How to use real paths in windows and not “This PC” as the parent folder?

So I’m used to Linux and work with proper paths all the time. I use windows only for Visual Studio and some games that wouldn’t run on wine or mono otherwise.
There is this annoying “feature” that makes file paths in the explorer show up for example as

This PC/Desktop

instead of “C:/Users/<username>/Desktop” and hitting the parent folder button will not bring me to my user directory, but instead to this This PC virtual folder, which I guess is nice, but I wouldn’t want that unless I go up from C:/ or click it directly. The same is true for Documents, Music, Downloads and other such directories. They all appear under this “This PC” folder and it’s really annoying me.

Is there some way to change that behavior?

co.combinatorics – Optimization problem on sums of differences between real numbers with combinatorial constraints

We are given a sequence $S_n$ of $n$ points on a straight line $L$, whose coordinates are denoted by $x_1, x_2, ldots, x_n$ in non-decreasing order (i.e., the corresponding non-decreasing ordered sequence of Euclidean distances between each point of $S_n$ and an arbitrarily chosen point of $L$). Let $D$ be equal to $max_{i,jin (n)} |x_i-x_j|$.

We denote by $tau$ a threshold point on $L$ maximizing the following sum $R(tau)$:

$$R(tau):=left(sum_{substack{1le i<j<kle n: \ x_i,x_j<tau,~~x_k>tau}}
left(sum_{substack{1le i<j<kle n: \ x_i<tau,~~x_j,x_k>tau}}

Given three distinct points with indices $i, j$ and $k$, we define

$$R(i,j,k):={sum_{1le i<j<kle n}


$$R'(tau):=left(sum_{substack{1le i<j<kle n: \ x_i,x_j<tau,~~x_kgetau}}
left(sum_{substack{1le i<j<kle n: \ x_i<tau,~~x_j,x_kgetau}}

Question: What is the minimum value for the ratio $frac{R(tau)}{R'(tau)}$ over all possible sequences $S_n$, asymptotically for $ntoinfty$?

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linear algebra – Reference on classifying real subspaces of complex vector spaces (based on restricted complex structure)

Every complex vector space can also been seen as real vector space. If we now choose a real subspace, it may not be a complex subspace (in particular, if it is of odd real dimension).

If the complex vector space was equipped with an inner product (for example, a Hilbert space), we can restrict the imaginary unit (also known as linear complex structure) to any real subspace using the orthogonal projection. We can then classify the types of real subspaces based on the spectrum of this “restricted complex structure”. In particular, if the restricted complex structure squares to minus identity, i.e., is itself a complex structure, the real subspace is also a complex subspace. In general, the spectrum encodes how being a complex subspace is violated.

I worked this out for myself, but I’m confident that this is standard material in linear algebra of complex vector spaces. However, the standard introductory text books that I checked do not discuss real subspaces of complex vector spaces and their classification in the above way.

Do you know of a standard reference that I could cite when discussing this (in particular, the above mentioned classification based on the spectrum of the restricted complex structure)?

checking free positions to fill real entries

general vector spaces

how do you check "free positions to fill real entries" in the subspace"

linear algebra – Given a chain of commuting matrices over the complex numbers, can we build one over the real numbers?

Suppose we have two $ntimes n$ matrices $A$ and $B$ with entries in $mathbb{R}$, and two non-scalar matrices $X$ and $Y$ with entries in $mathbb{C}$, such that $AX=XA$, $XY=YX$, and $BY=YB$.

Is it necessarily the case that there exist non-scalar matrices $X’$ and $Y’$ with entries in $mathbb{R}$ such that $AX’=X’A$, $X’Y’=Y’X’$, and $BY’=Y’B$?

(Here “non-scalar” just means that the matrices aren’t scalar multiples of the identity matrix.)

real analysis – $lim_{trightarrow 0}U(t)f=f$ implies the convergence at each point?

I am new here and I would like to see if you can help me.
My question is if the limit $limlimits_{trightarrow 0}U(t)f=f$ imply the convergence at each point?

It seems to me that the statement is true, but I don’t know how to write it. Can you please give me some help?

real analysis – Unboundedness of anti-derivative

Note that I got this question from just messing around with a couple of functions and don’t know if it is actually true.

Let $f$ be a function such that $limlimits_{x to a}|x-a|f(x)=L$, and $0<L<infty$. Additionally let $f$ be integrable in a region, $B$, where $a$ is an accumulation point of $B$.
Then we know that,
$$forall epsilon >0 exists delta >0 : 0<|x-a|<delta implies ||x-a|f(x)-L|<epsilon.$$
Fix $epsilon$ such that $L>epsilon$, then this means that,

Then let $F$, the anti-derivative of $f$ be defined as,
$$F(x)=int_{d}^{x}f(t) mathrm{d}t + k,$$
for some $d in B$ and $k in mathbb{R}$. Then I aim to show that $limlimits_{x to a}F(x)=infty$.

