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Proving an inequality between $(1, frac{alpha}{2})$-Hölder norms of two functions
$begingroup$
I have to prove that, given $fin C^{1, frac{alpha}{2}}([a, b])$, such that $|f|_{infty}<L$ for some $L$, $fgeq0$ and $alphain(0, 1)$, there exists a constant $K$ such that
$$
|e^f-1|_{1, frac{alpha}{2}}leq K|f|_{1, frac{alpha}{2}}.
$$
where $|cdot|_{1, frac{alpha}{2}}$ is the usual $(1, frac{alpha}{2})$-Hölder norm.
This is not a real exercise. It is only a thing to prove to deduce the second inequality in Theorem 3.1, (i), by Proposition 3.1, in the paper https://arxiv.org/pdf/1402.2467. In this case $f=C_T$.
My attempt. We have
$$
|e^f-1|_{1, frac{alpha}{2}}=sup_{[a, b]}|e^f-1|+sup_{[a, b]}|f'e^f|+sup_{xneq y}frac{|f'(x)e^{f(x)}-f'(y)e^{f(y)}|}{|x-y|^{frac{alpha}{2}}},
$$
$$
|f|_{1, frac{alpha}{2}}=sup_{[a, b]}|f|+sup_{[a, b]}|f'|+sup_{xneq y}frac{|f'(x)-f'(y)|}{|x-y|^{frac{alpha}{2}}}.
$$
For simplicity, we call the $sup$ in the $|e^f-1|_{1, frac{alpha}{2}}$ expression with $S_1$ and the $sup$ in the $|f|_{1, frac{alpha}{2}}$ expression with $S_2$. Observe that
$$
|e^f-1|_{1, frac{alpha}{2}}leq sup_{[a, b]}|e^f-1|+sup_{[a, b]}|f'|sup_{[a, b]}|e^f|+S_1leqsup_{[a, b]}|e^f-1|+|f|_{1, frac{alpha}{2}}sup_{[a, b]}|e^f|+S_1.
$$
Now, by the arguments in Another definition of a Holder norm, the norms $|e^f-1|_{1, frac{alpha}{2}}$ and
$$
|e^f-1|^*_{1, frac{alpha}{2}}:=sup_{[a, b]}|e^f-1|+S_1
$$
are equivalent, that is there exist two constants $c, C>0$ such that
$$
c|e^f-1|_{1, frac{alpha}{2}}leq|e^f-1|^*_{1, frac{alpha}{2}}leq C|e^f-1|_{1, frac{alpha}{2}}.
$$
In particular, for $c=C=frac{1}{2}$, we have
$$
|e^f-1|^*_{1, frac{alpha}{2}}leqfrac{1}{2}|e^f-1|_{1, frac{alpha}{2}}.
$$
Then
$$
|e^f-1|_{1, frac{alpha}{2}}leqfrac{1}{2}|e^f-1|_{1, frac{alpha}{2}}+|f|_{1, frac{alpha}{2}}sup_{[a, b]}|e^f|,
$$
that is
$$
|e^f-1|_{1, frac{alpha}{2}}leq2sup_{[a, b]}|e^f||f|_{1, frac{alpha}{2}}=K|f|_{1, frac{alpha}{2}}
$$
where $K=2sup_{[a, b]}|e^f|$. Is this proof right?
Thank You
functional-analysis inequality holder-spaces
asked Jan 19 at 15:46
JejiJeji
34318
$endgroup$
|
show 1 more comment
$begingroup$
I have to prove that, given $fin C^{1, frac{alpha}{2}}([a, b])$, such that $|f|_{infty}<L$ for some $L$, $fgeq0$ and $alphain(0, 1)$, there exists a constant $K$ such that
$$
|e^f-1|_{1, frac{alpha}{2}}leq K|f|_{1, frac{alpha}{2}}.
$$
where $|cdot|_{1, frac{alpha}{2}}$ is the usual $(1, frac{alpha}{2})$-Hölder norm.
This is not a real exercise. It is only a thing to prove to deduce the second inequality in Theorem 3.1, (i), by Proposition 3.1, in the paper https://arxiv.org/pdf/1402.2467. In this case $f=C_T$.
My attempt. We have
$$
|e^f-1|_{1, frac{alpha}{2}}=sup_{[a, b]}|e^f-1|+sup_{[a, b]}|f'e^f|+sup_{xneq y}frac{|f'(x)e^{f(x)}-f'(y)e^{f(y)}|}{|x-y|^{frac{alpha}{2}}},
$$
$$
|f|_{1, frac{alpha}{2}}=sup_{[a, b]}|f|+sup_{[a, b]}|f'|+sup_{xneq y}frac{|f'(x)-f'(y)|}{|x-y|^{frac{alpha}{2}}}.
$$
For simplicity, we call the $sup$ in the $|e^f-1|_{1, frac{alpha}{2}}$ expression with $S_1$ and the $sup$ in the $|f|_{1, frac{alpha}{2}}$ expression with $S_2$. Observe that
$$
|e^f-1|_{1, frac{alpha}{2}}leq sup_{[a, b]}|e^f-1|+sup_{[a, b]}|f'|sup_{[a, b]}|e^f|+S_1leqsup_{[a, b]}|e^f-1|+|f|_{1, frac{alpha}{2}}sup_{[a, b]}|e^f|+S_1.
