1
vote
1answer
78 views

Hölder's inequality. Understanding proof?!

I know how most standard textbooks show that $||f*g||_r \le ||f||_p||g||_q$ with $\frac{1}{r}+1=\frac{1}{p}+ \frac{1}{q}$, but I found a book where the hint $|f(x-y)g(y)|\le ...
3
votes
1answer
79 views

Bounding for convolution convergence

Suppose $f\in L^p(\mathbb{R})$ and $K\in L^1(\mathbb{R})$ with $\int_\mathbb{R}K(x)dx=1$. Define $$K_t(x)=\dfrac{1}{t}K\left(\dfrac{x}{t}\right)$$ I'm trying to prove that $\lim_{t\rightarrow ...
0
votes
1answer
61 views

Give an example of $f\in L^1$, $g\in L^{\infty}$, such that $f*g\notin C_0$ (meaning $\lim_{|x|\to\infty}(f*g)(x)\neq0$)

Give an example of $f\in L^1$, $g\in L^{\infty}$, such that $f*g\notin C_0$ (meaning $\lim_{|x|\to\infty}(f*g)(x)\neq0$) Here's a theorem from my real analysis book: Assume $1\le p\le \infty$ and ...
4
votes
1answer
116 views

Convolution of an $L^1$ function and a function that tends to $0$ results in a function that tends to $0$

I'm trying to solve the following problem in review for a test, but have only partly succeeded: Let $K \in L^1(\mathbb{R})$ and $f$ be a bounded, measurable function on $\mathbb{R}$, with ...
-2
votes
2answers
60 views

Lebesgue conduct integral

I. Suppose $f\in \mathcal{L^1}(R^n),g\in \mathcal{L^1}(R^n)$, then conduct integral $f*g$ is defined as $f*g(x)= \int_{R^n}f(x-y)g(y)dy$ for all $x$. My task is to prove following statements. (1) ...
2
votes
1answer
65 views

Convolution of $L_1$ functions is $L_1$

Let $f,g\in L_1(\mathbb R, m)$ where m is the Lebesgue measure. Prove that: a) $f(x-t)g(t) \in L_1(\mathbb R, m)$ as a fuction of $t$ almost all $x$ b) $h\in L_1(\mathbb R, m)$ where $h=(f \ast ...