Uniformly Lipschitz sets are bounded.











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Let $mathcal{F}subset C([a,b],G)$ $(a,bin mathbb{R},Gsubset mathbb{R}^d)$ be uniformly Lipschitz continuous. Show that $mathcal{F}$ is equicontinuous. Is such a set bounded in $C([a,b],G)$?



I have already shown the first statement and I was hoping someone could help me with the question, if such a set is bounded? Thanks.






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    Let $mathcal{F}subset C([a,b],G)$ $(a,bin mathbb{R},Gsubset mathbb{R}^d)$ be uniformly Lipschitz continuous. Show that $mathcal{F}$ is equicontinuous. Is such a set bounded in $C([a,b],G)$?



    I have already shown the first statement and I was hoping someone could help me with the question, if such a set is bounded? Thanks.






    functional-analysis







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $mathcal{F}subset C([a,b],G)$ $(a,bin mathbb{R},Gsubset mathbb{R}^d)$ be uniformly Lipschitz continuous. Show that $mathcal{F}$ is equicontinuous. Is such a set bounded in $C([a,b],G)$?



      I have already shown the first statement and I was hoping someone could help me with the question, if such a set is bounded? Thanks.






      functional-analysis







      share|cite|improve this question













      Let $mathcal{F}subset C([a,b],G)$ $(a,bin mathbb{R},Gsubset mathbb{R}^d)$ be uniformly Lipschitz continuous. Show that $mathcal{F}$ is equicontinuous. Is such a set bounded in $C([a,b],G)$?



      I have already shown the first statement and I was hoping someone could help me with the question, if such a set is bounded? Thanks.






      functional-analysis




      functional-analysis






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      asked Dec 2 '17 at 20:47









      Katakuri

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          Such a set does not need to be bounded. For example, let $mathcal F$ be the set of all constant functions. This set is uniformly Lipschitz continuous but not bounded in $C([a,b],G)$.






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            Such a set does not need to be bounded. For example, let $mathcal F$ be the set of all constant functions. This set is uniformly Lipschitz continuous but not bounded in $C([a,b],G)$.






            share|cite|improve this answer

























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              Such a set does not need to be bounded. For example, let $mathcal F$ be the set of all constant functions. This set is uniformly Lipschitz continuous but not bounded in $C([a,b],G)$.






              share|cite|improve this answer























                up vote
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                Such a set does not need to be bounded. For example, let $mathcal F$ be the set of all constant functions. This set is uniformly Lipschitz continuous but not bounded in $C([a,b],G)$.






                share|cite|improve this answer












                Such a set does not need to be bounded. For example, let $mathcal F$ be the set of all constant functions. This set is uniformly Lipschitz continuous but not bounded in $C([a,b],G)$.







                share|cite|improve this answer












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                answered Nov 16 at 16:25









                Davide Giraudo

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