$$
Now, by the arguments in Another definition of a Holder norm, the norms $|e^f-1|_{1, frac{alpha}{2}}$ and
$$
|e^f-1|^*_{1, frac{alpha}{2}}:=sup_{[a, b]}|e^f-1|+S_1
$$
are equivalent, that is there exist two constants $c, C>0$ such that
$$
c|e^f-1|_{1, frac{alpha}{2}}leq|e^f-1|^*_{1, frac{alpha}{2}}leq C|e^f-1|_{1, frac{alpha}{2}}.
$$
In particular, for $c=C=frac{1}{2}$, we have
$$
|e^f-1|^*_{1, frac{alpha}{2}}leqfrac{1}{2}|e^f-1|_{1, frac{alpha}{2}}.
$$
Then
$$
|e^f-1|_{1, frac{alpha}{2}}leqfrac{1}{2}|e^f-1|_{1, frac{alpha}{2}}+|f|_{1, frac{alpha}{2}}sup_{[a, b]}|e^f|,
$$
that is
$$
|e^f-1|_{1, frac{alpha}{2}}leq2sup_{[a, b]}|e^f||f|_{1, frac{alpha}{2}}=K|f|_{1, frac{alpha}{2}}
$$
where $K=2sup_{[a, b]}|e^f|$. Is this proof right?
Thank You
functional-analysis inequality holder-spaces
asked Jan 19 at 15:46
JejiJeji
34318
$endgroup$
$begingroup$
There must be a missing assumption here, something like $|f|_infty le C$ for some $C$. Otherwise the sequence of functions $f_n(x) = n, , n ge 1$ is a counterexample.
$endgroup$
– Hans Engler
Jan 19 at 15:57
$begingroup$
@HansEngler Yes, sorry. There is also the assumption that $|f|_{infty}leq C$ for some $C$. Is it now correct?
$endgroup$
– Jeji
Jan 19 at 17:28
$begingroup$
please edit your post accordingly. As to your proof attempt, you cannot assume that $c = C = frac{1}{2}$.
$endgroup$
– Hans Engler
Jan 19 at 21:54
$begingroup$
@HansEngler I modifed my post. Why can’t I assume $c=C=frac{1}{2}$? Then, how can I prove it?
$endgroup$
– Jeji
Jan 20 at 13:35
1
$begingroup$
The constants $c$ and $C$ are only known to exist. You cannot assume that they have specific values. You also need to replace $e^{f'(x)}$ etc. with $e^{f(x)}$ in $S_1$. Finally, please explain where this problem is coming from. It appears to be too complicated for an exercise.
$endgroup$
– Hans Engler
Jan 20 at 16:55
|
show 1 more comment
$begingroup$
I have to prove that, given $fin C^{1, frac{alpha}{2}}([a, b])$, such that $|f|_{infty}<L$ for some $L$, $fgeq0$ and $alphain(0, 1)$, there exists a constant $K$ such that
$$
|e^f-1|_{1, frac{alpha}{2}}leq K|f|_{1, frac{alpha}{2}}.
$$
where $|cdot|_{1, frac{alpha}{2}}$ is the usual $(1, frac{alpha}{2})$-Hölder norm.
This is not a real exercise. It is only a thing to prove to deduce the second inequality in Theorem 3.1, (i), by Proposition 3.1, in the paper https://arxiv.org/pdf/1402.2467. In this case $f=C_T$.
My attempt. We have
$$
|e^f-1|_{1, frac{alpha}{2}}=sup_{[a, b]}|e^f-1|+sup_{[a, b]}|f'e^f|+sup_{xneq y}frac{|f'(x)e^{f(x)}-f'(y)e^{f(y)}|}{|x-y|^{frac{alpha}{2}}},
$$
$$
|f|_{1, frac{alpha}{2}}=sup_{[a, b]}|f|+sup_{[a, b]}|f'|+sup_{xneq y}frac{|f'(x)-f'(y)|}{|x-y|^{frac{alpha}{2}}}.
$$
For simplicity, we call the $sup$ in the $|e^f-1|_{1, frac{alpha}{2}}$ expression with $S_1$ and the $sup$ in the $|f|_{1, frac{alpha}{2}}$ expression with $S_2$. Observe that
$$
|e^f-1|_{1, frac{alpha}{2}}leq sup_{[a, b]}|e^f-1|+sup_{[a, b]}|f'|sup_{[a, b]}|e^f|+S_1leqsup_{[a, b]}|e^f-1|+|f|_{1, frac{alpha}{2}}sup_{[a, b]}|e^f|+S_1.
$$
Now, by the arguments in Another definition of a Holder norm, the norms $|e^f-1|_{1, frac{alpha}{2}}$ and
$$
|e^f-1|^*_{1, frac{alpha}{2}}:=sup_{[a, b]}|e^f-1|+S_1
$$
are equivalent, that is there exist two constants $c, C>0$ such that
$$
c|e^f-1|_{1, frac{alpha}{2}}leq|e^f-1|^*_{1, frac{alpha}{2}}leq C|e^f-1|_{1, frac{alpha}{2}}.
