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I'm stuck with the following problem: Let K be a real-valued function such that: $\int_{-\infty}^{\infty} |K(t)|dt < \infty$ and $\int_{-\infty}^{\infty} K(t)dt = 1 $. Let f be a continuous function. Then we define: $f_{n} (x) = \int_{-\infty}^{\infty} f(x+ t/n)K(t)dt $. I want to prove that the sequence ${f_{n}}$ uniformly converges to f in any interval of the form $[a,b]$. I've thought about integration by parts, but you aren't given any hypothesis on the differentiability of either $K$ nor $f$. Any ideas?

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    Just as a side note, K kind of looks like the Dirac Delta. I'm not an expert on the topic, but maybe there are useful results in that context that can help with the problem – 2017-02-14
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    K is not the dirac delta since the dirac delta is not a function in the common sense and even if you look on the dirac delta as a function, it is infinity a point. Hence not real valued. – 2017-02-14
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    Is $f$ assumed to be bounded? What makes the integral defining $f_n$ converge? – 2017-02-14
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    Yes, I forgot to mention it – 2017-02-14

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Use the fact that $f(x) = \displaystyle \int_{-\infty}^\infty f(x) K(t) \, dt$. It follows that $$ |f_n(x) - f(x)| \le \int_{-\infty}^\infty |f(x+t/n) - f(x)| |K(t)| \, dt.$$

Can you proceed from here?

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    I think so. Given epsilon > 0, since f is continuous you can find a large enough N such that for all n > N, |f(x+t/n) - f(x)| < epsilon . Then the epsilon gets out of the integral sign and the resulting integral is convergent by hipothesis. Thanks a lot! – 2017-02-14
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    It is a little more complicated than that because $f(x+t/n)$ may not be close to $f(x)$ for all $x$, rather just for $t$ inside a bounded interval. When $t$ is large you can use the fact that $|f(x+t/n) - f(x)|$ is bounded. – 2017-02-14