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visits member for 3 years, 6 months
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Jun
1
comment Please verify this definition of locally convex vector space
Sorry, I misread your question. Note that every normed space is locally convex.
Jun
1
comment Please verify this definition of locally convex vector space
Can you include the definition of locally convex vector space in your question please?
Jun
1
comment $V$ is open , then $V=\{x\in \mathbb R:f(x)>0\}$ for some continuous function $f$
@Etienne Too bad : ) Thank you for your comment!
May
31
comment $V$ is open , then $V=\{x\in \mathbb R:f(x)>0\}$ for some continuous function $f$
@Etienne Would it also work to take the convolution of the characteristic function of $V$ with a bump function?
May
31
comment Compact Operator <=> Separable Range
@Freeze_S Compact operators have much nicer properties than arbitrary operators. You can think of them as behaving a bit like operators on finite dimensional vector spaces.
May
31
comment Compact Operator <=> Separable Range
It's true that compact implies separable range. The other direction I suspect does not hold. Why not try to find a counterexample?
May
28
comment How to write this as one matrix?
What is the notation $(\cdot, \cdot)$?
May
28
comment Show that $Z_i\sim N(\mu_i,V_{ii})$
Why the downvote?
May
28
comment Show that $E(Z)=\mu, Cov(Z)=V$
Why the downvote?
May
27
comment Prove or disprove: If L : V → U and M : U → V are linear mappings such that (M ◦ L)(x) = x for all x ∈ V, then M is onto.
I think you're absolutely right, it's obvious and there is nothing to prove. Given any $x$ in $V$, $Lx$ maps to it.
May
27
comment Math Major: How to read textbooks in better style or method ? And how to select best books?
This is the second time that I felt frustrated that my reading is far too slow and I read your answer and then didn't feel so bad anymore. Wonderful.
May
27
comment Diagonal convergence
Oh I see, it's in the title: $x_n \subseteq x_r$.
May
27
comment Diagonal convergence
Is $x_r$ a subsequence of $x_n$? I assume $f \in C(X)$?
May
26
comment If limit of $f(x)$ exists and the limit of $f(x)g(x)$ exists, then does the limit of $g(x)$ exist?
I'm not only giving the idea, I'm even giving a hint on solving the exam question (which the OP doesn't even ask for)
May
26
comment If limit of $f(x)$ exists and the limit of $f(x)g(x)$ exists, then does the limit of $g(x)$ exist?
@SanathDevalapurkar Let me copy and paste the question for you: "Whats the idea behind this question:".
May
26
comment If limit of $f(x)$ exists and the limit of $f(x)g(x)$ exists, then does the limit of $g(x)$ exist?
@SanathDevalapurkar No actually. Read the last sentence.
May
26
comment Question on finitely presented algebra
Are you asking for a proof of the statement?
May
26
comment How to figure whether it is a compact operator
Dear @kahen, thank you for your comment!
May
26
comment Dimension of vector space (count 0 or not)
Try to think of different bases of $\mathbb R^3$: there are many different bases, all of which consist of $3$ vectors but none of them contain the zero vector.
May
26
comment Dimension of vector space (count 0 or not)
@1234 No, the $0$ vector is strictly required by the definition of a vector space. If a space does not contain $0$ then it's not a vector space. But any set of vectors that form a basis for the space cannot contain $0$ otherwise they are not a basis.