Mixing
Are linear transformations between infinite dimensional vector spaces always differentiable?
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In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).
linear-algebra differential-geometry continuity linear-transformations
asked Jan 29 at 15:28
McNuggets666McNuggets666
725411
$endgroup$
add a comment |
$begingroup$
In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).
linear-algebra differential-geometry continuity linear-transformations
asked Jan 29 at 15:28
McNuggets666McNuggets666
725411
$endgroup$
add a comment |
$begingroup$
In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).
linear-algebra differential-geometry continuity linear-transformations
asked Jan 29 at 15:28
McNuggets666McNuggets666
725411
$endgroup$
In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).
linear-algebra differential-geometry continuity linear-transformations
linear-algebra differential-geometry continuity linear-transformations
asked Jan 29 at 15:28
McNuggets666McNuggets666
725411
asked Jan 29 at 15:28
McNuggets666McNuggets666
725411
asked Jan 29 at 15:28
McNuggets666McNuggets666
725411
asked Jan 29 at 15:28
McNuggets666McNuggets666
725411
asked Jan 29 at 15:28
McNuggets666McNuggets666
725411
725411
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
The differentiation operator is linear but not bounded in $mathcal{L}^2(mathbb{R})$ (which is infinite dimensional over $mathbb{R}$). Hence, it's not continuous as all continuous linear operators in Banach spaces are bounded. To see that it's not bounded, consider what differentiation does to the sequence of functions $$f_n(x) = sqrt{n}, e^{-n^2 x^2}$$
Also, it matters how differentiation is defined. For another example which is kind of different, consider conjugation in complex numbers where we view $mathbb{C}$ as a $2$-dimensional vector space over $mathbb{R}$. Conjugation is then linear and continuous because $f(a+bi)=a-bi$ but as a map from $mathbb{C}$ to $mathbb{C}$ it's not differentiable because it fails to satisfy the Cauchy-Riemann equations.
answered Jan 29 at 15:31
stressed outstressed out
6,5831939
$endgroup$
$begingroup$
Maybe it's important to add that bounded in infinite dimensional analysis means bounded in a ball and not bounded in the whole domain.
$endgroup$
– Nick A.
Jan 29 at 16:18
$begingroup$
@NickA. I'm afraid you're wrong. Bounded here means bounded in the operator norm and by definition, $T: X to X$ is a bounded linear operator iff $|T(u)| leq M | u |$ for all $u in X$ and some positive constant $M$. See Royden's Real Analysis, chapter 13, section 2. It makes sense because intuitively, It's not difficult to see that the size of a ball in a linear space doesn't matter because things can be scaled. So, there's no fixed radius really.
$endgroup$
– stressed out
Jan 29 at 16:27
1
$begingroup$
I don't see how what we are saying contradicts each other :). I just thought that for someone new to these areas maybe it is useful to comment on the difference on terminology.
$endgroup$
– Nick A.
Jan 29 at 16:34
$begingroup$
@NickA. I didn't say we contradicted each other. What I intended to say was that your interpretation of bounded was wrong probably because the definition works for all $u in X$, not just some particular vectors in a specific ball. So, indeed, it's about the whole domain. Or am I mistaken or missing something?
$endgroup$
– stressed out
Jan 29 at 16:37
$begingroup$
If I am not mistaken, if a linear function is bounded on some closed ball (of finite radius) then it is bounded in **all balls ** (through translations and rescaling) and therefore to the ball of radius $1$, which is the usual definition (you probably have to use the triangle inequality somewhere). If you are not convinced I will provide a proof as a separate answer :).
$endgroup$
– Nick A.
Jan 29 at 16:43
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
The differentiation operator is linear but not bounded in $mathcal{L}^2(mathbb{R})$ (which is infinite dimensional over $mathbb{R}$). Hence, it's not continuous as all continuous linear operators in Banach spaces are bounded. To see that it's not bounded, consider what differentiation does to the sequence of functions $$f_n(x) = sqrt{n}, e^{-n^2 x^2}$$
Also, it matters how differentiation is defined. For another example which is kind of different, consider conjugation in complex numbers where we view $mathbb{C}$ as a $2$-dimensional vector space over $mathbb{R}$. Conjugation is then linear and continuous because $f(a+bi)=a-bi$ but as a map from $mathbb{C}$ to $mathbb{C}$ it's not differentiable because it fails to satisfy the Cauchy-Riemann equations.
answered Jan 29 at 15:31
stressed outstressed out
6,5831939
$endgroup$
$begingroup$
Maybe it's important to add that bounded in infinite dimensional analysis means bounded in a ball and not bounded in the whole domain.
