Show $||f||_pleq ||f||_infty$
$begingroup$
On $C([0,1])$ define $||f||_p=(int_0^1|f(t)|^pdt)^frac{1}{p}$ for
$1leq p<infty$ and $||f||_infty=sup_{tin[0,1]}|f(t)|$. Show that
$||f||_pleq ||f||_infty$
My approach is the following
begin{align*}
||f||_p^p&=int_0^1|f(t)|^pdt\
&leq|1-0|cdotsup_{tin[0,1]}|f(t)|^p\
&=(sup_{tin[0,1]}|f(t)])^p \
&=||f||_infty^p
end{align*}
However I was told, this is not good enough - that I need to consider some "special case", but I don't see what I need?
functional-analysis inequality lp-spaces
asked Dec 11 '18 at 10:41
Kane JKane J
715
$endgroup$
add a comment |
$begingroup$
On $C([0,1])$ define $||f||_p=(int_0^1|f(t)|^pdt)^frac{1}{p}$ for
$1leq p<infty$ and $||f||_infty=sup_{tin[0,1]}|f(t)|$. Show that
$||f||_pleq ||f||_infty$
My approach is the following
begin{align*}
||f||_p^p&=int_0^1|f(t)|^pdt\
&leq|1-0|cdotsup_{tin[0,1]}|f(t)|^p\
&=(sup_{tin[0,1]}|f(t)])^p \
&=||f||_infty^p
end{align*}
However I was told, this is not good enough - that I need to consider some "special case", but I don't see what I need?
functional-analysis inequality lp-spaces
asked Dec 11 '18 at 10:41
Kane JKane J
715
$endgroup$
1
$begingroup$
This looks OK. Your best option is to ask the person who made this commentary in the first place. The best thing would have been to ask immediately; don't be afraid, if you don't understand something, don't pretend you understood, ask. (I also pretend that I understood sometimes. It's never good)
$endgroup$
– Giuseppe Negro
Dec 11 '18 at 10:56
add a comment |
$begingroup$
On $C([0,1])$ define $||f||_p=(int_0^1|f(t)|^pdt)^frac{1}{p}$ for
$1leq p<infty$ and $||f||_infty=sup_{tin[0,1]}|f(t)|$. Show that
$||f||_pleq ||f||_infty$
My approach is the following
begin{align*}
||f||_p^p&=int_0^1|f(t)|^pdt\
&leq|1-0|cdotsup_{tin[0,1]}|f(t)|^p\
&=(sup_{tin[0,1]}|f(t)])^p \
&=||f||_infty^p
end{align*}
However I was told, this is not good enough - that I need to consider some "special case", but I don't see what I need?
functional-analysis inequality lp-spaces
asked Dec 11 '18 at 10:41
Kane JKane J
715
$endgroup$
On $C([0,1])$ define $||f||_p=(int_0^1|f(t)|^pdt)^frac{1}{p}$ for
$1leq p<infty$ and $||f||_infty=sup_{tin[0,1]}|f(t)|$. Show that
$||f||_pleq ||f||_infty$
My approach is the following
begin{align*}
||f||_p^p&=int_0^1|f(t)|^pdt\
&leq|1-0|cdotsup_{tin[0,1]}|f(t)|^p\
&=(sup_{tin[0,1]}|f(t)])^p \
&=||f||_infty^p
end{align*}
However I was told, this is not good enough - that I need to consider some "special case", but I don't see what I need?
functional-analysis inequality lp-spaces
functional-analysis inequality lp-spaces
asked Dec 11 '18 at 10:41
Kane JKane J
715
asked Dec 11 '18 at 10:41
Kane JKane J
715
edited Dec 11 '18 at 10:50
Kane J
asked Dec 11 '18 at 10:41
Kane JKane J
715
asked Dec 11 '18 at 10:41
Kane JKane J
715
asked Dec 11 '18 at 10:41
Kane JKane J
715
715
1
$begingroup$
This looks OK. Your best option is to ask the person who made this commentary in the first place. The best thing would have been to ask immediately; don't be afraid, if you don't understand something, don't pretend you understood, ask. (I also pretend that I understood sometimes. It's never good)
$endgroup$
– Giuseppe Negro
Dec 11 '18 at 10:56
add a comment |
1
$begingroup$
This looks OK. Your best option is to ask the person who made this commentary in the first place. The best thing would have been to ask immediately; don't be afraid, if you don't understand something, don't pretend you understood, ask. (I also pretend that I understood sometimes. It's never good)
$endgroup$
– Giuseppe Negro
Dec 11 '18 at 10:56
1
1
$begingroup$
This looks OK. Your best option is to ask the person who made this commentary in the first place. The best thing would have been to ask immediately; don't be afraid, if you don't understand something, don't pretend you understood, ask. (I also pretend that I understood sometimes. It's never good)
$endgroup$
– Giuseppe Negro
Dec 11 '18 at 10:56
$begingroup$
This looks OK. Your best option is to ask the person who made this commentary in the first place. The best thing would have been to ask immediately; don't be afraid, if you don't understand something, don't pretend you understood, ask. (I also pretend that I understood sometimes. It's never good)
$endgroup$
– Giuseppe Negro
Dec 11 '18 at 10:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is almost correct (except that $int_0^q$ should be $int_0^1$). The only problem is that you are acting as if $lVert frVert_infty=suplvert frvert$. Instead, $lVert frVert_infty$ is the essential supremum of $lvert frvert$. Take that into account.
answered Dec 11 '18 at 10:48
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
$begingroup$
The assignment says $fin C([0,1])$, so that's an actual supremum. It's actually a maximum.
