Nemis L.
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 Mar20 comment The function $f(z)=|z|^2$ is only differentiable at the origin $f$ is complex differentiable if and only if the real and imaginary parts satisfy the Cauchy--Riemann equations. Feb17 comment Preimage of Legendre-Fenchel transform In the non-convex case it is exactly the same. But I would rather have some more direct statements .. Feb17 comment Preimage of Legendre-Fenchel transform Yes, I am interested in the case where $X$ may be nonreflexive, but also reflexive $X$ are of interest. Feb17 asked Preimage of Legendre-Fenchel transform Feb11 comment Prove that $f$ is identically zero. Note that you can solve the differential equation: $df/f^2 = dx$. Integrate to get $f(x) = 1/[1/f(y) - (x-y)]$. for any $x,y$. Jan25 comment $\sum (\sqrt{x_k-k^2}-k)^2=0$ implies $x_k=2k^2$? If you want to explore possible complex solutions, it may be fruitful to consider polar coordinate parameterizations of $X = x-1$ and $Y = y-4$, rewrite RHS as $(X+Y+5)/2$, and note that each square root has two branches ($X^{1/2} = |X|^{1/2}e^{i\theta + i k\pi})$, $k=0,1$), giving 4 possibilities on the LHS. I have no idea if you will find a solution, though. Jan24 comment $\sum (\sqrt{x_k-k^2}-k)^2=0$ implies $x_k=2k^2$? I think the problem implicitly assumes that $x_k\geq k^2$. If ${x_k - k^2}<0$, you get two imaginary roots, and it is ambiguous which to choose from the notation. What is the context of this problem? Jan22 comment Questions about coerciveness and convexity No, linear space means a set such that linear combinations of elements are in the set. Here, a linear combination of two functions is again a function. But this function may not be coercive! Jan22 answered Questions about coerciveness and convexity Jan18 comment Book recommendation for Measure Theory I think most of this is covered by Bartle's book. It is a compact Wiley Classics book. Dec29 comment Rational and irrational numbers There is no minimal value $\epsilon$. Dec29 comment Proving differentiability $(e^{hx}-1)/h$ is clearly bounded ... Dec21 awarded Caucus Dec18 comment Help with proof that $H^{1}(\mathbb{R})$ is closed under multiplication. It is clear to me that the statement cannot be generalized to any $H^1(R^n)$, as seen by studying singularities on the form $|x|^\alpha$ with $\alpha<0$. Dec18 comment Help with proof that $H^{1}(\mathbb{R})$ is closed under multiplication. Is it really true? If $u,v \in L^2(R)$ then $uv \in L^1(R)$ which is not a subset of $L^2(R)$. Dec13 comment Use differentials to estimate the error in volume of the box. You will obtain that the estimated error $dv$ will differ from the difference $v - v_\text{max}$ or $v - v_\text{min}$. Why is this? Dec13 comment Use differentials to estimate the error in volume of the box. I guess the maximum volume is the volume obtained when all lenghts are the longest -- add 1 % to all lengths. Similar for the minimum volume. Dec13 comment Use differentials to estimate the error in volume of the box. $dv$ has units of volume. $dv/v = 0.9 \%$. Dec10 comment Convergence of sequence of $L^{p}$ function Is $\Omega$ a bounded set? Dec10 comment Integrate $\int_1^\infty\lfloor x\rfloor e^{-x}\,dx$ The summation can be done by summing $\sum_n e^{-\beta n}$ using geometric series, then differentiating at $\beta = 1$.