$$
In particular, for $c=C=frac{1}{2}$, we have
$$
|e^f-1|^*_{1, frac{alpha}{2}}leqfrac{1}{2}|e^f-1|_{1, frac{alpha}{2}}.
$$
Then
$$
|e^f-1|_{1, frac{alpha}{2}}leqfrac{1}{2}|e^f-1|_{1, frac{alpha}{2}}+|f|_{1, frac{alpha}{2}}sup_{[a, b]}|e^f|,
$$
that is
$$
|e^f-1|_{1, frac{alpha}{2}}leq2sup_{[a, b]}|e^f||f|_{1, frac{alpha}{2}}=K|f|_{1, frac{alpha}{2}}
$$
where $K=2sup_{[a, b]}|e^f|$. Is this proof right?
Thank You
functional-analysis inequality holder-spaces
asked Jan 19 at 15:46
JejiJeji
34318
$endgroup$
I have to prove that, given $fin C^{1, frac{alpha}{2}}([a, b])$, such that $|f|_{infty}<L$ for some $L$, $fgeq0$ and $alphain(0, 1)$, there exists a constant $K$ such that
$$
|e^f-1|_{1, frac{alpha}{2}}leq K|f|_{1, frac{alpha}{2}}.
$$
where $|cdot|_{1, frac{alpha}{2}}$ is the usual $(1, frac{alpha}{2})$-Hölder norm.
This is not a real exercise. It is only a thing to prove to deduce the second inequality in Theorem 3.1, (i), by Proposition 3.1, in the paper https://arxiv.org/pdf/1402.2467. In this case $f=C_T$.
My attempt. We have
$$
|e^f-1|_{1, frac{alpha}{2}}=sup_{[a, b]}|e^f-1|+sup_{[a, b]}|f'e^f|+sup_{xneq y}frac{|f'(x)e^{f(x)}-f'(y)e^{f(y)}|}{|x-y|^{frac{alpha}{2}}},
$$
$$
|f|_{1, frac{alpha}{2}}=sup_{[a, b]}|f|+sup_{[a, b]}|f'|+sup_{xneq y}frac{|f'(x)-f'(y)|}{|x-y|^{frac{alpha}{2}}}.
$$
For simplicity, we call the $sup$ in the $|e^f-1|_{1, frac{alpha}{2}}$ expression with $S_1$ and the $sup$ in the $|f|_{1, frac{alpha}{2}}$ expression with $S_2$. Observe that
$$
|e^f-1|_{1, frac{alpha}{2}}leq sup_{[a, b]}|e^f-1|+sup_{[a, b]}|f'|sup_{[a, b]}|e^f|+S_1leqsup_{[a, b]}|e^f-1|+|f|_{1, frac{alpha}{2}}sup_{[a, b]}|e^f|+S_1.
$$
Now, by the arguments in Another definition of a Holder norm, the norms $|e^f-1|_{1, frac{alpha}{2}}$ and
$$
|e^f-1|^*_{1, frac{alpha}{2}}:=sup_{[a, b]}|e^f-1|+S_1
$$
are equivalent, that is there exist two constants $c, C>0$ such that
$$
c|e^f-1|_{1, frac{alpha}{2}}leq|e^f-1|^*_{1, frac{alpha}{2}}leq C|e^f-1|_{1, frac{alpha}{2}}.
$$
In particular, for $c=C=frac{1}{2}$, we have
$$
|e^f-1|^*_{1, frac{alpha}{2}}leqfrac{1}{2}|e^f-1|_{1, frac{alpha}{2}}.
$$
Then
$$
|e^f-1|_{1, frac{alpha}{2}}leqfrac{1}{2}|e^f-1|_{1, frac{alpha}{2}}+|f|_{1, frac{alpha}{2}}sup_{[a, b]}|e^f|,
$$
that is
$$
|e^f-1|_{1, frac{alpha}{2}}leq2sup_{[a, b]}|e^f||f|_{1, frac{alpha}{2}}=K|f|_{1, frac{alpha}{2}}
$$
where $K=2sup_{[a, b]}|e^f|$. Is this proof right?
Thank You
functional-analysis inequality holder-spaces
functional-analysis inequality holder-spaces
asked Jan 19 at 15:46
JejiJeji
34318
asked Jan 19 at 15:46
JejiJeji
34318
edited Jan 21 at 0:54
Jeji
asked Jan 19 at 15:46
JejiJeji
34318
asked Jan 19 at 15:46
JejiJeji
34318
asked Jan 19 at 15:46
JejiJeji
34318
34318
$begingroup$
There must be a missing assumption here, something like $|f|_infty le C$ for some $C$. Otherwise the sequence of functions $f_n(x) = n, , n ge 1$ is a counterexample.
$endgroup$
– Hans Engler
Jan 19 at 15:57
$begingroup$
@HansEngler Yes, sorry. There is also the assumption that $|f|_{infty}leq C$ for some $C$. Is it now correct?
