# Convolution powers tend to zero

Let $\mu$ be a probability measure on $[0,\infty)$ that is not degenerated ($\mu(0) < 1$) and $f$ be a bounded function on $[0,\infty)$. Show that pointwise $$f * \mu^{*n} \rightarrow 0$$ where $*$ denotes convolution $$(f * \mu)(t) = \int_0^t f(t-u) d \mu(u)$$ and $$\mu^{*n} = \mu * \dots * \mu\quad n\text{ times}$$

Edit: Added that $\mu$ is not degenerated. Edit2: Clarified that $f$ is defined on the positive reals.

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sorry, trying to understand the question, what is $()^{*n}$? –  Seyhmus Güngören Sep 4 '12 at 16:27
oh thanks, I edited the question for clarification. –  Haderlump Sep 4 '12 at 16:33

Okay, here is my attempt, please correct me if I'm wrong. Let's assume that $$\nu(t) := \lim_{k\rightarrow\infty} \sum_{n=0}^k \mu^{*n}(t)$$ defines a measure that is finite for bounded sets.
Then for every $t \ge 0$ $$\lim_{k\rightarrow\infty} \sum_{n=0}^k (f*\mu^{*n})(t)= (f*\nu)(t)$$ is finite, and hence $$\lim_{n\rightarrow\infty} (f*\mu^{*n})(t) = 0$$ For the special case $f \equiv 1$ this means $$\lim_{n\rightarrow\infty} \mu^{*n}(t) = 0$$