Kind of passage to the limit in the sense of distributions











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Suppose $B$ is a ball in $mathbb{R}^{n}$ with $n>1$, and $f$ a locally integrable function. Suppose $F$ is a closed set with empty interior and $M>0$ such that the distribution defined by $f$ satisfies
begin{equation}
int_{Bsetminus F}f(x-t)phi(t)dtleq M
end{equation}
for all test function $phi$. We know that if we had
$$f(x)leq M$$ for all $xin Bsetminus F$, we could get the same inequality by passing to the limit (supposing that $f$ is continuous); is it possible, in the same way, to circumvent around $F$ in (1) and prove that (1) holds for integrals over $B$?










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    Suppose $B$ is a ball in $mathbb{R}^{n}$ with $n>1$, and $f$ a locally integrable function. Suppose $F$ is a closed set with empty interior and $M>0$ such that the distribution defined by $f$ satisfies
    begin{equation}
    int_{Bsetminus F}f(x-t)phi(t)dtleq M
    end{equation}
    for all test function $phi$. We know that if we had
    $$f(x)leq M$$ for all $xin Bsetminus F$, we could get the same inequality by passing to the limit (supposing that $f$ is continuous); is it possible, in the same way, to circumvent around $F$ in (1) and prove that (1) holds for integrals over $B$?










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      up vote
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      down vote

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      Suppose $B$ is a ball in $mathbb{R}^{n}$ with $n>1$, and $f$ a locally integrable function. Suppose $F$ is a closed set with empty interior and $M>0$ such that the distribution defined by $f$ satisfies
      begin{equation}
      int_{Bsetminus F}f(x-t)phi(t)dtleq M
      end{equation}
      for all test function $phi$. We know that if we had
      $$f(x)leq M$$ for all $xin Bsetminus F$, we could get the same inequality by passing to the limit (supposing that $f$ is continuous); is it possible, in the same way, to circumvent around $F$ in (1) and prove that (1) holds for integrals over $B$?










      share|cite|improve this question













      Suppose $B$ is a ball in $mathbb{R}^{n}$ with $n>1$, and $f$ a locally integrable function. Suppose $F$ is a closed set with empty interior and $M>0$ such that the distribution defined by $f$ satisfies
      begin{equation}
      int_{Bsetminus F}f(x-t)phi(t)dtleq M
      end{equation}
      for all test function $phi$. We know that if we had
      $$f(x)leq M$$ for all $xin Bsetminus F$, we could get the same inequality by passing to the limit (supposing that $f$ is continuous); is it possible, in the same way, to circumvent around $F$ in (1) and prove that (1) holds for integrals over $B$?







      real-analysis integration distribution-theory






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      asked 9 hours ago









      M. Rahmat

      292211




      292211



























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