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Space with continuous bounded derivatives of all functions is complete
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I'm doing the following exercise:
Let $I=(a,b)subset mathbbR$, and we call $H^k(I)$ the space of continuous functions on $I$ such that its derivatives are all bounded $(f,f',f'',dots)$. We define now $$||f||_k=sup.$$
Prove that with the distance defined above $(||f-g||_k\ $ $with f, gin H^k(I))$ the space $H^k(I)$ is complete.
Okay, so now I have two questions.
First one: I've proved that $C_b(I)$, the space of all continuous bounded functions on $I$ is complete. Obviuosly, $H^k(I)subset C_b(I)$, so if $C_b(I)$ is complete, $H^k(I)$ has to be complete to. What's wrong with my way of seeing this?
Obviously I must be doing something wrong, so the one stated above is not the correct answer. That leads me to think that maybe we can prove this exercise using or following a not so different path we followed while proving $C_b(I)$ is complete. But I get stuck all the time. Here's my try:
Let $f_n$ be a Cauchy sequence in $C_b(I)$. Fot every $xin I$, we have that the sequence of real numbers $f_n(x)$ is a Cauchy sequence and it has to be convergent. Let's call its limit $f(x)$. We have then that $f_n(x)to f(x)$. So, now we have to see that $f_n$ converges uniformly to $f(x)$.
Now I don't now how to move on from here. I'll appreciate any hint.
Thanks for your time.
calculus differential-equations derivatives differential-topology
asked Jul 24 at 17:08
Relure
1,948833
add a comment |Â
up vote
1
down vote
favorite
I'm doing the following exercise:
Let $I=(a,b)subset mathbbR$, and we call $H^k(I)$ the space of continuous functions on $I$ such that its derivatives are all bounded $(f,f',f'',dots)$. We define now $$||f||_k=sup.$$
Prove that with the distance defined above $(||f-g||_k\ $ $with f, gin H^k(I))$ the space $H^k(I)$ is complete.
Okay, so now I have two questions.
First one: I've proved that $C_b(I)$, the space of all continuous bounded functions on $I$ is complete. Obviuosly, $H^k(I)subset C_b(I)$, so if $C_b(I)$ is complete, $H^k(I)$ has to be complete to. What's wrong with my way of seeing this?
Obviously I must be doing something wrong, so the one stated above is not the correct answer. That leads me to think that maybe we can prove this exercise using or following a not so different path we followed while proving $C_b(I)$ is complete. But I get stuck all the time. Here's my try:
Let $f_n$ be a Cauchy sequence in $C_b(I)$. Fot every $xin I$, we have that the sequence of real numbers $f_n(x)$ is a Cauchy sequence and it has to be convergent. Let's call its limit $f(x)$. We have then that $f_n(x)to f(x)$. So, now we have to see that $f_n$ converges uniformly to $f(x)$.
Now I don't now how to move on from here. I'll appreciate any hint.
Thanks for your time.
calculus differential-equations derivatives differential-topology
asked Jul 24 at 17:08
Relure
1,948833
1
A hint: If $f_n$ converge in the $lVertcdotrVert_k$ metric to some $f$, then the derivatives $f_n$, $f'_n$, ..., $f^(k)_n$ converge in the $C_b(I)$-norm to $f$, $f'$, ..., $f^(k)$ (can you prove that)?
– user539887
Jul 24 at 18:45
@user539887 Hi, I've been trying to solve it using this approach, but I can't prove that the derivatives converge. I've seen that $f_n$ converges in the $||cdot ||_k$ metric to a $f$ already. I'd really appreciate you to enlighten me uo in an answer, because I'm a bit frustrated with this one.
– Relure
Jul 26 at 17:15
Just start from the derivative of the highest order: $f^(k)_n$ uniformly converge to some $g$, and $f^(k-1)_n$ uniformly converge to some $h$. It follows then that $h'equiv g$. And you repeat this approach for derivatives of orders $k-1$, $k-2$, etc.