$endgroup$
– Nick A.
Jan 29 at 16:18
$begingroup$
@NickA. I'm afraid you're wrong. Bounded here means bounded in the operator norm and by definition, $T: X to X$ is a bounded linear operator iff $|T(u)| leq M | u |$ for all $u in X$ and some positive constant $M$. See Royden's Real Analysis, chapter 13, section 2. It makes sense because intuitively, It's not difficult to see that the size of a ball in a linear space doesn't matter because things can be scaled. So, there's no fixed radius really.
$endgroup$
– stressed out
Jan 29 at 16:27
1
$begingroup$
I don't see how what we are saying contradicts each other :). I just thought that for someone new to these areas maybe it is useful to comment on the difference on terminology.
$endgroup$
– Nick A.
Jan 29 at 16:34
$begingroup$
@NickA. I didn't say we contradicted each other. What I intended to say was that your interpretation of bounded was wrong probably because the definition works for all $u in X$, not just some particular vectors in a specific ball. So, indeed, it's about the whole domain. Or am I mistaken or missing something?
$endgroup$
– stressed out
Jan 29 at 16:37
$begingroup$
If I am not mistaken, if a linear function is bounded on some closed ball (of finite radius) then it is bounded in **all balls ** (through translations and rescaling) and therefore to the ball of radius $1$, which is the usual definition (you probably have to use the triangle inequality somewhere). If you are not convinced I will provide a proof as a separate answer :).
$endgroup$
– Nick A.
Jan 29 at 16:43
|
show 1 more comment
$begingroup$
The differentiation operator is linear but not bounded in $mathcal{L}^2(mathbb{R})$ (which is infinite dimensional over $mathbb{R}$). Hence, it's not continuous as all continuous linear operators in Banach spaces are bounded. To see that it's not bounded, consider what differentiation does to the sequence of functions $$f_n(x) = sqrt{n}, e^{-n^2 x^2}$$
Also, it matters how differentiation is defined. For another example which is kind of different, consider conjugation in complex numbers where we view $mathbb{C}$ as a $2$-dimensional vector space over $mathbb{R}$. Conjugation is then linear and continuous because $f(a+bi)=a-bi$ but as a map from $mathbb{C}$ to $mathbb{C}$ it's not differentiable because it fails to satisfy the Cauchy-Riemann equations.
answered Jan 29 at 15:31
stressed outstressed out
6,5831939
$endgroup$
$begingroup$
Maybe it's important to add that bounded in infinite dimensional analysis means bounded in a ball and not bounded in the whole domain.
$endgroup$
– Nick A.
Jan 29 at 16:18
$begingroup$
@NickA. I'm afraid you're wrong. Bounded here means bounded in the operator norm and by definition, $T: X to X$ is a bounded linear operator iff $|T(u)| leq M | u |$ for all $u in X$ and some positive constant $M$. See Royden's Real Analysis, chapter 13, section 2. It makes sense because intuitively, It's not difficult to see that the size of a ball in a linear space doesn't matter because things can be scaled. So, there's no fixed radius really.
$endgroup$
– stressed out
Jan 29 at 16:27
1
$begingroup$
I don't see how what we are saying contradicts each other :). I just thought that for someone new to these areas maybe it is useful to comment on the difference on terminology.
$endgroup$
– Nick A.
Jan 29 at 16:34
$begingroup$
@NickA. I didn't say we contradicted each other. What I intended to say was that your interpretation of bounded was wrong probably because the definition works for all $u in X$, not just some particular vectors in a specific ball. So, indeed, it's about the whole domain. Or am I mistaken or missing something?
$endgroup$
– stressed out
Jan 29 at 16:37
$begingroup$
If I am not mistaken, if a linear function is bounded on some closed ball (of finite radius) then it is bounded in **all balls ** (through translations and rescaling) and therefore to the ball of radius $1$, which is the usual definition (you probably have to use the triangle inequality somewhere). If you are not convinced I will provide a proof as a separate answer :).
$endgroup$
– Nick A.
