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Compute $d(0,B)$ where $B=fin mathcal C([0,1])mid f(0)=0, int_0^1 f=1$
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Let $(mathcal C([0,1]),|cdot |_infty )$ the normed space where $|cdot |_infty $ is the supremum norm.
1) Compute $$d(0,B):=inf_substackfneq 0 \ fin Bd(0,f)$$ where $$B=leftfin mathcal C([0,1])mid f(0)=0text and int_0^1 f=1right.$$
2) Is there $fin B$ s.t. $d(0,B)=|f|_infty $ ?
Attempt
1) The only thing I know is that if $fin B$, then $$1=int_0^1 fleq int_0^1 |f|leq |f|_infty,$$
and thus $d(0,B)geq 1$. I guess the norm it $1$, but I can't find a sequence $(f_n)$ in $B$ s.t. $|f_n|_infty to 1$.
2) No idea.
functional-analysis
edited Jul 15 at 11:42
Robert Z
84.2k955123
asked Jul 15 at 11:24
user330587
821310
add a comment |Â
up vote
1
down vote
favorite
Let $(mathcal C([0,1]),|cdot |_infty )$ the normed space where $|cdot |_infty $ is the supremum norm.
1) Compute $$d(0,B):=inf_substackfneq 0 \ fin Bd(0,f)$$ where $$B=leftfin mathcal C([0,1])mid f(0)=0text and int_0^1 f=1right.$$
2) Is there $fin B$ s.t. $d(0,B)=|f|_infty $ ?
Attempt
1) The only thing I know is that if $fin B$, then $$1=int_0^1 fleq int_0^1 |f|leq |f|_infty,$$
and thus $d(0,B)geq 1$. I guess the norm it $1$, but I can't find a sequence $(f_n)$ in $B$ s.t. $|f_n|_infty to 1$.
2) No idea.
functional-analysis
edited Jul 15 at 11:42
Robert Z
84.2k955123
asked Jul 15 at 11:24
user330587
821310
If $fin E$ then $int_0^1f(x)dx=1$. What is the condition for $f(0)$?
– Robert Z
Jul 15 at 11:28
Sorry, I edited f(0)=0
– user330587
Jul 15 at 11:30
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $(mathcal C([0,1]),|cdot |_infty )$ the normed space where $|cdot |_infty $ is the supremum norm.
1) Compute $$d(0,B):=inf_substackfneq 0 \ fin Bd(0,f)$$ where $$B=leftfin mathcal C([0,1])mid f(0)=0text and int_0^1 f=1right.$$
2) Is there $fin B$ s.t. $d(0,B)=|f|_infty $ ?
Attempt
1) The only thing I know is that if $fin B$, then $$1=int_0^1 fleq int_0^1 |f|leq |f|_infty,$$
and thus $d(0,B)geq 1$. I guess the norm it $1$, but I can't find a sequence $(f_n)$ in $B$ s.t. $|f_n|_infty to 1$.
2) No idea.
functional-analysis
edited Jul 15 at 11:42
Robert Z
84.2k955123
asked Jul 15 at 11:24
user330587
821310
Let $(mathcal C([0,1]),|cdot |_infty )$ the normed space where $|cdot |_infty $ is the supremum norm.
1) Compute $$d(0,B):=inf_substackfneq 0 \ fin Bd(0,f)$$ where $$B=leftfin mathcal C([0,1])mid f(0)=0text and int_0^1 f=1right.$$
2) Is there $fin B$ s.t. $d(0,B)=|f|_infty $ ?
Attempt
1) The only thing I know is that if $fin B$, then $$1=int_0^1 fleq int_0^1 |f|leq |f|_infty,$$
and thus $d(0,B)geq 1$. I guess the norm it $1$, but I can't find a sequence $(f_n)$ in $B$ s.t. $|f_n|_infty to 1$.
2) No idea.
functional-analysis
edited Jul 15 at 11:42
Robert Z
84.2k955123
asked Jul 15 at 11:24
user330587
821310
edited Jul 15 at 11:42
Robert Z
84.2k955123
edited Jul 15 at 11:42
Robert Z
84.2k955123
edited Jul 15 at 11:42
Robert Z
84.2k955123
84.2k955123
asked Jul 15 at 11:24
user330587
821310
asked Jul 15 at 11:24
user330587
821310
asked Jul 15 at 11:24
user330587
821310
821310
If $fin E$ then $int_0^1f(x)dx=1$. What is the condition for $f(0)$?
