Calculate the derivative of $f(x) = | x|^2.$
Suppose we have a normed space $E$ with the induced inner product and a map $f:Eto R$ such that $f(x) = |x|^2.$ I want to compute the derivative of this map.
We have that
$$f(x+h)=|x|^2+|h|^2+2langle x,hrangle=f(x)+o(|h|)+df_{x}(h).$$
Clearly the derivative $df_x(h) = 2langle x,hrangle$ is linear in $h.$ We also want to show that $df_x(h)$ is continuous in $h$. This can be done by simply observing that $$|df_x(h)|leq 2|x|cdot |h|$$ and therefore
$$|df_x|leq 2|x|$$ implying that $df_x$ is continuous in $h.$
I also want to show that the derivative $df_x$ is $C^{1}.$ For this we have to show that the map $xto df_x$ is continuous. Therefore observe that for each $hin E$ we have that,
$$|df_x(h)-df_y(h)|=2|langle x-y,hrangle|leq |x-y|cdot2|h|.$$
I am not sure how to proceed after this step. Any hints will be much appreciated.
real-analysis functional-analysis
add a comment |
Suppose we have a normed space $E$ with the induced inner product and a map $f:Eto R$ such that $f(x) = |x|^2.$ I want to compute the derivative of this map.
We have that
$$f(x+h)=|x|^2+|h|^2+2langle x,hrangle=f(x)+o(|h|)+df_{x}(h).$$
Clearly the derivative $df_x(h) = 2langle x,hrangle$ is linear in $h.$ We also want to show that $df_x(h)$ is continuous in $h$. This can be done by simply observing that $$|df_x(h)|leq 2|x|cdot |h|$$ and therefore
$$|df_x|leq 2|x|$$ implying that $df_x$ is continuous in $h.$
I also want to show that the derivative $df_x$ is $C^{1}.$ For this we have to show that the map $xto df_x$ is continuous. Therefore observe that for each $hin E$ we have that,
$$|df_x(h)-df_y(h)|=2|langle x-y,hrangle|leq |x-y|cdot2|h|.$$
I am not sure how to proceed after this step. Any hints will be much appreciated.
real-analysis functional-analysis
1
What is a inner product induced by a norm? Most norms don't come from inner products.
– José Carlos Santos
Nov 26 at 17:43
1
Oh I think I meant the other way around.
– Hello_World
Nov 26 at 18:01
If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
– Gibbs
Nov 26 at 18:15
add a comment |
Suppose we have a normed space $E$ with the induced inner product and a map $f:Eto R$ such that $f(x) = |x|^2.$ I want to compute the derivative of this map.
We have that
$$f(x+h)=|x|^2+|h|^2+2langle x,hrangle=f(x)+o(|h|)+df_{x}(h).$$
Clearly the derivative $df_x(h) = 2langle x,hrangle$ is linear in $h.$ We also want to show that $df_x(h)$ is continuous in $h$. This can be done by simply observing that $$|df_x(h)|leq 2|x|cdot |h|$$ and therefore
$$|df_x|leq 2|x|$$ implying that $df_x$ is continuous in $h.$
I also want to show that the derivative $df_x$ is $C^{1}.$ For this we have to show that the map $xto df_x$ is continuous. Therefore observe that for each $hin E$ we have that,
$$|df_x(h)-df_y(h)|=2|langle x-y,hrangle|leq |x-y|cdot2|h|.$$
I am not sure how to proceed after this step. Any hints will be much appreciated.
real-analysis functional-analysis
Suppose we have a normed space $E$ with the induced inner product and a map $f:Eto R$ such that $f(x) = |x|^2.$ I want to compute the derivative of this map.
We have that
$$f(x+h)=|x|^2+|h|^2+2langle x,hrangle=f(x)+o(|h|)+df_{x}(h).$$
Clearly the derivative $df_x(h) = 2langle x,hrangle$ is linear in $h.$ We also want to show that $df_x(h)$ is continuous in $h$. This can be done by simply observing that $$|df_x(h)|leq 2|x|cdot |h|$$ and therefore
$$|df_x|leq 2|x|$$ implying that $df_x$ is continuous in $h.$
I also want to show that the derivative $df_x$ is $C^{1}.$ For this we have to show that the map $xto df_x$ is continuous. Therefore observe that for each $hin E$ we have that,
$$|df_x(h)-df_y(h)|=2|langle x-y,hrangle|leq |x-y|cdot2|h|.$$
I am not sure how to proceed after this step. Any hints will be much appreciated.
real-analysis functional-analysis
real-analysis functional-analysis
edited Nov 26 at 18:18
Martin Sleziak
44.6k7115270
44.6k7115270
asked Nov 26 at 17:41
Hello_World
3,84721630
3,84721630
1
What is a inner product induced by a norm? Most norms don't come from inner products.
– José Carlos Santos
Nov 26 at 17:43
1
Oh I think I meant the other way around.
– Hello_World
Nov 26 at 18:01
If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
– Gibbs
Nov 26 at 18:15
add a comment |
1
What is a inner product induced by a norm? Most norms don't come from inner products.
– José Carlos Santos
Nov 26 at 17:43
1
Oh I think I meant the other way around.
– Hello_World
Nov 26 at 18:01
If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
– Gibbs
Nov 26 at 18:15
1
1
What is a inner product induced by a norm? Most norms don't come from inner products.
– José Carlos Santos
Nov 26 at 17:43
What is a inner product induced by a norm? Most norms don't come from inner products.
– José Carlos Santos
Nov 26 at 17:43
1
1
Oh I think I meant the other way around.
– Hello_World
Nov 26 at 18:01
Oh I think I meant the other way around.
– Hello_World
Nov 26 at 18:01
If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
– Gibbs
Nov 26 at 18:15
If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
– Gibbs
Nov 26 at 18:15
add a comment |
2 Answers
2
active
oldest
votes
Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
$$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
and apply the chain rule.
add a comment |
I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.
add a comment |
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2 Answers
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2 Answers
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Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
$$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
and apply the chain rule.
add a comment |
Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
$$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
and apply the chain rule.
add a comment |
Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
$$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
and apply the chain rule.
Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
$$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
and apply the chain rule.
answered Nov 26 at 18:50
Martín-Blas Pérez Pinilla
34.1k42771
34.1k42771
add a comment |
add a comment |
I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.
add a comment |
I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.
add a comment |
I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.
I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.
answered Nov 26 at 18:21
Will M.
2,377314
2,377314
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add a comment |
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1
What is a inner product induced by a norm? Most norms don't come from inner products.
– José Carlos Santos
Nov 26 at 17:43
1
Oh I think I meant the other way around.
– Hello_World
Nov 26 at 18:01
If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
– Gibbs
Nov 26 at 18:15