$endgroup$
– Jeji
Jan 19 at 17:28
$begingroup$
please edit your post accordingly. As to your proof attempt, you cannot assume that $c = C = frac{1}{2}$.
$endgroup$
– Hans Engler
Jan 19 at 21:54
$begingroup$
@HansEngler I modifed my post. Why can’t I assume $c=C=frac{1}{2}$? Then, how can I prove it?
$endgroup$
– Jeji
Jan 20 at 13:35
1
$begingroup$
The constants $c$ and $C$ are only known to exist. You cannot assume that they have specific values. You also need to replace $e^{f'(x)}$ etc. with $e^{f(x)}$ in $S_1$. Finally, please explain where this problem is coming from. It appears to be too complicated for an exercise.
$endgroup$
– Hans Engler
Jan 20 at 16:55
|
show 1 more comment
$begingroup$
There must be a missing assumption here, something like $|f|_infty le C$ for some $C$. Otherwise the sequence of functions $f_n(x) = n, , n ge 1$ is a counterexample.
$endgroup$
– Hans Engler
Jan 19 at 15:57
$begingroup$
@HansEngler Yes, sorry. There is also the assumption that $|f|_{infty}leq C$ for some $C$. Is it now correct?
$endgroup$
– Jeji
Jan 19 at 17:28
$begingroup$
please edit your post accordingly. As to your proof attempt, you cannot assume that $c = C = frac{1}{2}$.
$endgroup$
– Hans Engler
Jan 19 at 21:54
$begingroup$
@HansEngler I modifed my post. Why can’t I assume $c=C=frac{1}{2}$? Then, how can I prove it?
$endgroup$
– Jeji
Jan 20 at 13:35
1
$begingroup$
The constants $c$ and $C$ are only known to exist. You cannot assume that they have specific values. You also need to replace $e^{f'(x)}$ etc. with $e^{f(x)}$ in $S_1$. Finally, please explain where this problem is coming from. It appears to be too complicated for an exercise.
$endgroup$
– Hans Engler
Jan 20 at 16:55
$begingroup$
There must be a missing assumption here, something like $|f|_infty le C$ for some $C$. Otherwise the sequence of functions $f_n(x) = n, , n ge 1$ is a counterexample.
$endgroup$
– Hans Engler
Jan 19 at 15:57
$begingroup$
There must be a missing assumption here, something like $|f|_infty le C$ for some $C$. Otherwise the sequence of functions $f_n(x) = n, , n ge 1$ is a counterexample.
$endgroup$
– Hans Engler
Jan 19 at 15:57
$begingroup$
@HansEngler Yes, sorry. There is also the assumption that $|f|_{infty}leq C$ for some $C$. Is it now correct?
$endgroup$
– Jeji
Jan 19 at 17:28
$begingroup$
@HansEngler Yes, sorry. There is also the assumption that $|f|_{infty}leq C$ for some $C$. Is it now correct?
$endgroup$
– Jeji
Jan 19 at 17:28
$begingroup$
please edit your post accordingly. As to your proof attempt, you cannot assume that $c = C = frac{1}{2}$.
$endgroup$
– Hans Engler
Jan 19 at 21:54
$begingroup$
please edit your post accordingly. As to your proof attempt, you cannot assume that $c = C = frac{1}{2}$.
$endgroup$
– Hans Engler
Jan 19 at 21:54
$begingroup$
@HansEngler I modifed my post. Why can’t I assume $c=C=frac{1}{2}$? Then, how can I prove it?
$endgroup$
– Jeji
Jan 20 at 13:35
$begingroup$
@HansEngler I modifed my post. Why can’t I assume $c=C=frac{1}{2}$? Then, how can I prove it?
$endgroup$
– Jeji
Jan 20 at 13:35
1
1
$begingroup$
The constants $c$ and $C$ are only known to exist. You cannot assume that they have specific values. You also need to replace $e^{f'(x)}$ etc. with $e^{f(x)}$ in $S_1$. Finally, please explain where this problem is coming from. It appears to be too complicated for an exercise.
$endgroup$
– Hans Engler
Jan 20 at 16:55
$begingroup$
The constants $c$ and $C$ are only known to exist. You cannot assume that they have specific values. You also need to replace $e^{f'(x)}$ etc. with $e^{f(x)}$ in $S_1$. Finally, please explain where this problem is coming from. It appears to be too complicated for an exercise.
$endgroup$
– Hans Engler
Jan 20 at 16:55
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The estimate is correct for all $(1,alpha)$ Holder norms.
We can start out as in the OP's attempt to solve the problem. Recall that
$$
|f|_{1, alpha}=sup_{[a, b]}|f|+sup_{[a, b]}|f'|+sup_{xneq y}frac{|f'(x)-f'(y)|}{|x-y|^{alpha}} = A_0(f) + A_1(f) + A_2(f)
$$
Similarly
$$
|e^f-1|_{1, alpha}=sup_{[a, b]}|e^f - 1|+sup_{[a, b]}|f'e^f|+sup_{xneq y}frac{|f'(x)e^{f(x)}-f'(y)e^{f(y)}|}{|x-y|^{alpha}} =: B_0 + B_1 + B_2
$$
From the answer to the question quoted in your post, we know that there is a constant $C_0$ such that
$$
boxed{A_1(f) le C_0 A_0(f)^gamma A_2(f)^{1-gamma}}
$$
where $gamma = frac{alpha}{1+alpha}$.
