Ir al contenido principal

## Due May 3

### Problem 1

Let $\gamma$ be the lower semicircle of radius $N$ around the origin, and $\xi > 0.$ Then
$\displaystyle \int_\gamma f(z) dz = 2\pi i\text{Res}_{z=-it} f(z) = i e^{-2\pi t\xi},$
where $f(z)$ is the function defined by
$f(z) = \dfrac{1}{\pi} \dfrac{z}{z^2+t^2} e^{-2\pi iz\xi}.$

### Problem 2

If $f\in L^1(\mathbb R)$ and diferentiable at $x\in\mathbb R$, then the limit
$\displaystyle \lim_{t\to 0} \int_{|y|\ge t} \frac{f(x-y)}{y} dy$
exists. (Hint: Use the identity, for any $\delta_n>0$,
$\displaystyle \int_{t\le|y|<\delta_n} \frac{f(x-y)}{y} dy = \int_{t\le|y|<\delta_n} \frac{f(x-y) - f(x)}{y} dy + \int_{t\le|y|<\delta_n} \frac{f(x)}{y} dy,$
and take $\delta_n\to 0$.)

### Problem 3

If $f_n, g_n$ are sequences in $C_c^\infty(\mathbb R)$ that converge in $L^1(\mathbb R)$ to $f$, then $Hf_n$ and $Hg_n$ converge in measure to the same function $Hf$.

### Problem 4

Let $K\in L^2(\mathbb R)$ satisfy
1. $\hat K \in L^\infty(\mathbb R)$; and
2. there exists a constant $A>0$ such that $|K'(x)| \le \dfrac{A}{|x|^2}$ for any $x\in\mathbb R$.
Then the operator $Tf = K*f$ is weakly bounded on $L^1(\mathbb R)$.

### Problem 5

We can replace hypothesis (2) from the previous problem by
• there exists a constant $A>0$ such that $\displaystyle \int_{|x|\ge 2|y|} | K(x - y) - K(x)| dx \le A$ for any $y\in\mathbb R$.