$f in C_{00}(mathbb{R^p},mathbb{C})$. $ mapsto f_t in L_infty(mathbb{R}^p, mathcal B_p, lambda_p, mathbb{C})$...












1












$begingroup$


Continuing from here



Let $f_t(x):=f(x+t)$



Consider $f mapsto f_t$ which is a linear, isometric bijection from $L_infty(mathbb{R}^p, mathcal B_p, lambda_p, mathbb{C})to L_infty(mathbb{R}^p, mathcal B_p, lambda_p, mathbb{C})$ for every $t in mathbb{R}^p$




How can I show that for $f in C_{00}(mathbb{R^p},mathbb{C})$ (continuous and compact support) the mapping $mathbb{R}^pni t mapsto f_t in L_infty(mathbb{R}^p, mathcal B_p, lambda_p, mathbb{C})$ is uniformly continuous?











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$endgroup$

















    1












    $begingroup$


    Continuing from here



    Let $f_t(x):=f(x+t)$



    Consider $f mapsto f_t$ which is a linear, isometric bijection from $L_infty(mathbb{R}^p, mathcal B_p, lambda_p, mathbb{C})to L_infty(mathbb{R}^p, mathcal B_p, lambda_p, mathbb{C})$ for every $t in mathbb{R}^p$




    How can I show that for $f in C_{00}(mathbb{R^p},mathbb{C})$ (continuous and compact support) the mapping $mathbb{R}^pni t mapsto f_t in L_infty(mathbb{R}^p, mathcal B_p, lambda_p, mathbb{C})$ is uniformly continuous?











    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Continuing from here



      Let $f_t(x):=f(x+t)$



      Consider $f mapsto f_t$ which is a linear, isometric bijection from $L_infty(mathbb{R}^p, mathcal B_p, lambda_p, mathbb{C})to L_infty(mathbb{R}^p, mathcal B_p, lambda_p, mathbb{C})$ for every $t in mathbb{R}^p$




      How can I show that for $f in C_{00}(mathbb{R^p},mathbb{C})$ (continuous and compact support) the mapping $mathbb{R}^pni t mapsto f_t in L_infty(mathbb{R}^p, mathcal B_p, lambda_p, mathbb{C})$ is uniformly continuous?











      share|cite|improve this question









      $endgroup$




      Continuing from here



      Let $f_t(x):=f(x+t)$



      Consider $f mapsto f_t$ which is a linear, isometric bijection from $L_infty(mathbb{R}^p, mathcal B_p, lambda_p, mathbb{C})to L_infty(mathbb{R}^p, mathcal B_p, lambda_p, mathbb{C})$ for every $t in mathbb{R}^p$




      How can I show that for $f in C_{00}(mathbb{R^p},mathbb{C})$ (continuous and compact support) the mapping $mathbb{R}^pni t mapsto f_t in L_infty(mathbb{R}^p, mathcal B_p, lambda_p, mathbb{C})$ is uniformly continuous?








      real-analysis analysis lp-spaces






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      asked Jan 5 at 13:56









      user626880user626880

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          $begingroup$

          This follows from the fact that a continuous function with compact support is uniformly continuous. For a fixed $varepsilon$, there exists $delta$ such that if $s,tinmathbb R^p$ satisfy $lVert t-srVertltdelta$, then $leftlvert f(t)-f(s)rightrvertltvarepsilon$.



          For any $xinmathbb R^p$ and $s,t$ satisfying $lVert t-srVertltdelta$, the following inequality holds
          $$
          leftlvert f(x+t)-f(x+s)rightrvertltvarepsilon
          $$

          hence
          $$leftlVert f_t-f_srightrVert_infty ... $$






          share|cite|improve this answer









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            $begingroup$

            This follows from the fact that a continuous function with compact support is uniformly continuous. For a fixed $varepsilon$, there exists $delta$ such that if $s,tinmathbb R^p$ satisfy $lVert t-srVertltdelta$, then $leftlvert f(t)-f(s)rightrvertltvarepsilon$.



            For any $xinmathbb R^p$ and $s,t$ satisfying $lVert t-srVertltdelta$, the following inequality holds
            $$
            leftlvert f(x+t)-f(x+s)rightrvertltvarepsilon
            $$

            hence
            $$leftlVert f_t-f_srightrVert_infty ... $$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              This follows from the fact that a continuous function with compact support is uniformly continuous. For a fixed $varepsilon$, there exists $delta$ such that if $s,tinmathbb R^p$ satisfy $lVert t-srVertltdelta$, then $leftlvert f(t)-f(s)rightrvertltvarepsilon$.



              For any $xinmathbb R^p$ and $s,t$ satisfying $lVert t-srVertltdelta$, the following inequality holds
              $$
              leftlvert f(x+t)-f(x+s)rightrvertltvarepsilon
              $$

              hence
              $$leftlVert f_t-f_srightrVert_infty ... $$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                This follows from the fact that a continuous function with compact support is uniformly continuous. For a fixed $varepsilon$, there exists $delta$ such that if $s,tinmathbb R^p$ satisfy $lVert t-srVertltdelta$, then $leftlvert f(t)-f(s)rightrvertltvarepsilon$.



                For any $xinmathbb R^p$ and $s,t$ satisfying $lVert t-srVertltdelta$, the following inequality holds
                $$
                leftlvert f(x+t)-f(x+s)rightrvertltvarepsilon
                $$

                hence
                $$leftlVert f_t-f_srightrVert_infty ... $$






                share|cite|improve this answer









                $endgroup$



                This follows from the fact that a continuous function with compact support is uniformly continuous. For a fixed $varepsilon$, there exists $delta$ such that if $s,tinmathbb R^p$ satisfy $lVert t-srVertltdelta$, then $leftlvert f(t)-f(s)rightrvertltvarepsilon$.



                For any $xinmathbb R^p$ and $s,t$ satisfying $lVert t-srVertltdelta$, the following inequality holds
                $$
                leftlvert f(x+t)-f(x+s)rightrvertltvarepsilon
                $$

                hence
                $$leftlVert f_t-f_srightrVert_infty ... $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 5 at 14:07









                Davide GiraudoDavide Giraudo

                127k17154268




                127k17154268






























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