The lebesgue-measure tag has no usage guidance.

**5**

votes

**1**answer

248 views

### Dominated convergence to characteristic function

Let $\phi_m(x):=\chi_{[0,1]} * \chi_{[0,1]} *...* \chi_{[0,1]}$ be the m -times convolution (so $m+1$ characteristic functions are involved).
Then the Fourier transform of this function is given by
...

**6**

votes

**0**answers

80 views

### orthogonal projector onto the set of convex functions

Let $\Omega\subset \mathbb R^d$ be an open, convex domain, and consider the Hilbert space $L^2(\Omega)$. Each sum of convex functions is convex, hence the subset $Conv(\Omega)$ of all convex functions ...

**3**

votes

**0**answers

146 views

### trace-class embeddings

There is a classical theorem of Riesz-Kolmogorov that characterizes compact embedding in $L^p$-spaces of some subspace of them. A generalization to arbitrary metric spaces has been recently obtained ...

**1**

vote

**0**answers

115 views

### Is the implication ($f$ is Riemann integrable over $D_1$ and $D_2$) $\Rightarrow $ ($f$ is Riemann integrable over $D=D_1\cup D_2$) true?

Let $D_1,D_2$ be a bounded subset of $\mathbb{R}^n$ and $\partial D_1,\partial D_2$ are both of Lebesgue measure zero (that is to say: $D_1,D_2$ are Jordan measurable).
Also, let $f:D_1\cup ...

**1**

vote

**0**answers

85 views

### Convergence of solutions of the volterra integral equation with convergent kernels

Consider the following Volterra integral equation
$$
g(t) = \int_0^t K_n(t,s)w_n(s) ds
$$
where g(t) and K_n(t,s) are continuous and $K_n(t,s)\geq K_{n+1}(t,s)$ for all $t,s$. Moreover, $K_n(t,s)$ ...

**1**

vote

**0**answers

94 views

### On the proof of a $W^{2,p}$ estimate - regularity on eliptic PDE

I see this proof on http://www.math.uiowa.edu/~lwang/cccalderon.pdf and I couldn't understand what he did. If $||f||_{L^p(B_{4})} = \delta$ is small and the measure $|\{ x \in B_1; ...

**1**

vote

**0**answers

166 views

### Relationship between weak Lp and strong Lq topologies for q<p

Specificaly:
Does convergence in $L^{\frac{1}{2}}$ imply weak $L^2$ convergence?
Having a limit in $L^{\frac{1}{2}}$ topology and a limit in weak $L^2$ topology whether these are always equal? If ...

**0**

votes

**0**answers

48 views

### Strong Lebesgue regularity lemma

I have troubles in understanding haw the Strong Lebesgue regularity lemma implies the Rademacher differentiation theorem. An informal proof was given by Tao on his blog: ...