is a famous result due to Szego that characterizes the extreme points of the closed unit ball in $H^{\infty}(\mathbb{D})$ (where recall that $H^{\infty}(\mathbb{D})$ denotes the space of all ...
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• 7,830
Accepted

### For a given $k \in \mathbb{N}$, define the mapping $A: X \to X$ by the rule $(Af)(x) = x^k f(x).$

To show that $A$ is linear means to show that $$A(\alpha f+\beta g)=\alpha Af+\beta Ag$$ or, equivalently, that $$x^k(\alpha f(x)+\beta g(x))=\alpha x^kf(x)+\beta x^kg(x).$$ This identity clearly ...
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### Kernel dimension is preserved under uniform convergence of bounded operators on a Hilbert space

The conclusion does not follow the opposite way. Assume $T$ is the operator defined by $Te_n=2^{-n}e_n,$ where $\{e_n\}_{n=1}^\infty$ is an orthonormal basis. For fixed $m$ choose a sequence of ...
• 32.7k
1 vote
Accepted

### I want to prove that $T$ belongs to $B(X)$, meaning it is a bounded linear operator.

I think that $f \in X^*$ should be a bounded linear operator as well. I'm assuming it is a operator of this kind: $f:X\to\mathbb{K}$ where $\mathbb{K}$ is a field equal to $\mathbb{R}$ or $\mathbb{C}$...
Accepted

### If $X$ is a normed space and $x \neq 0$ and $f(y)x = 0$ for every $f \in X^*$, then $y = 0$.

Hint: By Hahn-Banach there exists $f \in X^*$ such that $f(y) = \|y\|$
1 vote
Accepted

### Supremum Norm on Space of Continuous Functions

The issue is that there are unbounded continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(0)=0$, so there are elements of $C$ which are unbounded. As you stated in your question, the ...
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• 441
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1 vote
Accepted

### Does the closure of product of two ideals satisfy $\overline{I_1I_2}=\overline{I_1}\ \overline{I_2}$.

It suffices to show $\bar{I_1}\bar{I_2}$ is closed. Thus, it suffices to prove that, if $I_1, I_2 \subset A$ are two closed ideals, then $I_1I_2$ is closed. Recall that any closed ideal in a $C^\ast$-...
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### Closed range of $T^*T$ for Hilbert space map.
First suppose that $T$ is injective and has dense range. To show that $R(T)$ is closed, suppose that $Tx_n \to y$ in $H$. Then $T^* Tx_n \to T^* y \in R(T^*T)$ so that $T^*y =T^*Tx$ for some \$x \in ...