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"Nothing is more important than to see the sources of invention which, in my opinion, are more interesting than the inventions themselves."

--Gottfried von Leibniz (1646-1716)


1d
comment Uniform Boundedness Principle for $L^p( \mathbb{R})$
No. You would need to find a single $g$ such that $\bigl(\int f_{n_k}g\bigr) $ is unbounded. You might find it easier to prove your assertion directly, using the theorem named in your title.
1d
comment weak convergent sequence in $L^p(\mathbb{R})$ with $(1\leq p < \infty)$ implies norm is bounded
In a normed space, a weakly convergent sequence is norm bounded. If $X$ is a Banach space, then a weak* convergent sequence in $X^*$ is norm bounded.
1d
comment Continuity of Thomae's Function at Irrationals
See this.
2d
comment Why does this proof fail?[convergence of infinite sums]
Here, you're assuming $\sum_{n=1}^\infty x_n$ is defined at the start (that is, that it's convergent).
2d
comment There is no continuous mapping from $L^1([0,1])$ onto $L^\infty([0,1])$
It's fine, but might need a bit more detail for "and we can approximate $g$ with $Tf_{n_k}$ since $T$ is continuous". (You could show more generally that the continuous image of a separable space is separable.)
2d
revised When is an operator on $\ell_1$ the dual of an operator on $c_0$?
added 112 characters in body
Aug
25
comment Let $M$ ba any infinite set and $A$, any countable set. Prove $M\sim A\cup M$
See this.
Aug
25
comment Limit of a (pseudo) non-increasing sequence
For your second question, consider the sequence $$ 0,1,\textstyle{1\over2},0,\textstyle{1\over3},\textstyle{2\over3}, 1,\textstyle{3\over4},\textstyle{2\over4},\textstyle{1\over4},0,\textstyle{1\ove‌​r 5},\ldots $$
Aug
25
comment How to simplify $\frac{(3x^{3/2}y^3)} {(x^2y^{-1/2})}^{-2}$?
@user1534664 $-6-(-1/2)=-6+1/2=-12/2+1/2=(-12+1)/2=-11/2$. Also $y^{-11/2}={1\over y^{11/2}}$.
Aug
25
comment proving that to every $\delta$ exist $n\in$ $\mathbb{N} $
$\sin (0.5\delta)$ is positive within your range of $\delta$. So, you need to argue why there is an $n$ with $n> 1/(2\pi\sin(0.5\delta))$.
Aug
25
answered Is this set measurable?
Aug
25
comment Is this set measurable?
Actually, $A\cup\Bbb Q$ would be a counterexample for any non-measurable $A$.
Aug
25
comment Is this set measurable?
What if you took a non-measurable subset $A$ of the irrationals and set $E=A\cup\Bbb Q$?
Aug
25
comment Probability of picking two letters from the word MATHEMATICAL
The textbook's answer is wrong.
Aug
25
answered When is an operator on $\ell_1$ the dual of an operator on $c_0$?
Aug
24
comment When is an operator on $\ell_1$ the dual of an operator on $c_0$?
If $(x_n^*)$ is weak* null in $\ell_1$ then $Se_i=x_i^*$ gives an operator from $\ell_1$ to $\ell_1$. One can define a bounded operator from $c_0$ to $c_0$ via $T x=(x_n^* x)$. One then has $T^*=S$, no?
Aug
24
comment When is an operator on $\ell_1$ the dual of an operator on $c_0$?
Some thoughts: In general, a bounded operator between dual spaces is an adjoint operator iff it is weak*-weak* continuous. Also of use may be the fact that operators from $\ell_1$ to a Banach space $X$ are characterized by the bounded sequences in $X$ (given a bounded sequence $(x_i)$ in $X$, map $e_i$ to $x_i$). So in order to be an adjoint operator, $(T e_i)$ must be weak* null. An example: define $T:\ell_1\rightarrow\ell_1$ via $Te_1=e_1$ and $Te_j=e_1+e_j$, $j>1$. $T$ is continuous, in fact an isomorphism, but not weak*-weak* continuous.
Aug
24
answered Example of a rearrangement that diverges
Aug
24
comment Example of a rearrangement that diverges
It might be nasty to do so for what I wrote. But, take something like $S_1-1+S_2-{1\over2}+S_3-{1\over3}+\cdots$ where $S_1={1\over2}+{1\over4} $, $S_2={1\over 6}+{1\over 8}+{1\over 10}+{1\over12}$, $S_3={1\over14}+\cdots+{1\over28}$, $\ldots$. Each block $S_n$ is larger than $1/4$ (if I calculated things correctly). The series diverges since its partial sums are not Cauchy.
Aug
24
comment If function $f$ has zero value and positive derivative at both endpoints, then $f''(\eta)=f(\eta)$ for some $\eta$
$f'(0)\cdot f'(\pi)$ is negative for your $f$.