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Feb
4
comment Convergence in distribution - X/Y
No. Almost the same argument works for $\Omega = [1,2]$ with the uniform distribution, $X$ the identity, $Y(\omega) = 3-\omega$ and $X_n = Y_n = X$
Feb
4
answered Convergence in distribution - X/Y
Feb
4
reviewed Approve When can a function have its variables seperated
Feb
4
answered Colon and equals under product operator
Feb
4
comment Minimum of the Schatten 1-norm
What type of lower bound do you expect? For $A = B$ (which is possible by your assumptions), you have $\|A-B\|_1 = 0$.
Feb
4
answered Sobolev space interpolation
Feb
4
answered Find the matrix representation
Feb
3
answered affine hull of a sum of sets
Feb
2
answered If $G \cong \mathbb Z/3\mathbb Z$, then $\mathbb R[G] \cong \mathbb R \times \mathbb C$
Feb
1
comment Continuous but not compact operator on $L^2(0,\infty)$
Fubini, as I said. For $g(x,y) = \frac 2{x^{3/2}}y^{1/2}|f(y)|^2$, we have $$ \int_0^\infty \int_0^x g(x,y)\, dy \,dx = \int_0^\infty \int_y^\infty g(x,y)\, dx\, dy $$
Feb
1
comment Continuous but not compact operator on $L^2(0,\infty)$
In the line above that I did two steps at once, sry. On one side, we have $$ \int_0^x y^{-1/2}\, dy = 2x^{1/2}$$ giving $$ \int_0^\infty \frac1{x^2} \cdot 2x^{1/2} \int_0^x y^{1/2}|f(y)|^2\, dy\, dx $$ Now we use Fubini.
Feb
1
comment Continuous but not compact operator on $L^2(0,\infty)$
$$\int_y^\infty \frac 2{x^{3/2}}\, dx = \left[-\frac 4{x^{1/2}}\right]_y^\infty = \frac 4{y^{1/2}}$$
Feb
1
answered Continuous but not compact operator on $L^2(0,\infty)$
Feb
1
comment Homography with line correspondences
Could you provide some context? What is $H$, $\ell_i$, $(\cdot)'$, $x$?
Jan
29
comment What is the primitive function of $xe^{x^2+2x}$?
@AmineMarzouki No. The derivative of $\frac 1{2x+2} \exp(x^2 + 2x)$ is $-\frac 2{(2x+2)^2}\exp(x^2 + 2x) + \exp(x^2 + 2x)$.
Jan
27
comment Show that every algebraic subset of $\Bbb A^2(\Bbb R)$ is equal to $V(F)$ for some $F∈\mathbb R[X,Y]$.
@user152715 Added something.
Jan
27
revised Show that every algebraic subset of $\Bbb A^2(\Bbb R)$ is equal to $V(F)$ for some $F∈\mathbb R[X,Y]$.
added 674 characters in body
Jan
27
answered Show that every algebraic subset of $\Bbb A^2(\Bbb R)$ is equal to $V(F)$ for some $F∈\mathbb R[X,Y]$.
Jan
27
answered What is a generalized stochastic process? I've found two different definitions. Are they equivalent?
Jan
24
answered prove unique existence of a solution of a Cauchy problem