– user539887
Jul 26 at 18:01
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm doing the following exercise:
Let $I=(a,b)subset mathbbR$, and we call $H^k(I)$ the space of continuous functions on $I$ such that its derivatives are all bounded $(f,f',f'',dots)$. We define now $$||f||_k=sup.$$
Prove that with the distance defined above $(||f-g||_k\ $ $with f, gin H^k(I))$ the space $H^k(I)$ is complete.
Okay, so now I have two questions.
First one: I've proved that $C_b(I)$, the space of all continuous bounded functions on $I$ is complete. Obviuosly, $H^k(I)subset C_b(I)$, so if $C_b(I)$ is complete, $H^k(I)$ has to be complete to. What's wrong with my way of seeing this?
Obviously I must be doing something wrong, so the one stated above is not the correct answer. That leads me to think that maybe we can prove this exercise using or following a not so different path we followed while proving $C_b(I)$ is complete. But I get stuck all the time. Here's my try:
Let $f_n$ be a Cauchy sequence in $C_b(I)$. Fot every $xin I$, we have that the sequence of real numbers $f_n(x)$ is a Cauchy sequence and it has to be convergent. Let's call its limit $f(x)$. We have then that $f_n(x)to f(x)$. So, now we have to see that $f_n$ converges uniformly to $f(x)$.
Now I don't now how to move on from here. I'll appreciate any hint.
Thanks for your time.
calculus differential-equations derivatives differential-topology
asked Jul 24 at 17:08
Relure
1,948833
I'm doing the following exercise:
Let $I=(a,b)subset mathbbR$, and we call $H^k(I)$ the space of continuous functions on $I$ such that its derivatives are all bounded $(f,f',f'',dots)$. We define now $$||f||_k=sup.$$
Prove that with the distance defined above $(||f-g||_k\ $ $with f, gin H^k(I))$ the space $H^k(I)$ is complete.
Okay, so now I have two questions.
First one: I've proved that $C_b(I)$, the space of all continuous bounded functions on $I$ is complete. Obviuosly, $H^k(I)subset C_b(I)$, so if $C_b(I)$ is complete, $H^k(I)$ has to be complete to. What's wrong with my way of seeing this?
Obviously I must be doing something wrong, so the one stated above is not the correct answer. That leads me to think that maybe we can prove this exercise using or following a not so different path we followed while proving $C_b(I)$ is complete. But I get stuck all the time. Here's my try:
Let $f_n$ be a Cauchy sequence in $C_b(I)$. Fot every $xin I$, we have that the sequence of real numbers $f_n(x)$ is a Cauchy sequence and it has to be convergent. Let's call its limit $f(x)$. We have then that $f_n(x)to f(x)$. So, now we have to see that $f_n$ converges uniformly to $f(x)$.
Now I don't now how to move on from here. I'll appreciate any hint.
Thanks for your time.
calculus differential-equations derivatives differential-topology
asked Jul 24 at 17:08
Relure
1,948833
edited Jul 24 at 18:03
asked Jul 24 at 17:08
Relure
1,948833
asked Jul 24 at 17:08
Relure
1,948833
asked Jul 24 at 17:08
Relure
1,948833
1,948833
1
A hint: If $f_n$ converge in the $lVertcdotrVert_k$ metric to some $f$, then the derivatives $f_n$, $f'_n$, ..., $f^(k)_n$ converge in the $C_b(I)$-norm to $f$, $f'$, ..., $f^(k)$ (can you prove that)?
– user539887
Jul 24 at 18:45
@user539887 Hi, I've been trying to solve it using this approach, but I can't prove that the derivatives converge. I've seen that $f_n$ converges in the $||cdot ||_k$ metric to a $f$ already. I'd really appreciate you to enlighten me uo in an answer, because I'm a bit frustrated with this one.