Finding $K_0$ such that $B_0 le K_0 A_0(f)$
Set $K_0 = e^L$. Since $0 le f(x) le L$ on $[a,b]$, it follows that
$$
0 le e^{f(x)} - 1 = e^{f(x)} - e^0 = e^xi f(x) le K_0 f(x) le K_0 A_0
$$
for all $x$. Take the supremum on the left to deduce $B_0 le K_0 A_0$.
Note that also $A_0(f) le K_0$.
Finding $K_1$ such that $B_1 le K_1 A_1(f)$
Set $K_1 = e^L$. Then, since $0 le f(x) le L$, for $x in [a,b]$
$$
|f'(x) e^{f(x)}| le e^L |f'(x)| = K_1 |f'(x)| le K_1 A_1, .
$$
This implies $B_1 le K_1 A_1$,
Finding $K_2$ such that $B_2 le K_2 A_2(f)$
Set $K_2 = K_0 + 2 C_0 K_0^{1 + alpha}$ where $C_0$ is from the boxed inequality. Let $a le x < y le b$. Then
$$
|f'(x)e^{f(x)} - f'(y)e^{f(y)}| le |f'(x)e^{f(x)} - f'(x)e^{f(y)}| +
|f'(x)e^{f(y)} - f'(y)e^{f(y)}|
$$
The second term may be estimated by
$$
dots le |f'(x) - f'(y)| e^{f(y)} le A_2(f) |x-y|^alpha K_0 , .
$$
The first term on the right may be estimated, using the mean value theorem, by
$$
dots le |f'(x)||e^{f(x)} - e^{f(y)}| = |f'(x)||x-y||f'(zeta) e^{f(zeta)}|
le |x-y| A_1^2(f) K_0
$$
We can alternatively estimate the first term on the right hand side by
$$
dots le 2 K_0 A_1(f)
$$
Raise the first estimate to the power $alpha$ and the second to the power $1 - alpha$ and multiply them together. The result is
$$
|f'(x)e^{f(x)} - f'(x)e^{f(y)}| le 2|x-y|^alpha A_1^{1 + alpha}(f) K_0 le 2 C_0 K_0 A_2(f)A_0(f)^alpha |x-y|^alpha , .
$$
We may replace $A_0(f)^alpha$ by $K_0^alpha$. Combining the estimates and dividing by $|x-y|^alpha$, we obtain
$$
|frac{|f'(x)e^{f(x)} - f'(y)e^{f(y)}|}{|x-y|^alpha} le K_0A_2(f) + 2C_0 K_0^{1+alpha} A_2(f) = K_2 A_2(f) , .
$$
Taking the supremum we see that $B_2 le K_2 A_2(f)$.
Combining all three estimates implies
$$
B_0 + B_1 + B_2 le K(A_0(f) + A_1(f) + A_2(f))
$$
for some $K$.
answered Jan 20 at 18:18
Hans EnglerHans Engler
10.5k11836
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1 Answer
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$begingroup$
The estimate is correct for all $(1,alpha)$ Holder norms.
We can start out as in the OP's attempt to solve the problem. Recall that
$$
|f|_{1, alpha}=sup_{[a, b]}|f|+sup_{[a, b]}|f'|+sup_{xneq y}frac{|f'(x)-f'(y)|}{|x-y|^{alpha}} = A_0(f) + A_1(f) + A_2(f)
$$
Similarly
$$
|e^f-1|_{1, alpha}=sup_{[a, b]}|e^f - 1|+sup_{[a, b]}|f'e^f|+sup_{xneq y}frac{|f'(x)e^{f(x)}-f'(y)e^{f(y)}|}{|x-y|^{alpha}} =: B_0 + B_1 + B_2
$$
From the answer to the question quoted in your post, we know that there is a constant $C_0$ such that
$$
boxed{A_1(f) le C_0 A_0(f)^gamma A_2(f)^{1-gamma}}
$$
where $gamma = frac{alpha}{1+alpha}$.
Finding $K_0$ such that $B_0 le K_0 A_0(f)$
Set $K_0 = e^L$. Since $0 le f(x) le L$ on $[a,b]$, it follows that
$$
0 le e^{f(x)} - 1 = e^{f(x)} - e^0 = e^xi f(x) le K_0 f(x) le K_0 A_0
$$
for all $x$. Take the supremum on the left to deduce $B_0 le K_0 A_0$.
Note that also $A_0(f) le K_0$.
