dilation operator on $L^2(mathbbR)$ is continuous

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Prove the statement let $D:mathbbR^+rightarrow L^2(mathbbR)$ defined by $D(a)=f_a$ and $f_a(x)=frac1sqrtaf(fracxa)$, where $fin L^2(mathbbR)$ then the mapping $D$ is continuous on $mathbbR^+.$










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  • Why is this tagged as "wavelets" ? Plus, you should show some efforts if you want someone to help you.
    – nicomezi
    Aug 29 at 6:04















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Prove the statement let $D:mathbbR^+rightarrow L^2(mathbbR)$ defined by $D(a)=f_a$ and $f_a(x)=frac1sqrtaf(fracxa)$, where $fin L^2(mathbbR)$ then the mapping $D$ is continuous on $mathbbR^+.$










share|cite|improve this question























  • Why is this tagged as "wavelets" ? Plus, you should show some efforts if you want someone to help you.
    – nicomezi
    Aug 29 at 6:04













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up vote
0
down vote

favorite











Prove the statement let $D:mathbbR^+rightarrow L^2(mathbbR)$ defined by $D(a)=f_a$ and $f_a(x)=frac1sqrtaf(fracxa)$, where $fin L^2(mathbbR)$ then the mapping $D$ is continuous on $mathbbR^+.$










share|cite|improve this question















Prove the statement let $D:mathbbR^+rightarrow L^2(mathbbR)$ defined by $D(a)=f_a$ and $f_a(x)=frac1sqrtaf(fracxa)$, where $fin L^2(mathbbR)$ then the mapping $D$ is continuous on $mathbbR^+.$







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edited Sep 7 at 6:20

























asked Aug 29 at 5:59









kamalakkannan

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  • Why is this tagged as "wavelets" ? Plus, you should show some efforts if you want someone to help you.
    – nicomezi
    Aug 29 at 6:04

















  • Why is this tagged as "wavelets" ? Plus, you should show some efforts if you want someone to help you.
    – nicomezi
    Aug 29 at 6:04
















Why is this tagged as "wavelets" ? Plus, you should show some efforts if you want someone to help you.
– nicomezi
Aug 29 at 6:04





Why is this tagged as "wavelets" ? Plus, you should show some efforts if you want someone to help you.
– nicomezi
Aug 29 at 6:04











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Yes. You can approximate $f$ by a continuous function $g$ whose support is a compact subset of $(0,infty)$ and it is fairly starightforward to verify continuity of $a to g_a$. (The given function becomes a uniform limit of continuous functions).






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    1 Answer
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    Yes. You can approximate $f$ by a continuous function $g$ whose support is a compact subset of $(0,infty)$ and it is fairly starightforward to verify continuity of $a to g_a$. (The given function becomes a uniform limit of continuous functions).






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      Yes. You can approximate $f$ by a continuous function $g$ whose support is a compact subset of $(0,infty)$ and it is fairly starightforward to verify continuity of $a to g_a$. (The given function becomes a uniform limit of continuous functions).






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        up vote
        1
        down vote









        Yes. You can approximate $f$ by a continuous function $g$ whose support is a compact subset of $(0,infty)$ and it is fairly starightforward to verify continuity of $a to g_a$. (The given function becomes a uniform limit of continuous functions).






        share|cite|improve this answer












        Yes. You can approximate $f$ by a continuous function $g$ whose support is a compact subset of $(0,infty)$ and it is fairly starightforward to verify continuity of $a to g_a$. (The given function becomes a uniform limit of continuous functions).







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        answered Aug 29 at 6:07









        Kavi Rama Murthy

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