– Relure
Jul 26 at 17:15
Just start from the derivative of the highest order: $f^(k)_n$ uniformly converge to some $g$, and $f^(k-1)_n$ uniformly converge to some $h$. It follows then that $h'equiv g$. And you repeat this approach for derivatives of orders $k-1$, $k-2$, etc.
– user539887
Jul 26 at 18:01
add a comment |Â
1
A hint: If $f_n$ converge in the $lVertcdotrVert_k$ metric to some $f$, then the derivatives $f_n$, $f'_n$, ..., $f^(k)_n$ converge in the $C_b(I)$-norm to $f$, $f'$, ..., $f^(k)$ (can you prove that)?
– user539887
Jul 24 at 18:45
@user539887 Hi, I've been trying to solve it using this approach, but I can't prove that the derivatives converge. I've seen that $f_n$ converges in the $||cdot ||_k$ metric to a $f$ already. I'd really appreciate you to enlighten me uo in an answer, because I'm a bit frustrated with this one.
– Relure
Jul 26 at 17:15
Just start from the derivative of the highest order: $f^(k)_n$ uniformly converge to some $g$, and $f^(k-1)_n$ uniformly converge to some $h$. It follows then that $h'equiv g$. And you repeat this approach for derivatives of orders $k-1$, $k-2$, etc.
– user539887
Jul 26 at 18:01
1
1
A hint: If $f_n$ converge in the $lVertcdotrVert_k$ metric to some $f$, then the derivatives $f_n$, $f'_n$, ..., $f^(k)_n$ converge in the $C_b(I)$-norm to $f$, $f'$, ..., $f^(k)$ (can you prove that)?
– user539887
Jul 24 at 18:45
A hint: If $f_n$ converge in the $lVertcdotrVert_k$ metric to some $f$, then the derivatives $f_n$, $f'_n$, ..., $f^(k)_n$ converge in the $C_b(I)$-norm to $f$, $f'$, ..., $f^(k)$ (can you prove that)?
– user539887
Jul 24 at 18:45
@user539887 Hi, I've been trying to solve it using this approach, but I can't prove that the derivatives converge. I've seen that $f_n$ converges in the $||cdot ||_k$ metric to a $f$ already. I'd really appreciate you to enlighten me uo in an answer, because I'm a bit frustrated with this one.
– Relure
Jul 26 at 17:15
@user539887 Hi, I've been trying to solve it using this approach, but I can't prove that the derivatives converge. I've seen that $f_n$ converges in the $||cdot ||_k$ metric to a $f$ already. I'd really appreciate you to enlighten me uo in an answer, because I'm a bit frustrated with this one.
– Relure
Jul 26 at 17:15
Just start from the derivative of the highest order: $f^(k)_n$ uniformly converge to some $g$, and $f^(k-1)_n$ uniformly converge to some $h$. It follows then that $h'equiv g$. And you repeat this approach for derivatives of orders $k-1$, $k-2$, etc.
– user539887
Jul 26 at 18:01
Just start from the derivative of the highest order: $f^(k)_n$ uniformly converge to some $g$, and $f^(k-1)_n$ uniformly converge to some $h$. It follows then that $h'equiv g$. And you repeat this approach for derivatives of orders $k-1$, $k-2$, etc.
– user539887
Jul 26 at 18:01
add a comment |Â
1 Answer
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It's not true that a subspace of a complete space is complete. For example, we have that $mathbbQ subset mathbbR$, and we know $mathbbR$ is complete, but $mathbbQ$ is not.
Rather, we know that a closed subspace of a complete space is complete. However, this only holds if both the complete space and the subspace have the same norm. Using metric spaces instead of normed vector spaces (because I couldn't come up with a good example otherwise), in the discrete metric, $mathbbR$ is complete and $mathbbQ$ is closed in $mathbbR$, but $mathbbQ$ is not complete in the usual metric.
