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  • 0 posts edited
  • 1 helpful flag
  • 10 votes cast
Apr
27
revised What is the norm of $H^4(0, 1) \cap H_0^2(0, 1)$?
typo
Apr
27
answered What is the norm of $H^4(0, 1) \cap H_0^2(0, 1)$?
Apr
16
comment How can I split this into its' real and imaginary parts, and simplify?
Please tell us next time what your real goal is (here: estimating $| \sum \cos(n) |$). Posting your own thoughts and results is great, but if you don't tell us what you really want to do it's hard for us to determine a) what you want, as in your other question, and b) whether you took a wrong/overcomplicated turn somewhere already.
Apr
16
comment How can I split this into its' real and imaginary parts, and simplify?
Yes. You should start to think of complex numbers geometrically: If $z = x + iy$ is the result of the sum we have shown that $|z|$ is bounded, i.e. the distance between the point $(x, y)$ and the origin on the plane $\mathbb R^2$ is bounded. This also means that both $x$ and $y$ are bounded.
Apr
16
comment How can I split this into its' real and imaginary parts, and simplify?
@volcanomane See my edit.
Apr
16
revised How can I split this into its' real and imaginary parts, and simplify?
added 281 characters in body
Apr
16
answered How can I split this into its' real and imaginary parts, and simplify?
Apr
16
comment ((a ⇔ b) ⇒ c) ⇔ (a ⇔ (b ⇒ c)) tautology, contradiction, or neither?
Thanks, I already changed that.
Apr
16
comment Lipschitz condition not satisfied
Because your argument shows that the fraction goes to infinity for $u, v \to 0$. Therefore it cannot be bounded by any constant $L$ (constant in the sense that must be independent of $u$ and $v$).
Apr
16
answered ((a ⇔ b) ⇒ c) ⇔ (a ⇔ (b ⇒ c)) tautology, contradiction, or neither?
Apr
16
comment Lipschitz condition not satisfied
Yes, your reasoning is sufficient.
Apr
16
answered Lipschitz condition not satisfied
Apr
14
comment $X$ is inner product space then its completion is Hilbert space?
You can complete any metric space by the technique mentioned and the completion is unique up to an isometry (Wikipedia).
Apr
14
revised $X$ is inner product space then its completion is Hilbert space?
added 194 characters in body
Apr
14
answered $X$ is inner product space then its completion is Hilbert space?
Apr
10
comment How to write the below proof rigorously for : If $p$ is a boundary point of $S$, then $p$ is a lim point of $S$
I assumed that $p \not \in S$ (as the original poster judging from his definition). How else does this not conform with the definition of a limit point?
Apr
10
revised How to write the below proof rigorously for : If $p$ is a boundary point of $S$, then $p$ is a lim point of $S$
added 353 characters in body
Apr
10
answered How to write the below proof rigorously for : If $p$ is a boundary point of $S$, then $p$ is a lim point of $S$
Apr
2
comment What initial value do I have to take at the beginning?
Although I couldn't find a reason in your link, you're right of course — I missed the change of behavior of your curves at $y_0 = 0$. Even though everything went well here, I still think that parametrizing the initial value boundary of the domain for the construction of characteristic curves is the "safer" approach.
Apr
1
answered Finiteness condition