Mixing
Showing an inequality in the Banach space $C^1([a,b])$?
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Given $f in C^1([a,b]),$ how can it be shown that $$|f|_infty^2 le dfrac_2^2b-a + 2 |f|_2 |f'|_2?$$
I suppose I could use Cauchy-Schwarz, but how to handle the derivative? And I don't think Holder's applies...
This is a problem from an old Preliminary exam.
banach-spaces norm cauchy-schwarz-inequality
asked 2 days ago
Mathemagica
85922762
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up vote
0
down vote
favorite
Given $f in C^1([a,b]),$ how can it be shown that $$|f|_infty^2 le dfrac_2^2b-a + 2 |f|_2 |f'|_2?$$
I suppose I could use Cauchy-Schwarz, but how to handle the derivative? And I don't think Holder's applies...
This is a problem from an old Preliminary exam.
banach-spaces norm cauchy-schwarz-inequality
asked 2 days ago
Mathemagica
85922762
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given $f in C^1([a,b]),$ how can it be shown that $$|f|_infty^2 le dfrac_2^2b-a + 2 |f|_2 |f'|_2?$$
I suppose I could use Cauchy-Schwarz, but how to handle the derivative? And I don't think Holder's applies...
This is a problem from an old Preliminary exam.
banach-spaces norm cauchy-schwarz-inequality
asked 2 days ago
Mathemagica
85922762
Given $f in C^1([a,b]),$ how can it be shown that $$|f|_infty^2 le dfrac_2^2b-a + 2 |f|_2 |f'|_2?$$
I suppose I could use Cauchy-Schwarz, but how to handle the derivative? And I don't think Holder's applies...
This is a problem from an old Preliminary exam.
banach-spaces norm cauchy-schwarz-inequality
asked 2 days ago
Mathemagica
85922762
asked 2 days ago
Mathemagica
85922762
asked 2 days ago
Mathemagica
85922762
asked 2 days ago
Mathemagica
85922762
85922762
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
One idea is to let $xinarg min |f^2(x)|$, fix $yin [a,b]$, and apply the FTC to $f^2$:
$$
f(y)^2-f(x)^2=2int_x^yf(t)f'(t)dt,
$$
so that by Holder,
$$
sup_yin[a,b]f(y)^2le f(x)^2+2|f|_2|f'|_2.
$$
To deal with the $f(x)^2$ term,
$$
f(x)^2=frac1b-aint_a^bf(x)^2dtle frac1b-aint_a^bf(t)^2dt
$$
since $f(x)^2le f(t)^2$ for all $t$.
answered 2 days ago
user254433
2,072612
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
One idea is to let $xinarg min |f^2(x)|$, fix $yin [a,b]$, and apply the FTC to $f^2$:
$$
f(y)^2-f(x)^2=2int_x^yf(t)f'(t)dt,
$$
so that by Holder,
$$
sup_yin[a,b]f(y)^2le f(x)^2+2|f|_2|f'|_2.
$$
To deal with the $f(x)^2$ term,
$$
f(x)^2=frac1b-aint_a^bf(x)^2dtle frac1b-aint_a^bf(t)^2dt
$$
since $f(x)^2le f(t)^2$ for all $t$.
answered 2 days ago
user254433
2,072612
add a comment |Â
up vote
2
down vote
accepted
One idea is to let $xinarg min |f^2(x)|$, fix $yin [a,b]$, and apply the FTC to $f^2$:
$$
f(y)^2-f(x)^2=2int_x^yf(t)f'(t)dt,
$$
so that by Holder,
$$
sup_yin[a,b]f(y)^2le f(x)^2+2|f|_2|f'|_2.
$$
To deal with the $f(x)^2$ term,
$$
f(x)^2=frac1b-aint_a^bf(x)^2dtle frac1b-aint_a^bf(t)^2dt
$$
since $f(x)^2le f(t)^2$ for all $t$.
answered 2 days ago
user254433
2,072612
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
One idea is to let $xinarg min |f^2(x)|$, fix $yin [a,b]$, and apply the FTC to $f^2$:
$$
f(y)^2-f(x)^2=2int_x^yf(t)f'(t)dt,
$$
so that by Holder,
$$
sup_yin[a,b]f(y)^2le f(x)^2+2|f|_2|f'|_2.
$$
To deal with the $f(x)^2$ term,
$$
f(x)^2=frac1b-aint_a^bf(x)^2dtle frac1b-aint_a^bf(t)^2dt
$$
since $f(x)^2le f(t)^2$ for all $t$.
answered 2 days ago
user254433
2,072612
One idea is to let $xinarg min |f^2(x)|$, fix $yin [a,b]$, and apply the FTC to $f^2$:
$$
f(y)^2-f(x)^2=2int_x^yf(t)f'(t)dt,
$$
so that by Holder,
$$
sup_yin[a,b]f(y)^2le f(x)^2+2|f|_2|f'|_2.
$$
To deal with the $f(x)^2$ term,
$$
f(x)^2=frac1b-aint_a^bf(x)^2dtle frac1b-aint_a^bf(t)^2dt
$$
since $f(x)^2le f(t)^2$ for all $t$.
answered 2 days ago
user254433
2,072612
answered 2 days ago
user254433
2,072612
answered 2 days ago
user254433
2,072612
answered 2 days ago
user254433
2,072612
2,072612
add a comment |Â
add a comment |Â
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