Jan 29 at 16:43
|
show 1 more comment
$begingroup$
The differentiation operator is linear but not bounded in $mathcal{L}^2(mathbb{R})$ (which is infinite dimensional over $mathbb{R}$). Hence, it's not continuous as all continuous linear operators in Banach spaces are bounded. To see that it's not bounded, consider what differentiation does to the sequence of functions $$f_n(x) = sqrt{n}, e^{-n^2 x^2}$$
Also, it matters how differentiation is defined. For another example which is kind of different, consider conjugation in complex numbers where we view $mathbb{C}$ as a $2$-dimensional vector space over $mathbb{R}$. Conjugation is then linear and continuous because $f(a+bi)=a-bi$ but as a map from $mathbb{C}$ to $mathbb{C}$ it's not differentiable because it fails to satisfy the Cauchy-Riemann equations.
answered Jan 29 at 15:31
stressed outstressed out
6,5831939
$endgroup$
The differentiation operator is linear but not bounded in $mathcal{L}^2(mathbb{R})$ (which is infinite dimensional over $mathbb{R}$). Hence, it's not continuous as all continuous linear operators in Banach spaces are bounded. To see that it's not bounded, consider what differentiation does to the sequence of functions $$f_n(x) = sqrt{n}, e^{-n^2 x^2}$$
Also, it matters how differentiation is defined. For another example which is kind of different, consider conjugation in complex numbers where we view $mathbb{C}$ as a $2$-dimensional vector space over $mathbb{R}$. Conjugation is then linear and continuous because $f(a+bi)=a-bi$ but as a map from $mathbb{C}$ to $mathbb{C}$ it's not differentiable because it fails to satisfy the Cauchy-Riemann equations.
answered Jan 29 at 15:31
stressed outstressed out
6,5831939
edited Jan 29 at 16:11
answered Jan 29 at 15:31
stressed outstressed out
6,5831939
answered Jan 29 at 15:31
stressed outstressed out
6,5831939
answered Jan 29 at 15:31
stressed outstressed out
6,5831939
6,5831939
$begingroup$
Maybe it's important to add that bounded in infinite dimensional analysis means bounded in a ball and not bounded in the whole domain.
$endgroup$
– Nick A.
Jan 29 at 16:18
$begingroup$
@NickA. I'm afraid you're wrong. Bounded here means bounded in the operator norm and by definition, $T: X to X$ is a bounded linear operator iff $|T(u)| leq M | u |$ for all $u in X$ and some positive constant $M$. See Royden's Real Analysis, chapter 13, section 2. It makes sense because intuitively, It's not difficult to see that the size of a ball in a linear space doesn't matter because things can be scaled. So, there's no fixed radius really.
$endgroup$
– stressed out
Jan 29 at 16:27
1
$begingroup$
I don't see how what we are saying contradicts each other :). I just thought that for someone new to these areas maybe it is useful to comment on the difference on terminology.
$endgroup$
– Nick A.
Jan 29 at 16:34
$begingroup$
@NickA. I didn't say we contradicted each other. What I intended to say was that your interpretation of bounded was wrong probably because the definition works for all $u in X$, not just some particular vectors in a specific ball. So, indeed, it's about the whole domain. Or am I mistaken or missing something?
$endgroup$
– stressed out
Jan 29 at 16:37
$begingroup$
If I am not mistaken, if a linear function is bounded on some closed ball (of finite radius) then it is bounded in **all balls ** (through translations and rescaling) and therefore to the ball of radius $1$, which is the usual definition (you probably have to use the triangle inequality somewhere). If you are not convinced I will provide a proof as a separate answer :).
$endgroup$
– Nick A.
Jan 29 at 16:43
|
show 1 more comment
$begingroup$
Maybe it's important to add that bounded in infinite dimensional analysis means bounded in a ball and not bounded in the whole domain.
$endgroup$
– Nick A.
Jan 29 at 16:18
$begingroup$
@NickA. I'm afraid you're wrong. Bounded here means bounded in the operator norm and by definition, $T: X to X$ is a bounded linear operator iff $|T(u)| leq M | u |$ for all $u in X$ and some positive constant $M$. See Royden's Real Analysis, chapter 13, section 2. It makes sense because intuitively, It's not difficult to see that the size of a ball in a linear space doesn't matter because things can be scaled. So, there's no fixed radius really.