However, you have the advantage that $|f| leq |f|_k$ for any $f in H^k(mathbbR)$. Using this fact, your proof should start like this:
Let $(f_n)_n$ be a Cauchy sequence in $H^k(mathbbR)$. By the above inequality, $(f_n)_n$ is also a Cauchy sequence in $C_b(mathbbR)$, so by completeness, it converges to a function $f in C_b(mathbbR)$.
Now you must show that $f$ is in fact infinitely differentiable, and all its derivatives are bounded. This should follow from the fact that this is true of all the $f_n$ and from the uniform convergence. I'll leave this for you to figure out!
P.S. Are you sure that $|cdot|_k$ is in fact a norm? Are the norms of the derivatives guaranteed to be uniformly bounded? That is, can't the supremum be infinite, meaning the map isn't a norm?
answered Jul 24 at 17:58
Sambo
1,2461427
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
It's not true that a subspace of a complete space is complete. For example, we have that $mathbbQ subset mathbbR$, and we know $mathbbR$ is complete, but $mathbbQ$ is not.
Rather, we know that a closed subspace of a complete space is complete. However, this only holds if both the complete space and the subspace have the same norm. Using metric spaces instead of normed vector spaces (because I couldn't come up with a good example otherwise), in the discrete metric, $mathbbR$ is complete and $mathbbQ$ is closed in $mathbbR$, but $mathbbQ$ is not complete in the usual metric.
However, you have the advantage that $|f| leq |f|_k$ for any $f in H^k(mathbbR)$. Using this fact, your proof should start like this:
Let $(f_n)_n$ be a Cauchy sequence in $H^k(mathbbR)$. By the above inequality, $(f_n)_n$ is also a Cauchy sequence in $C_b(mathbbR)$, so by completeness, it converges to a function $f in C_b(mathbbR)$.
Now you must show that $f$ is in fact infinitely differentiable, and all its derivatives are bounded. This should follow from the fact that this is true of all the $f_n$ and from the uniform convergence. I'll leave this for you to figure out!
P.S. Are you sure that $|cdot|_k$ is in fact a norm? Are the norms of the derivatives guaranteed to be uniformly bounded? That is, can't the supremum be infinite, meaning the map isn't a norm?
answered Jul 24 at 17:58
Sambo
1,2461427
add a comment |Â
up vote
1
down vote
It's not true that a subspace of a complete space is complete. For example, we have that $mathbbQ subset mathbbR$, and we know $mathbbR$ is complete, but $mathbbQ$ is not.
Rather, we know that a closed subspace of a complete space is complete. However, this only holds if both the complete space and the subspace have the same norm. Using metric spaces instead of normed vector spaces (because I couldn't come up with a good example otherwise), in the discrete metric, $mathbbR$ is complete and $mathbbQ$ is closed in $mathbbR$, but $mathbbQ$ is not complete in the usual metric.
However, you have the advantage that $|f| leq |f|_k$ for any $f in H^k(mathbbR)$. Using this fact, your proof should start like this:
Let $(f_n)_n$ be a Cauchy sequence in $H^k(mathbbR)$. By the above inequality, $(f_n)_n$ is also a Cauchy sequence in $C_b(mathbbR)$, so by completeness, it converges to a function $f in C_b(mathbbR)$.
Now you must show that $f$ is in fact infinitely differentiable, and all its derivatives are bounded. This should follow from the fact that this is true of all the $f_n$ and from the uniform convergence. I'll leave this for you to figure out!
P.S. Are you sure that $|cdot|_k$ is in fact a norm? Are the norms of the derivatives guaranteed to be uniformly bounded? That is, can't the supremum be infinite, meaning the map isn't a norm?
answered Jul 24 at 17:58
Sambo
1,2461427
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It's not true that a subspace of a complete space is complete. For example, we have that $mathbbQ subset mathbbR$, and we know $mathbbR$ is complete, but $mathbbQ$ is not.
