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The function $K$ is defined as $K(f)(x)=\int_{\mathbb{R}^n}k(x,y)f(y)dy$ (with k previously defined). The thing is, I have some more hypotesis, and I'm asked to show that $K$ is uniformly continuous. The thing is that I'm having doubts about what this would mean.

It's that if $||f-g||_{L^p} < \delta$ then $||K(f) - K(g)||_{L^p} < \epsilon$?

It's that if $|x-y| < \delta$ then $|K(f)(x) - K(f)(y)| < \epsilon$?

Is any other?

Sorry if this seems like a stupid question

Silkking
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    I would say the first, that is, for all $\epsilon$ there is $\delta$ such that $|f-g|_p<\delta$, then $|K(f)-K(g)|_p<\epsilon$. The second is if you want to prove that the function $K(f)$ alone is uniformly continuous – Alessandro Dec 21 '20 at 15:14
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    While the meaning of uniform continuity of $K$ is indeed the first one, but I think the question actually might be to show that $Kf$ is uniformly continuous for each $f\in L^p.$ The reason for this is that under mild assumptions $K$ would be bounded and as it is linear the uniform continuity of $K$ would become a trivial statement. Every linear bounded operator is uniformly continuous. However, with a slightly restrictive assumption one can actually say that $kf$ is uniformly continuous for each $f.$ – Raghav Dec 21 '20 at 15:49
  • It was to show that $K:L^p \rightarrow L^p$ defined as above was well defined and uniformly continuous. So, I guess it would still be the 1st one? – Silkking Dec 21 '20 at 16:38
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    As WhoKnowsWho said, the definition of uniform continuity of $K$ itself is the first one. So if that is what they asked for, that is what you should give. – Paul Sinclair Dec 22 '20 at 00:05

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