– Robert Z
Jul 15 at 11:28
Sorry, I edited f(0)=0
– user330587
Jul 15 at 11:30
add a comment |Â
If $fin E$ then $int_0^1f(x)dx=1$. What is the condition for $f(0)$?
– Robert Z
Jul 15 at 11:28
Sorry, I edited f(0)=0
– user330587
Jul 15 at 11:30
If $fin E$ then $int_0^1f(x)dx=1$. What is the condition for $f(0)$?
– Robert Z
Jul 15 at 11:28
If $fin E$ then $int_0^1f(x)dx=1$. What is the condition for $f(0)$?
– Robert Z
Jul 15 at 11:28
Sorry, I edited f(0)=0
– user330587
Jul 15 at 11:30
Sorry, I edited f(0)=0
– user330587
Jul 15 at 11:30
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
1) Take $$g_n=begincases
na_n x& xin [0,1/n]\
a_n&xin [1/n,1]
endcases.$$
It's a sequence of $B$ s.t. $|g_n|_infty to 1$. Therefore, $d(0,B)leq 1$ and thus $d(0,B)=1$.
2) Suppose there is $fin B$ s.t. $1=d(0,B)=|f|_infty $. Then $$0leq int_0^1 (|f|_infty -f)=|f|_infty -int_0^1 f=1-1=0.$$
Since $|f|_infty -f$ is continuous and non negative, we get $|f|_infty -f=0$ and thus $f=|f|_infty $. Since $f(0)=0$, we get that $fequiv 0$ and thus $fnotin B$.
answered Jul 15 at 11:38
Surb
36.3k84274
You have the same inequality twice, you mean to flip one, no?
– JuliusL33t
Jul 15 at 11:58
I corrected it, thank you @JuliusL33t
– Surb
Jul 15 at 12:00
add a comment |Â
up vote
1
down vote
Hint: Take$$beginarrayrcccf_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&begincasesnleft(x-1+frac1nright)&text if x>1-frac1n\0&text otherwise.endcasesendarray$$Prove that $(forall ninmathbbN):f_nin B$ and compute $d(0,f_n)$.
answered Jul 15 at 11:34
José Carlos Santos
114k1698177
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
1) Take $$g_n=begincases
na_n x& xin [0,1/n]\
a_n&xin [1/n,1]
endcases.$$
It's a sequence of $B$ s.t. $|g_n|_infty to 1$. Therefore, $d(0,B)leq 1$ and thus $d(0,B)=1$.
2) Suppose there is $fin B$ s.t. $1=d(0,B)=|f|_infty $. Then $$0leq int_0^1 (|f|_infty -f)=|f|_infty -int_0^1 f=1-1=0.$$
Since $|f|_infty -f$ is continuous and non negative, we get $|f|_infty -f=0$ and thus $f=|f|_infty $. Since $f(0)=0$, we get that $fequiv 0$ and thus $fnotin B$.
answered Jul 15 at 11:38
Surb
36.3k84274
You have the same inequality twice, you mean to flip one, no?
– JuliusL33t
Jul 15 at 11:58
I corrected it, thank you @JuliusL33t
– Surb
Jul 15 at 12:00
add a comment |Â
up vote
2
down vote
accepted
1) Take $$g_n=begincases
na_n x& xin [0,1/n]\
a_n&xin [1/n,1]
endcases.$$
It's a sequence of $B$ s.t. $|g_n|_infty to 1$. Therefore, $d(0,B)leq 1$ and thus $d(0,B)=1$.
2) Suppose there is $fin B$ s.t. $1=d(0,B)=|f|_infty $. Then $$0leq int_0^1 (|f|_infty -f)=|f|_infty -int_0^1 f=1-1=0.$$
Since $|f|_infty -f$ is continuous and non negative, we get $|f|_infty -f=0$ and thus $f=|f|_infty $. Since $f(0)=0$, we get that $fequiv 0$ and thus $fnotin B$.
answered Jul 15 at 11:38
Surb
36.3k84274
You have the same inequality twice, you mean to flip one, no?
– JuliusL33t
Jul 15 at 11:58
I corrected it, thank you @JuliusL33t
– Surb
Jul 15 at 12:00
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
1) Take $$g_n=begincases
na_n x& xin [0,1/n]\
a_n&xin [1/n,1]
endcases.$$
It's a sequence of $B$ s.t. $|g_n|_infty to 1$. Therefore, $d(0,B)leq 1$ and thus $d(0,B)=1$.