Rather, we know that a closed subspace of a complete space is complete. However, this only holds if both the complete space and the subspace have the same norm. Using metric spaces instead of normed vector spaces (because I couldn't come up with a good example otherwise), in the discrete metric, $mathbbR$ is complete and $mathbbQ$ is closed in $mathbbR$, but $mathbbQ$ is not complete in the usual metric.
However, you have the advantage that $|f| leq |f|_k$ for any $f in H^k(mathbbR)$. Using this fact, your proof should start like this:
Let $(f_n)_n$ be a Cauchy sequence in $H^k(mathbbR)$. By the above inequality, $(f_n)_n$ is also a Cauchy sequence in $C_b(mathbbR)$, so by completeness, it converges to a function $f in C_b(mathbbR)$.
Now you must show that $f$ is in fact infinitely differentiable, and all its derivatives are bounded. This should follow from the fact that this is true of all the $f_n$ and from the uniform convergence. I'll leave this for you to figure out!
P.S. Are you sure that $|cdot|_k$ is in fact a norm? Are the norms of the derivatives guaranteed to be uniformly bounded? That is, can't the supremum be infinite, meaning the map isn't a norm?
answered Jul 24 at 17:58
Sambo
1,2461427
It's not true that a subspace of a complete space is complete. For example, we have that $mathbbQ subset mathbbR$, and we know $mathbbR$ is complete, but $mathbbQ$ is not.
Rather, we know that a closed subspace of a complete space is complete. However, this only holds if both the complete space and the subspace have the same norm. Using metric spaces instead of normed vector spaces (because I couldn't come up with a good example otherwise), in the discrete metric, $mathbbR$ is complete and $mathbbQ$ is closed in $mathbbR$, but $mathbbQ$ is not complete in the usual metric.
However, you have the advantage that $|f| leq |f|_k$ for any $f in H^k(mathbbR)$. Using this fact, your proof should start like this:
Let $(f_n)_n$ be a Cauchy sequence in $H^k(mathbbR)$. By the above inequality, $(f_n)_n$ is also a Cauchy sequence in $C_b(mathbbR)$, so by completeness, it converges to a function $f in C_b(mathbbR)$.
Now you must show that $f$ is in fact infinitely differentiable, and all its derivatives are bounded. This should follow from the fact that this is true of all the $f_n$ and from the uniform convergence. I'll leave this for you to figure out!
P.S. Are you sure that $|cdot|_k$ is in fact a norm? Are the norms of the derivatives guaranteed to be uniformly bounded? That is, can't the supremum be infinite, meaning the map isn't a norm?
answered Jul 24 at 17:58
Sambo
1,2461427
answered Jul 24 at 17:58
Sambo
1,2461427
answered Jul 24 at 17:58
Sambo
1,2461427
answered Jul 24 at 17:58
Sambo
1,2461427
1,2461427
add a comment |Â
add a comment |Â
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1
A hint: If $f_n$ converge in the $lVertcdotrVert_k$ metric to some $f$, then the derivatives $f_n$, $f'_n$, ..., $f^(k)_n$ converge in the $C_b(I)$-norm to $f$, $f'$, ..., $f^(k)$ (can you prove that)?
– user539887
Jul 24 at 18:45
@user539887 Hi, I've been trying to solve it using this approach, but I can't prove that the derivatives converge. I've seen that $f_n$ converges in the $||cdot ||_k$ metric to a $f$ already. I'd really appreciate you to enlighten me uo in an answer, because I'm a bit frustrated with this one.
– Relure
Jul 26 at 17:15
Just start from the derivative of the highest order: $f^(k)_n$ uniformly converge to some $g$, and $f^(k-1)_n$ uniformly converge to some $h$. It follows then that $h'equiv g$. And you repeat this approach for derivatives of orders $k-1$, $k-2$, etc.
– user539887
Jul 26 at 18:01