Finding $K_1$ such that $B_1 le K_1 A_1(f)$
Set $K_1 = e^L$. Then, since $0 le f(x) le L$, for $x in [a,b]$
$$
|f'(x) e^{f(x)}| le e^L |f'(x)| = K_1 |f'(x)| le K_1 A_1, .
$$
This implies $B_1 le K_1 A_1$,
Finding $K_2$ such that $B_2 le K_2 A_2(f)$
Set $K_2 = K_0 + 2 C_0 K_0^{1 + alpha}$ where $C_0$ is from the boxed inequality. Let $a le x < y le b$. Then
$$
|f'(x)e^{f(x)} - f'(y)e^{f(y)}| le |f'(x)e^{f(x)} - f'(x)e^{f(y)}| +
|f'(x)e^{f(y)} - f'(y)e^{f(y)}|
$$
The second term may be estimated by
$$
dots le |f'(x) - f'(y)| e^{f(y)} le A_2(f) |x-y|^alpha K_0 , .
$$
The first term on the right may be estimated, using the mean value theorem, by
$$
dots le |f'(x)||e^{f(x)} - e^{f(y)}| = |f'(x)||x-y||f'(zeta) e^{f(zeta)}|
le |x-y| A_1^2(f) K_0
$$
We can alternatively estimate the first term on the right hand side by
$$
dots le 2 K_0 A_1(f)
$$
Raise the first estimate to the power $alpha$ and the second to the power $1 - alpha$ and multiply them together. The result is
$$
|f'(x)e^{f(x)} - f'(x)e^{f(y)}| le 2|x-y|^alpha A_1^{1 + alpha}(f) K_0 le 2 C_0 K_0 A_2(f)A_0(f)^alpha |x-y|^alpha , .
$$
We may replace $A_0(f)^alpha$ by $K_0^alpha$. Combining the estimates and dividing by $|x-y|^alpha$, we obtain
$$
|frac{|f'(x)e^{f(x)} - f'(y)e^{f(y)}|}{|x-y|^alpha} le K_0A_2(f) + 2C_0 K_0^{1+alpha} A_2(f) = K_2 A_2(f) , .
$$
Taking the supremum we see that $B_2 le K_2 A_2(f)$.
Combining all three estimates implies
$$
B_0 + B_1 + B_2 le K(A_0(f) + A_1(f) + A_2(f))
$$
for some $K$.
answered Jan 20 at 18:18
Hans EnglerHans Engler
10.5k11836
$endgroup$
add a comment |
$begingroup$
The estimate is correct for all $(1,alpha)$ Holder norms.
We can start out as in the OP's attempt to solve the problem. Recall that
$$
|f|_{1, alpha}=sup_{[a, b]}|f|+sup_{[a, b]}|f'|+sup_{xneq y}frac{|f'(x)-f'(y)|}{|x-y|^{alpha}} = A_0(f) + A_1(f) + A_2(f)
$$
Similarly
$$
|e^f-1|_{1, alpha}=sup_{[a, b]}|e^f - 1|+sup_{[a, b]}|f'e^f|+sup_{xneq y}frac{|f'(x)e^{f(x)}-f'(y)e^{f(y)}|}{|x-y|^{alpha}} =: B_0 + B_1 + B_2
$$
From the answer to the question quoted in your post, we know that there is a constant $C_0$ such that
$$
boxed{A_1(f) le C_0 A_0(f)^gamma A_2(f)^{1-gamma}}
$$
where $gamma = frac{alpha}{1+alpha}$.
Finding $K_0$ such that $B_0 le K_0 A_0(f)$
Set $K_0 = e^L$. Since $0 le f(x) le L$ on $[a,b]$, it follows that
$$
0 le e^{f(x)} - 1 = e^{f(x)} - e^0 = e^xi f(x) le K_0 f(x) le K_0 A_0
$$
for all $x$. Take the supremum on the left to deduce $B_0 le K_0 A_0$.
Note that also $A_0(f) le K_0$.
Finding $K_1$ such that $B_1 le K_1 A_1(f)$
Set $K_1 = e^L$. Then, since $0 le f(x) le L$, for $x in [a,b]$
$$
|f'(x) e^{f(x)}| le e^L |f'(x)| = K_1 |f'(x)| le K_1 A_1, .
$$
This implies $B_1 le K_1 A_1$,
Finding $K_2$ such that $B_2 le K_2 A_2(f)$
Set $K_2 = K_0 + 2 C_0 K_0^{1 + alpha}$ where $C_0$ is from the boxed inequality. Let $a le x < y le b$. Then
$$
|f'(x)e^{f(x)} - f'(y)e^{f(y)}| le |f'(x)e^{f(x)} - f'(x)e^{f(y)}| +
|f'(x)e^{f(y)} - f'(y)e^{f(y)}|
$$
The second term may be estimated by
$$
dots le |f'(x) - f'(y)| e^{f(y)} le A_2(f) |x-y|^alpha K_0 , .
$$
The first term on the right may be estimated, using the mean value theorem, by
$$
dots le |f'(x)||e^{f(x)} - e^{f(y)}| = |f'(x)||x-y||f'(zeta) e^{f(zeta)}|
le |x-y| A_1^2(f) K_0
$$
We can alternatively estimate the first term on the right hand side by
$$
dots le 2 K_0 A_1(f)
$$
Raise the first estimate to the power $alpha$ and the second to the power $1 - alpha$ and multiply them together. The result is
$$
|f'(x)e^{f(x)} - f'(x)e^{f(y)}| le 2|x-y|^alpha A_1^{1 + alpha}(f) K_0 le 2 C_0 K_0 A_2(f)A_0(f)^alpha |x-y|^alpha , .
$$
We may replace $A_0(f)^alpha$ by $K_0^alpha$. Combining the estimates and dividing by $|x-y|^alpha$, we obtain
$$
|frac{|f'(x)e^{f(x)} - f'(y)e^{f(y)}|}{|x-y|^alpha} le K_0A_2(f) + 2C_0 K_0^{1+alpha} A_2(f) = K_2 A_2(f) , .
$$
Taking the supremum we see that $B_2 le K_2 A_2(f)$.
