Riesz lemma for $L^p$ space












1












$begingroup$


I need a proof for special case of Riesz lemma (when $varepsilon$ is 0):



If Y is a closed proper subspace of $L^p(mu)$ for some $1<p<infty$, then there exist $fin L^p(mu)$ such that $||f||=1$ and $||f-g||geq 1$ for every $gin Y$.



I know that Clarkson's inequality (uniform convexity) can be used.










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$endgroup$












  • $begingroup$
    This is an easy application of Hahn-Banach (hint: Every element of the dual of $L^p$ for $1<p<infty$ attains its norm...)
    $endgroup$
    – David C. Ullrich
    Jan 12 at 13:06










  • $begingroup$
    Can you give me further explanation how to apply that?
    $endgroup$
    – Hana
    Jan 16 at 11:08
















1












$begingroup$


I need a proof for special case of Riesz lemma (when $varepsilon$ is 0):



If Y is a closed proper subspace of $L^p(mu)$ for some $1<p<infty$, then there exist $fin L^p(mu)$ such that $||f||=1$ and $||f-g||geq 1$ for every $gin Y$.



I know that Clarkson's inequality (uniform convexity) can be used.










share|cite|improve this question









$endgroup$












  • $begingroup$
    This is an easy application of Hahn-Banach (hint: Every element of the dual of $L^p$ for $1<p<infty$ attains its norm...)
    $endgroup$
    – David C. Ullrich
    Jan 12 at 13:06










  • $begingroup$
    Can you give me further explanation how to apply that?
    $endgroup$
    – Hana
    Jan 16 at 11:08














1












1








1





$begingroup$


I need a proof for special case of Riesz lemma (when $varepsilon$ is 0):



If Y is a closed proper subspace of $L^p(mu)$ for some $1<p<infty$, then there exist $fin L^p(mu)$ such that $||f||=1$ and $||f-g||geq 1$ for every $gin Y$.



I know that Clarkson's inequality (uniform convexity) can be used.










share|cite|improve this question









$endgroup$




I need a proof for special case of Riesz lemma (when $varepsilon$ is 0):



If Y is a closed proper subspace of $L^p(mu)$ for some $1<p<infty$, then there exist $fin L^p(mu)$ such that $||f||=1$ and $||f-g||geq 1$ for every $gin Y$.



I know that Clarkson's inequality (uniform convexity) can be used.







functional-analysis banach-spaces lp-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 10:19









HanaHana

161




161












  • $begingroup$
    This is an easy application of Hahn-Banach (hint: Every element of the dual of $L^p$ for $1<p<infty$ attains its norm...)
    $endgroup$
    – David C. Ullrich
    Jan 12 at 13:06










  • $begingroup$
    Can you give me further explanation how to apply that?
    $endgroup$
    – Hana
    Jan 16 at 11:08


















  • $begingroup$
    This is an easy application of Hahn-Banach (hint: Every element of the dual of $L^p$ for $1<p<infty$ attains its norm...)
    $endgroup$
    – David C. Ullrich
    Jan 12 at 13:06










  • $begingroup$
    Can you give me further explanation how to apply that?
    $endgroup$
    – Hana
    Jan 16 at 11:08
















$begingroup$
This is an easy application of Hahn-Banach (hint: Every element of the dual of $L^p$ for $1<p<infty$ attains its norm...)
$endgroup$
– David C. Ullrich
Jan 12 at 13:06




$begingroup$
This is an easy application of Hahn-Banach (hint: Every element of the dual of $L^p$ for $1<p<infty$ attains its norm...)
$endgroup$
– David C. Ullrich
Jan 12 at 13:06












$begingroup$
Can you give me further explanation how to apply that?
$endgroup$
– Hana
Jan 16 at 11:08




$begingroup$
Can you give me further explanation how to apply that?
$endgroup$
– Hana
Jan 16 at 11:08










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