2) Suppose there is $fin B$ s.t. $1=d(0,B)=|f|_infty $. Then $$0leq int_0^1 (|f|_infty -f)=|f|_infty -int_0^1 f=1-1=0.$$
Since $|f|_infty -f$ is continuous and non negative, we get $|f|_infty -f=0$ and thus $f=|f|_infty $. Since $f(0)=0$, we get that $fequiv 0$ and thus $fnotin B$.
answered Jul 15 at 11:38
Surb
36.3k84274
1) Take $$g_n=begincases
na_n x& xin [0,1/n]\
a_n&xin [1/n,1]
endcases.$$
It's a sequence of $B$ s.t. $|g_n|_infty to 1$. Therefore, $d(0,B)leq 1$ and thus $d(0,B)=1$.
2) Suppose there is $fin B$ s.t. $1=d(0,B)=|f|_infty $. Then $$0leq int_0^1 (|f|_infty -f)=|f|_infty -int_0^1 f=1-1=0.$$
Since $|f|_infty -f$ is continuous and non negative, we get $|f|_infty -f=0$ and thus $f=|f|_infty $. Since $f(0)=0$, we get that $fequiv 0$ and thus $fnotin B$.
answered Jul 15 at 11:38
Surb
36.3k84274
edited Jul 15 at 11:59
answered Jul 15 at 11:38
Surb
36.3k84274
answered Jul 15 at 11:38
Surb
36.3k84274
answered Jul 15 at 11:38
Surb
36.3k84274
36.3k84274
You have the same inequality twice, you mean to flip one, no?
– JuliusL33t
Jul 15 at 11:58
I corrected it, thank you @JuliusL33t
– Surb
Jul 15 at 12:00
add a comment |Â
You have the same inequality twice, you mean to flip one, no?
– JuliusL33t
Jul 15 at 11:58
I corrected it, thank you @JuliusL33t
– Surb
Jul 15 at 12:00
You have the same inequality twice, you mean to flip one, no?
– JuliusL33t
Jul 15 at 11:58
You have the same inequality twice, you mean to flip one, no?
– JuliusL33t
Jul 15 at 11:58
I corrected it, thank you @JuliusL33t
– Surb
Jul 15 at 12:00
I corrected it, thank you @JuliusL33t
– Surb
Jul 15 at 12:00
add a comment |Â
up vote
1
down vote
Hint: Take$$beginarrayrcccf_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&begincasesnleft(x-1+frac1nright)&text if x>1-frac1n\0&text otherwise.endcasesendarray$$Prove that $(forall ninmathbbN):f_nin B$ and compute $d(0,f_n)$.
answered Jul 15 at 11:34
José Carlos Santos
114k1698177
add a comment |Â
up vote
1
down vote
Hint: Take$$beginarrayrcccf_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&begincasesnleft(x-1+frac1nright)&text if x>1-frac1n\0&text otherwise.endcasesendarray$$Prove that $(forall ninmathbbN):f_nin B$ and compute $d(0,f_n)$.
answered Jul 15 at 11:34
José Carlos Santos
114k1698177
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: Take$$beginarrayrcccf_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&begincasesnleft(x-1+frac1nright)&text if x>1-frac1n\0&text otherwise.endcasesendarray$$Prove that $(forall ninmathbbN):f_nin B$ and compute $d(0,f_n)$.
answered Jul 15 at 11:34
José Carlos Santos
114k1698177
Hint: Take$$beginarrayrcccf_ncolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&begincasesnleft(x-1+frac1nright)&text if x>1-frac1n\0&text otherwise.endcasesendarray$$Prove that $(forall ninmathbbN):f_nin B$ and compute $d(0,f_n)$.
answered Jul 15 at 11:34
José Carlos Santos
114k1698177
edited Jul 15 at 11:43
answered Jul 15 at 11:34
José Carlos Santos
114k1698177
answered Jul 15 at 11:34
José Carlos Santos
114k1698177
answered Jul 15 at 11:34
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
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If $fin E$ then $int_0^1f(x)dx=1$. What is the condition for $f(0)$?
– Robert Z
Jul 15 at 11:28
Sorry, I edited f(0)=0
– user330587
Jul 15 at 11:30