Combining all three estimates implies
$$
B_0 + B_1 + B_2 le K(A_0(f) + A_1(f) + A_2(f))
$$
for some $K$.
answered Jan 20 at 18:18
Hans EnglerHans Engler
10.5k11836
$endgroup$
add a comment |
$begingroup$
The estimate is correct for all $(1,alpha)$ Holder norms.
We can start out as in the OP's attempt to solve the problem. Recall that
$$
|f|_{1, alpha}=sup_{[a, b]}|f|+sup_{[a, b]}|f'|+sup_{xneq y}frac{|f'(x)-f'(y)|}{|x-y|^{alpha}} = A_0(f) + A_1(f) + A_2(f)
$$
Similarly
$$
|e^f-1|_{1, alpha}=sup_{[a, b]}|e^f - 1|+sup_{[a, b]}|f'e^f|+sup_{xneq y}frac{|f'(x)e^{f(x)}-f'(y)e^{f(y)}|}{|x-y|^{alpha}} =: B_0 + B_1 + B_2
$$
From the answer to the question quoted in your post, we know that there is a constant $C_0$ such that
$$
boxed{A_1(f) le C_0 A_0(f)^gamma A_2(f)^{1-gamma}}
$$
where $gamma = frac{alpha}{1+alpha}$.
Finding $K_0$ such that $B_0 le K_0 A_0(f)$
Set $K_0 = e^L$. Since $0 le f(x) le L$ on $[a,b]$, it follows that
$$
0 le e^{f(x)} - 1 = e^{f(x)} - e^0 = e^xi f(x) le K_0 f(x) le K_0 A_0
$$
for all $x$. Take the supremum on the left to deduce $B_0 le K_0 A_0$.
Note that also $A_0(f) le K_0$.
Finding $K_1$ such that $B_1 le K_1 A_1(f)$
Set $K_1 = e^L$. Then, since $0 le f(x) le L$, for $x in [a,b]$
$$
|f'(x) e^{f(x)}| le e^L |f'(x)| = K_1 |f'(x)| le K_1 A_1, .
$$
This implies $B_1 le K_1 A_1$,
Finding $K_2$ such that $B_2 le K_2 A_2(f)$
Set $K_2 = K_0 + 2 C_0 K_0^{1 + alpha}$ where $C_0$ is from the boxed inequality. Let $a le x < y le b$. Then
$$
|f'(x)e^{f(x)} - f'(y)e^{f(y)}| le |f'(x)e^{f(x)} - f'(x)e^{f(y)}| +
|f'(x)e^{f(y)} - f'(y)e^{f(y)}|
$$
The second term may be estimated by
$$
dots le |f'(x) - f'(y)| e^{f(y)} le A_2(f) |x-y|^alpha K_0 , .
$$
The first term on the right may be estimated, using the mean value theorem, by
$$
dots le |f'(x)||e^{f(x)} - e^{f(y)}| = |f'(x)||x-y||f'(zeta) e^{f(zeta)}|
le |x-y| A_1^2(f) K_0
$$
We can alternatively estimate the first term on the right hand side by
$$
dots le 2 K_0 A_1(f)
$$
Raise the first estimate to the power $alpha$ and the second to the power $1 - alpha$ and multiply them together. The result is
$$
|f'(x)e^{f(x)} - f'(x)e^{f(y)}| le 2|x-y|^alpha A_1^{1 + alpha}(f) K_0 le 2 C_0 K_0 A_2(f)A_0(f)^alpha |x-y|^alpha , .
$$
We may replace $A_0(f)^alpha$ by $K_0^alpha$. Combining the estimates and dividing by $|x-y|^alpha$, we obtain
$$
|frac{|f'(x)e^{f(x)} - f'(y)e^{f(y)}|}{|x-y|^alpha} le K_0A_2(f) + 2C_0 K_0^{1+alpha} A_2(f) = K_2 A_2(f) , .
$$
Taking the supremum we see that $B_2 le K_2 A_2(f)$.
Combining all three estimates implies
$$
B_0 + B_1 + B_2 le K(A_0(f) + A_1(f) + A_2(f))
$$
for some $K$.
answered Jan 20 at 18:18
Hans EnglerHans Engler
10.5k11836
$endgroup$
The estimate is correct for all $(1,alpha)$ Holder norms.
