1

Suppose that $f \in \mathcal{L}^p$ and $g \in \mathcal{L}^q$, and $p,q$ are conjugate exponents. Then prove that

(a) $h(x) = \int_{-\infty}^{\infty} f(t) g(x+t) \, dt$ defines a bounded continuous function on $\mathbb{R}$ that satisfies $\Vert h \Vert_\infty \leq \Vert f \Vert_p \Vert g \Vert_q$

(b) $h$ is differentiable, if either $f$ or $g$ is differentiable and find $h'$ in terms of $f'$ or $g'$.

Any help is appreciated.

Community
  • 1
user62089
  • 1,356
  • This is related. Also, your question is about convolution. Editing your title might help. – Julien Mar 18 '13 at 03:45
  • For a, it is enough to use Holder (for boundedness) and the dominated convergence theorem (for continuity). – Julien Mar 18 '13 at 03:57
  • @julien: Thanks for your tip. I was able to solve (a) by direct application of Holder's. And continuity because of the continuity of the translation. Any ideas for (b)? – user62089 Mar 18 '13 at 03:57
  • Yes, Holder works. That's what I meant...Sorry. – Julien Mar 18 '13 at 04:00
  • @julien: I figured it out. Thanks anyway. – user62089 Mar 18 '13 at 04:34
  • @julien: On a different note, I am looking for an example for a bounded function on an infinite measure space that is not close to any integrable simple function (A counterexample for simple integrable functions being dense in $\mathcal{L}^\infty$ on an infinite measure space). – user62089 Mar 18 '13 at 12:45
  • A simple integrable function $s$ has in particular $\lim_{\pm\infty}s=0$. Therefore, every uniform limit of such functions has the same property. So it suffices to take $f$ bounded without this property. E.g. $f(x)=1$. – Julien Mar 18 '13 at 13:52

1 Answers1

0

a) Holder is the answer.

b) if $g$ is differentiable, then $h'(x) = \int f(t)g'(x+t)dt$, otherwise let $t' = x + t$, then $h(x) = \int f(t' - x)g(t')dt'$ and therefore if $f$ is differentiable $h'(x) = -\int f'(t - x)g(t)dt$