$endgroup$
– Giuseppe Negro
Dec 11 '18 at 10:55
$begingroup$
You are right. I missed that. But then, other than the misprint that I mentioned, the proof is indeed correct.
$endgroup$
– José Carlos Santos
Dec 11 '18 at 10:56
add a comment |
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1 Answer
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$begingroup$
It is almost correct (except that $int_0^q$ should be $int_0^1$). The only problem is that you are acting as if $lVert frVert_infty=suplvert frvert$. Instead, $lVert frVert_infty$ is the essential supremum of $lvert frvert$. Take that into account.
answered Dec 11 '18 at 10:48
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
$begingroup$
The assignment says $fin C([0,1])$, so that's an actual supremum. It's actually a maximum.
$endgroup$
– Giuseppe Negro
Dec 11 '18 at 10:55
$begingroup$
You are right. I missed that. But then, other than the misprint that I mentioned, the proof is indeed correct.
$endgroup$
– José Carlos Santos
Dec 11 '18 at 10:56
add a comment |
$begingroup$
It is almost correct (except that $int_0^q$ should be $int_0^1$). The only problem is that you are acting as if $lVert frVert_infty=suplvert frvert$. Instead, $lVert frVert_infty$ is the essential supremum of $lvert frvert$. Take that into account.
answered Dec 11 '18 at 10:48
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
$begingroup$
The assignment says $fin C([0,1])$, so that's an actual supremum. It's actually a maximum.
$endgroup$
– Giuseppe Negro
Dec 11 '18 at 10:55
$begingroup$
You are right. I missed that. But then, other than the misprint that I mentioned, the proof is indeed correct.
$endgroup$
– José Carlos Santos
Dec 11 '18 at 10:56
add a comment |
$begingroup$
It is almost correct (except that $int_0^q$ should be $int_0^1$). The only problem is that you are acting as if $lVert frVert_infty=suplvert frvert$. Instead, $lVert frVert_infty$ is the essential supremum of $lvert frvert$. Take that into account.
answered Dec 11 '18 at 10:48
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
It is almost correct (except that $int_0^q$ should be $int_0^1$). The only problem is that you are acting as if $lVert frVert_infty=suplvert frvert$. Instead, $lVert frVert_infty$ is the essential supremum of $lvert frvert$. Take that into account.
answered Dec 11 '18 at 10:48
José Carlos SantosJosé Carlos Santos
158k22126228
answered Dec 11 '18 at 10:48
José Carlos SantosJosé Carlos Santos
158k22126228
answered Dec 11 '18 at 10:48
José Carlos SantosJosé Carlos Santos
158k22126228
answered Dec 11 '18 at 10:48
José Carlos SantosJosé Carlos Santos
158k22126228
158k22126228
$begingroup$
The assignment says $fin C([0,1])$, so that's an actual supremum. It's actually a maximum.
$endgroup$
– Giuseppe Negro
Dec 11 '18 at 10:55
$begingroup$
You are right. I missed that. But then, other than the misprint that I mentioned, the proof is indeed correct.
$endgroup$
– José Carlos Santos
Dec 11 '18 at 10:56
add a comment |
$begingroup$
The assignment says $fin C([0,1])$, so that's an actual supremum. It's actually a maximum.
$endgroup$
– Giuseppe Negro
Dec 11 '18 at 10:55
$begingroup$
You are right. I missed that. But then, other than the misprint that I mentioned, the proof is indeed correct.
$endgroup$
– José Carlos Santos
Dec 11 '18 at 10:56
$begingroup$
The assignment says $fin C([0,1])$, so that's an actual supremum. It's actually a maximum.
$endgroup$
– Giuseppe Negro
Dec 11 '18 at 10:55
$begingroup$
The assignment says $fin C([0,1])$, so that's an actual supremum. It's actually a maximum.
$endgroup$
– Giuseppe Negro
Dec 11 '18 at 10:55
$begingroup$
You are right. I missed that. But then, other than the misprint that I mentioned, the proof is indeed correct.
$endgroup$
– José Carlos Santos
Dec 11 '18 at 10:56
$begingroup$
You are right. I missed that. But then, other than the misprint that I mentioned, the proof is indeed correct.
$endgroup$
– José Carlos Santos
Dec 11 '18 at 10:56
add a comment |
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$begingroup$
This looks OK. Your best option is to ask the person who made this commentary in the first place. The best thing would have been to ask immediately; don't be afraid, if you don't understand something, don't pretend you understood, ask. (I also pretend that I understood sometimes. It's never good)
$endgroup$
– Giuseppe Negro
Dec 11 '18 at 10:56