Now we know that $dfrac{1}{|x-a|}$ is integrable everywhere except $x=a$. Additionally $$int_{y}^{a}dfrac{1}{|x-a|} mathrm{d}x,$$
is unbounded and goes to $infty$ for $y neq a$.
Then taking the left hand side of the inequality above and integrating from $d<a$ to $x$ where, where $x$ is in $B$, we get that,
$$(L-epsilon)int_{d}^{x}frac{1}{|t-a|} mathrm{d}t+k<int_{d}^{x}f(t) mathrm{d}t+k=F(x).$$
Now taking the limit as $x$ approaches $a$ on both sides we achieve the desired result. That is $limlimits_{x to a}F(x)=infty$.

$textbf{My first question is:}$

Is this proof valid? Initially I thought that $L>1$ but it seems this ‘proof’ doesn’t use this assumption anywhere so I omitted it. It seems intuitively correct to assume this as the area under $dfrac{1}{|x-a|}$ is unbounded if we take one of the bounds as $a$, so the area under $f(x)$ must be unbounded as well.

$textbf{Second question (and the one I’m more concerned with):}$

Supposing the fact that this is true. Then couldn’t I supposedly assume the following proposition.

Let $g(x)$ and $f(x)$ have the following properties described in the proof, that is as $x$ approaches $a$, $G(x)$ and $F(x)$ (their respective anti-derivatives) are unbounded and approach $infty$.

Then if we take the limit of the ratio of $g(x)$ and $f(x)$, ($G(x), g(x) neq 0$),
$$lim_{x to a} frac{f(x)}{g(x)}.$$
Then integrate the numerator and denominator to obtain a supposedly different limit, that is,
$$lim_{x to a} frac{F(x)}{G(x)}.$$
However since we know that $F(x)$ and $G(x)$ both go to $infty$ as $x$ approaches $a$ we can apply L’Hospital’s rule and supposedly achieve the following,
$$lim_{x to a} frac{F(x)}{G(x)}=lim_{x to a} frac{f(x)}{g(x)}.$$
That is to say we achieve a sort of “inverse” L’Hopital’s rule if you will, assuming that the functions that are being concerned obey the properties laid out earlier.

$textbf{Third and final question:}$

The function $dfrac{1}{|x-a|}$ was chosen somewhat arbitrarily since it has that nature of having an unbounded area. Is there any function perhaps smaller than this but also has an unbounded area? Then perhaps this condition would capture a wider array of functions.

The reason for not being able to assume that solely $f(x)$ goes to infinity as $x$ goes to $a$ is because counter-examples exist for this, that is $ln|x|$, and a more strict unboundedness needed to be placed.

real analysis – Vector-valued interpolation for sublinear operators

Grafakos in his $textit{Classical Fourier Analysis}$ formulates (see Exercise 4.5.2 therein) the following vector-valued version of the Riesz-Thorin interpolation theorem.


Let $1le p_0, q_0,p_1,q_1, r_0, s_0, r_1, s_1leinfty$ and $thetain(0,1)$ satisfy
frac{1-theta}{p_0}+frac{theta}{p_1}&=frac{1}{p},qquad frac{1-theta}{q_0}+frac{theta}{q_1}=frac{1}{q},\
frac{1-theta}{r_0}+frac{theta}{r_1}&=frac{1}{r},qquad frac{1-theta}{s_0}+frac{theta}{s_1}=frac{1}{s},

and let $T$ be a linear operator mapping $L^{p_0}(mathbb{R}^n, ell^{r_0})$ to $L^{q_0}(mathbb{R}^n, ell^{s_0})$ and $L^{p_1}(mathbb{R}^n, ell^{r_1})$ to $L^{q_1}(mathbb{R}^n, ell^{s_1})$.
Then $T$ maps $L^{p}(mathbb{R}^n, ell^{r})$ to $L^{q}(mathbb{R}^n, ell^{s})$.

My $textbf{question}$ is whether we can replace the assumption that $T$ is linear in the above theorem with the weaker assumption that $T$ is sublinear? In the scalar-valued case one can do it and the relevant interpolation theorem goes by the name of Marcinkiewicz-Zygmund.

I would appreciate any hints or perhaps a reference to suitable literature.

Do republicans seriously think that creationism is real?

Since creation is part of Islamic, Christian and Judaic faith, republicans of those religious backgrounds, like their democrat counterparts, do believe in creationism, however, most republicans believe in evolution due to its pervasive scientific credibility.

Not sure what the self-answer was intended to accomplish,but it is not an accurate statement; Some of the people running your country hold scientifically ignorant beliefs, but they are increasingly rare.