11h
comment Lp spaces are nested but then why is 1/x square summable but not summable?
It requires that the measure space be finite.
22h
comment Prove (xn) does not converge to real number x
Can you show the sequence $(-1)^n$ does not converge?
1d
comment $ l^{\infty}$ is not compact.
Well, if $\ell_\infty$ were compact, then your set would be as well.
1d
comment $ l^{\infty}$ is not compact.
You're OK: A closed subset of a compact set is ...
1d
comment Does $\int_0^{\infty}\frac{x\hspace{1mm}dx}{x^3+1}$ converge?
$0$ is a problem here...
1d
comment Is a Riemann integrable function always measurable?
Your function is Lebesgue measurable (and Lebesgue integrable). As you show, a Riemann integrable function need not be Borel measurable. This is also addressed here.
2d
comment Showing a series is not the fourier series of a riemann integrable function.
If you assume $f$ is Riemann integrable and is represented by your series, then the right hand side of Bessel's inequality is finite by assumption (since $f^2$ would be Riemann integrable), but the left hand side, $\sum_{n=1}^\infty 1/(\sqrt n)^2$, is infinite. (So, no need to evaluate the integral; you're assuming it's finite.)
2d
comment Showing a series is not the fourier series of a riemann integrable function.
You could show that Bessel's Inequality would not hold.
Oct
22
awarded Enlightened
Oct
19
comment Why ${1\over n+1}+ \cdot\cdot\cdot +{1\over 2n}>{n\over 2n} $?
Each term is at least $1/(2n)$.
Oct
18
comment Is there a basis for the continuous functions space?
Yes. In fact, every vector space has a Hamel Basis. For the space in question, see this.
Oct
17
comment How can this series be calculated?
See this for some ideas.
Oct
17
comment Different ways of approximation of $\pi$
u r using 2 mny abbr
Oct
15
comment Normed space and convex hull of closed subset
If you draw a picture of $S$ and typical line segments with left endpoint in $\{0\}\times[0,1]$ and right endpoint on the nonnegative $x$-axis, you should see it.
Oct
13
comment Separable dual space implies existence of complete series
If $x=\sum_{i=1}^m\alpha_i f_i$, then by linearity $f_n^* x=\sum_{i=1}^m\alpha_i f_n^* f_i $. (Actually, by continuity of $f_n^*$, this holds for $x$ expressed as an infinite sum as well.)
Oct
13
comment Separable dual space implies existence of complete series
Yes. It follows that $f_n^*$ agrees with the coefficient functional of $f_n$ on a dense subset of $c_0$ (the finite linear combinations of the $f_n$). So $f_n^*$ is indeed the coefficient functional of $f_n$.
Oct
13
comment Separable dual space implies existence of complete series
Sorry, $0$ if $n\ne m$, $1$ if $n=m$. (Should have been a "$\delta$"; it's edited now.)
Oct
13
comment Separable dual space implies existence of complete series
You can verify directly that $f_n^* f_m=\delta_{n,m}$.
Oct
13
comment Separable dual space implies existence of complete series
Consider the summing basis of $c_0$, $(f_n)$ defined by $f_n=\sum_{i=1}^n e_i$. The coefficient functionals are given by $f_n^*=e_{n}-e_{n+1}$. Their closed linear span does not contain $e_1$.
Oct
13
comment Mazing questions of Compacted sets
$\{0\}\cup\{1/n\mid n\in\Bbb N\}$ has exactly one limit point. Add a similar set "around $2$".
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