We can start out as in the OP's attempt to solve the problem. Recall that
$$
|f|_{1, alpha}=sup_{[a, b]}|f|+sup_{[a, b]}|f'|+sup_{xneq y}frac{|f'(x)-f'(y)|}{|x-y|^{alpha}} = A_0(f) + A_1(f) + A_2(f)
$$
Similarly
$$
|e^f-1|_{1, alpha}=sup_{[a, b]}|e^f - 1|+sup_{[a, b]}|f'e^f|+sup_{xneq y}frac{|f'(x)e^{f(x)}-f'(y)e^{f(y)}|}{|x-y|^{alpha}} =: B_0 + B_1 + B_2
$$
From the answer to the question quoted in your post, we know that there is a constant $C_0$ such that
$$
boxed{A_1(f) le C_0 A_0(f)^gamma A_2(f)^{1-gamma}}
$$
where $gamma = frac{alpha}{1+alpha}$.
Finding $K_0$ such that $B_0 le K_0 A_0(f)$
Set $K_0 = e^L$. Since $0 le f(x) le L$ on $[a,b]$, it follows that
$$
0 le e^{f(x)} - 1 = e^{f(x)} - e^0 = e^xi f(x) le K_0 f(x) le K_0 A_0
$$
for all $x$. Take the supremum on the left to deduce $B_0 le K_0 A_0$.
Note that also $A_0(f) le K_0$.
Finding $K_1$ such that $B_1 le K_1 A_1(f)$
Set $K_1 = e^L$. Then, since $0 le f(x) le L$, for $x in [a,b]$
$$
|f'(x) e^{f(x)}| le e^L |f'(x)| = K_1 |f'(x)| le K_1 A_1, .
$$
This implies $B_1 le K_1 A_1$,
Finding $K_2$ such that $B_2 le K_2 A_2(f)$
Set $K_2 = K_0 + 2 C_0 K_0^{1 + alpha}$ where $C_0$ is from the boxed inequality. Let $a le x < y le b$. Then
$$
|f'(x)e^{f(x)} - f'(y)e^{f(y)}| le |f'(x)e^{f(x)} - f'(x)e^{f(y)}| +
|f'(x)e^{f(y)} - f'(y)e^{f(y)}|
$$
The second term may be estimated by
$$
dots le |f'(x) - f'(y)| e^{f(y)} le A_2(f) |x-y|^alpha K_0 , .
$$
The first term on the right may be estimated, using the mean value theorem, by
$$
dots le |f'(x)||e^{f(x)} - e^{f(y)}| = |f'(x)||x-y||f'(zeta) e^{f(zeta)}|
le |x-y| A_1^2(f) K_0
$$
We can alternatively estimate the first term on the right hand side by
$$
dots le 2 K_0 A_1(f)
$$
Raise the first estimate to the power $alpha$ and the second to the power $1 - alpha$ and multiply them together. The result is
$$
|f'(x)e^{f(x)} - f'(x)e^{f(y)}| le 2|x-y|^alpha A_1^{1 + alpha}(f) K_0 le 2 C_0 K_0 A_2(f)A_0(f)^alpha |x-y|^alpha , .
$$
We may replace $A_0(f)^alpha$ by $K_0^alpha$. Combining the estimates and dividing by $|x-y|^alpha$, we obtain
$$
|frac{|f'(x)e^{f(x)} - f'(y)e^{f(y)}|}{|x-y|^alpha} le K_0A_2(f) + 2C_0 K_0^{1+alpha} A_2(f) = K_2 A_2(f) , .
$$
Taking the supremum we see that $B_2 le K_2 A_2(f)$.
Combining all three estimates implies
$$
B_0 + B_1 + B_2 le K(A_0(f) + A_1(f) + A_2(f))
$$
for some $K$.
answered Jan 20 at 18:18
Hans EnglerHans Engler
10.5k11836
answered Jan 20 at 18:18
Hans EnglerHans Engler
10.5k11836
answered Jan 20 at 18:18
Hans EnglerHans Engler
10.5k11836
answered Jan 20 at 18:18
Hans EnglerHans Engler
10.5k11836
10.5k11836
add a comment |
add a comment |
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$begingroup$
There must be a missing assumption here, something like $|f|_infty le C$ for some $C$. Otherwise the sequence of functions $f_n(x) = n, , n ge 1$ is a counterexample.
$endgroup$
– Hans Engler
Jan 19 at 15:57
$begingroup$
@HansEngler Yes, sorry. There is also the assumption that $|f|_{infty}leq C$ for some $C$. Is it now correct?
$endgroup$
– Jeji
Jan 19 at 17:28
$begingroup$
please edit your post accordingly. As to your proof attempt, you cannot assume that $c = C = frac{1}{2}$.
$endgroup$
– Hans Engler
Jan 19 at 21:54
$begingroup$
@HansEngler I modifed my post. Why can’t I assume $c=C=frac{1}{2}$? Then, how can I prove it?
$endgroup$
– Jeji
Jan 20 at 13:35
1
$begingroup$
The constants $c$ and $C$ are only known to exist. You cannot assume that they have specific values. You also need to replace $e^{f'(x)}$ etc. with $e^{f(x)}$ in $S_1$. Finally, please explain where this problem is coming from. It appears to be too complicated for an exercise.
$endgroup$
– Hans Engler
Jan 20 at 16:55