$endgroup$
– stressed out
Jan 29 at 16:27
1
$begingroup$
I don't see how what we are saying contradicts each other :). I just thought that for someone new to these areas maybe it is useful to comment on the difference on terminology.
$endgroup$
– Nick A.
Jan 29 at 16:34
$begingroup$
@NickA. I didn't say we contradicted each other. What I intended to say was that your interpretation of bounded was wrong probably because the definition works for all $u in X$, not just some particular vectors in a specific ball. So, indeed, it's about the whole domain. Or am I mistaken or missing something?
$endgroup$
– stressed out
Jan 29 at 16:37
$begingroup$
If I am not mistaken, if a linear function is bounded on some closed ball (of finite radius) then it is bounded in **all balls ** (through translations and rescaling) and therefore to the ball of radius $1$, which is the usual definition (you probably have to use the triangle inequality somewhere). If you are not convinced I will provide a proof as a separate answer :).
$endgroup$
– Nick A.
Jan 29 at 16:43
$begingroup$
Maybe it's important to add that bounded in infinite dimensional analysis means bounded in a ball and not bounded in the whole domain.
$endgroup$
– Nick A.
Jan 29 at 16:18
$begingroup$
Maybe it's important to add that bounded in infinite dimensional analysis means bounded in a ball and not bounded in the whole domain.
$endgroup$
– Nick A.
Jan 29 at 16:18
$begingroup$
@NickA. I'm afraid you're wrong. Bounded here means bounded in the operator norm and by definition, $T: X to X$ is a bounded linear operator iff $|T(u)| leq M | u |$ for all $u in X$ and some positive constant $M$. See Royden's Real Analysis, chapter 13, section 2. It makes sense because intuitively, It's not difficult to see that the size of a ball in a linear space doesn't matter because things can be scaled. So, there's no fixed radius really.
$endgroup$
– stressed out
Jan 29 at 16:27
$begingroup$
@NickA. I'm afraid you're wrong. Bounded here means bounded in the operator norm and by definition, $T: X to X$ is a bounded linear operator iff $|T(u)| leq M | u |$ for all $u in X$ and some positive constant $M$. See Royden's Real Analysis, chapter 13, section 2. It makes sense because intuitively, It's not difficult to see that the size of a ball in a linear space doesn't matter because things can be scaled. So, there's no fixed radius really.
$endgroup$
– stressed out
Jan 29 at 16:27
1
1
$begingroup$
I don't see how what we are saying contradicts each other :). I just thought that for someone new to these areas maybe it is useful to comment on the difference on terminology.
$endgroup$
– Nick A.
Jan 29 at 16:34
$begingroup$
I don't see how what we are saying contradicts each other :). I just thought that for someone new to these areas maybe it is useful to comment on the difference on terminology.
$endgroup$
– Nick A.
Jan 29 at 16:34
$begingroup$
@NickA. I didn't say we contradicted each other. What I intended to say was that your interpretation of bounded was wrong probably because the definition works for all $u in X$, not just some particular vectors in a specific ball. So, indeed, it's about the whole domain. Or am I mistaken or missing something?
$endgroup$
– stressed out
Jan 29 at 16:37
$begingroup$
@NickA. I didn't say we contradicted each other. What I intended to say was that your interpretation of bounded was wrong probably because the definition works for all $u in X$, not just some particular vectors in a specific ball. So, indeed, it's about the whole domain. Or am I mistaken or missing something?
$endgroup$
– stressed out
Jan 29 at 16:37
$begingroup$
If I am not mistaken, if a linear function is bounded on some closed ball (of finite radius) then it is bounded in **all balls ** (through translations and rescaling) and therefore to the ball of radius $1$, which is the usual definition (you probably have to use the triangle inequality somewhere). If you are not convinced I will provide a proof as a separate answer :).
$endgroup$
– Nick A.
Jan 29 at 16:43
$begingroup$
If I am not mistaken, if a linear function is bounded on some closed ball (of finite radius) then it is bounded in **all balls ** (through translations and rescaling) and therefore to the ball of radius $1$, which is the usual definition (you probably have to use the triangle inequality somewhere). If you are not convinced I will provide a proof as a separate answer :).
$endgroup$
– Nick A.
Jan 29 at 16:43
|
show 1 more comment
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