## The set $D_v=\{ x\in \mathbb{R}^N: D_vf(x) \text{ exists } \}$ is measurable.

Let $$f:\mathbb{R}^n\to \mathbb{R}$$ be Lipsticz function. The directional derivative in the direction of $$\mathbf{v}$$ is given by $$D_\mathbf{v}f(\mathbf{x})=\lim_{t\to 0}\dfrac{f(\mathbf{x}+t\mathbf{v})-f(\mathbf{x})}{t}.$$ Define a set $$D_\mathbf{v}=\{ \mathbf{x}\in \mathbb{R}^n:D_\mathbf{v}f(\mathbf{x})\text{ exists } \} .$$ Show that $$D_\mathbf{v}$$ is measurable.

Posted on Categories proxies

## Volume of manifolds embedded in $\mathbb{R}^n$

Let $$N$$ be a closed, connected, oriented hypersurface of $$\mathbb{R}^n$$. Such a manifold inherits a volume form from the usual volume from on $$\mathbb{R}^n$$ and has an associated volume given by integration of the inherited volume form over $$N$$.

I would like to know if there is a proof of the following intuivitely obvious (I think) fact: let $$X$$ be a vector field on $$\mathbb{R}^n$$ everywhere transverse to the hypersurface. Let $$\psi_\epsilon^X$$ be the flow of such a vector field for a small time $$\epsilon$$ (we can choose $$\epsilon$$ small enough so that $$\psi_\epsilon^X(N)$$ is diffeomorphic to $$N$$). Then the volume of $$\psi_\epsilon^X(N)$$ is not equal to the volume of $$N$$.

Does such a result exist? I assume, if so, it would not generalise to other manifolds/volumes other than the canonical one?

## Classifying Radon partitions in $\mathbb{R}^n$ whose affine hull is $\mathbb{R}^n$

Specifically, I want to determine all distinct “types” of Radon partitions of $$n+2$$ points in $$\mathbb{R}^n$$ for which the affine hull is all of $$\mathbb{R}^n$$. This is a homework question, so I’m primarily looking for advice on getting started.

Radon’s theorem states that any set of $$n+2$$ points in $$\mathbb{R}^n$$ can be partitioned into two disjoint sets with intersecting convex hulls.

The affine hull of a set $$S$$ is given by $$\mbox{aff}(S)=\left\{\sum_{i=1}^k\alpha_ix_i:k>0,x_i\in S,\alpha_i\in\mathbb{R},\sum_{i=1}^k\alpha_i=1\right\}.$$

So, the simplest case is when $$n=1$$, and clearly the affine hull of any Radon partition of $$n+2=3$$ points in $$\mathbb{R}$$ is all of $$\mathbb{R}$$. When $$n=2$$, the $$n+2=4$$ points can be partitioned as a triple and a singleton or two pairs of points. In the case of the former, the convex hull of the triple must contain the singleton. In the case of the latter, the pairs of points must form intersecting line segments. But the former is the only “type” of partition I want to consider, as it’s affine hull is in fact $$\mathbb{R}^2$$, but the affine hull of the partitions in the latter is not all of $$\mathbb{R}^2$$. So I might conjecture that the full classification requires that the convex hull of one of the partitions must be fully contained in the convex hull of the other partition, but I’m not sure if I even believe this intuitively.

Posted on Categories proxies

## Extendind a a function defined in $\mathbb{R}^n\setminus\{0\}$ to a continuous function defined in $\mathbb{R}^{n}$.

Let $$g: \mathbb{R}^n\setminus\{0\} \to \mathbb{R}$$ be a function of class $$C^{1}$$ and suppose that there is $$M > 0$$ such that $$\left|\frac{\partial}{\partial x_{i}}g(x)\right| \leq M.$$ Prove that if $$n \geq 2$$ then $$g$$ can be extended to a continuous function defined in $$\mathbb{R}^n$$. Show that if $$n = 1$$ the statement is false.

My attempt.

I want define the extension $$\bar{g}: \mathbb{R}^n \to \mathbb{R}$$ by $$\bar{g}(x) = g(x)$$ if $$x \in \mathbb{R}^{n}\setminus\{0$$} and $$\displaystyle \bar{g}(0) = \lim_{x \to 0}g(x)$$. Thus $$\lim_{x \to 0} \bar{g}(x) = \lim_{x \to 0}g(x) = \bar{g}(0)$$ and so, $$\bar{g}$$ is continuous. So the question is reduced to prove that $$\displaystyle \lim_{x \to 0}g(x)$$ exists. Thus, I must show that $$\forall \epsilon > 0, \exists \delta > 0\text{ s.t. } \Vert X \Vert < \delta \Longrightarrow |g(x)|<\epsilon.$$

The hypothesis $$\left|\frac{\partial}{\partial x_{i}}g(x)\right| \leq M$$ seems necessary for get $$|g(x)-g(y)| \leq M|x-y|$$ using the Means Value Inequality. This almost solves the problem, because if $$g(0) = 0$$ we can write $$|g(x)| \leq M|x|.$$ But $$g(0) = 0$$ doesnt make sense. I’m stuck here.

Also, I cannot see the whe is necessary $$n \geq 2$$, where I use this in the demonstration and why it fails when $$n=1$$.

Posted on Categories proxies

## Reference for compact embedding between (weighted) Holder space on $\mathbb{R}^n$

Suppose $$0<\alpha<\beta<1$$, and $$\Omega$$ is a bounded subset of $$\mathbb{R}^n$$. Then the Holder space $$C^{\beta}(\Omega)$$ is compactly embedded into $$C^{\alpha}(\Omega)$$. But if $$\Omega=\mathbb{R}^n$$, then the compact embedding is not true.

However, if we consider the weaker weighted Holder space $$C^{\alpha, -\delta}(\mathbb{R}^n)$$ (for any $$\delta>0$$) instead of $$C^{\alpha}(\mathbb{R}^n)$$. Then is $$C^{\beta}(\mathbb{R}^n)$$ compactly embedded to $$C^{\alpha, -\delta}(\mathbb{R}^n)$$?

Here $$\|f\|_{C^{\alpha, -\delta}}=\|(1+|\cdot|^2)^{-\frac{\delta}{2}}f\|_{C^{\alpha}}.$$

I could not find a precise reference from some books on functional analysis. Any comment is welcome.

## $U=\{x\in \mathbb{R^n} :g_i(x)\leq 0 : \text{for} \quad i=1,…,m \}$ is closed

Let $$U=\{x\in \mathbb{R^n} :g_i(x)\leq 0 : \text{for} \quad i=1,…,m \}$$ be a subset of $$\mathbb{R^n}$$.

with $$(g_i)_{1\leq i \leq n}$$ are convex functions from $$\mathbb{R^n}$$ in $$\mathbb{R}$$.

we want to prove that $$U$$ is a closed set.

Posted on Categories proxies

## Connected and homogeneous $T_2$-space not homeomorphic to a subset of $\mathbb{R}^n$

What is an example of an connected and homogeneous $$T_2$$-space $$(X,\tau)$$ with $$|X|=2^{\aleph_0}$$ such that for no $$n\in\mathbb{R}$$ the space $$(X,\tau)$$ is homeomorphic to a subspace of $$\mathbb{R}^n$$?