Kernel of linear mapping from infinitely differentiable function to their derivatives
From S.L Linear Algebra:
Let $V$ be the vector space of functions which have derivatives of all
orders, and let $D:V rightarrow V$ be the derivative. What is the kernel of $D$?
This seems to be a very simple question, with very simple answer that I've constructed, though I'm not aware of how sufficient would it be.
I'm assuming, that "vector space of functions which have derivatives of all orders" is basically a function space of infinitely differentiable functions over some arbitrary field $K$ (since function space over a field $K$ is a vector space).
From basic calculus, we are aware of Fermat's interior extremum theorem, and that every point that has derivative of $0$ (stationary point) is local extremum (either maxima, minima or a saddle point).
Hence, the kernel of infinitely differentiable functions under $D$ is a space spanned by all local extremum's of infinitely differentiable functions.
Is this highly simple answer sufficient? or should I add something more.
Thank you!
calculus linear-algebra derivatives linear-transformations
edited Nov 28 '18 at 10:51
José Carlos Santos
151k22123224
asked Nov 28 '18 at 10:38
ShellRox
25328
add a comment |
From S.L Linear Algebra:
Let $V$ be the vector space of functions which have derivatives of all
orders, and let $D:V rightarrow V$ be the derivative. What is the kernel of $D$?
This seems to be a very simple question, with very simple answer that I've constructed, though I'm not aware of how sufficient would it be.
I'm assuming, that "vector space of functions which have derivatives of all orders" is basically a function space of infinitely differentiable functions over some arbitrary field $K$ (since function space over a field $K$ is a vector space).
From basic calculus, we are aware of Fermat's interior extremum theorem, and that every point that has derivative of $0$ (stationary point) is local extremum (either maxima, minima or a saddle point).
Hence, the kernel of infinitely differentiable functions under $D$ is a space spanned by all local extremum's of infinitely differentiable functions.
Is this highly simple answer sufficient? or should I add something more.
Thank you!
calculus linear-algebra derivatives linear-transformations
edited Nov 28 '18 at 10:51
José Carlos Santos
151k22123224
asked Nov 28 '18 at 10:38
ShellRox
25328
@DavidC.Ullrich Thanks, it was supposed to say "a function in the kernel has the zero function as derivative", of course. I was trying to make a different point, but that incorrect formulation certainly didn't help... :-)
– StackTD
Nov 28 '18 at 14:49
add a comment |
From S.L Linear Algebra:
Let $V$ be the vector space of functions which have derivatives of all
orders, and let $D:V rightarrow V$ be the derivative. What is the kernel of $D$?
This seems to be a very simple question, with very simple answer that I've constructed, though I'm not aware of how sufficient would it be.
I'm assuming, that "vector space of functions which have derivatives of all orders" is basically a function space of infinitely differentiable functions over some arbitrary field $K$ (since function space over a field $K$ is a vector space).
From basic calculus, we are aware of Fermat's interior extremum theorem, and that every point that has derivative of $0$ (stationary point) is local extremum (either maxima, minima or a saddle point).
Hence, the kernel of infinitely differentiable functions under $D$ is a space spanned by all local extremum's of infinitely differentiable functions.
Is this highly simple answer sufficient? or should I add something more.
Thank you!
calculus linear-algebra derivatives linear-transformations
edited Nov 28 '18 at 10:51
José Carlos Santos
151k22123224
asked Nov 28 '18 at 10:38
ShellRox
25328
From S.L Linear Algebra:
Let $V$ be the vector space of functions which have derivatives of all
orders, and let $D:V rightarrow V$ be the derivative. What is the kernel of $D$?
This seems to be a very simple question, with very simple answer that I've constructed, though I'm not aware of how sufficient would it be.
I'm assuming, that "vector space of functions which have derivatives of all orders" is basically a function space of infinitely differentiable functions over some arbitrary field $K$ (since function space over a field $K$ is a vector space).
From basic calculus, we are aware of Fermat's interior extremum theorem, and that every point that has derivative of $0$ (stationary point) is local extremum (either maxima, minima or a saddle point).
Hence, the kernel of infinitely differentiable functions under $D$ is a space spanned by all local extremum's of infinitely differentiable functions.
Is this highly simple answer sufficient? or should I add something more.
Thank you!
calculus linear-algebra derivatives linear-transformations
calculus linear-algebra derivatives linear-transformations
edited Nov 28 '18 at 10:51
José Carlos Santos
151k22123224
asked Nov 28 '18 at 10:38
ShellRox
25328
edited Nov 28 '18 at 10:51
José Carlos Santos
151k22123224
asked Nov 28 '18 at 10:38
ShellRox
25328
edited Nov 28 '18 at 10:51
José Carlos Santos
151k22123224
edited Nov 28 '18 at 10:51
José Carlos Santos
151k22123224
edited Nov 28 '18 at 10:51
José Carlos Santos
151k22123224
151k22123224
asked Nov 28 '18 at 10:38
ShellRox
25328
asked Nov 28 '18 at 10:38
ShellRox
25328
asked Nov 28 '18 at 10:38
ShellRox
25328
25328
@DavidC.Ullrich Thanks, it was supposed to say "a function in the kernel has the zero function as derivative", of course. I was trying to make a different point, but that incorrect formulation certainly didn't help... :-)
– StackTD
Nov 28 '18 at 14:49
add a comment |
@DavidC.Ullrich Thanks, it was supposed to say "a function in the kernel has the zero function as derivative", of course. I was trying to make a different point, but that incorrect formulation certainly didn't help... :-)
– StackTD
Nov 28 '18 at 14:49
@DavidC.Ullrich Thanks, it was supposed to say "a function in the kernel has the zero function as derivative", of course. I was trying to make a different point, but that incorrect formulation certainly didn't help... :-)
– StackTD
Nov 28 '18 at 14:49
@DavidC.Ullrich Thanks, it was supposed to say "a function in the kernel has the zero function as derivative", of course. I was trying to make a different point, but that incorrect formulation certainly didn't help... :-)
– StackTD
Nov 28 '18 at 14:49
add a comment |
1 Answer
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I suppose that it is implicit here that $V$ is the space of all fonctions $fcolonmathbb{R}longrightarrowmathbb R$ which have derivatives of all orders. Then$$ker D=left{fin V,middle|,f'=0right}.$$But the functions $f$ such that $f'$ is the null function are the constant functions and only those functions.
answered Nov 28 '18 at 10:42
José Carlos Santos
151k22123224
Oh, I should've utilized the definition of kernel more appropriately... Hence every point of the function's derivative is stationary, not just some specific one. Thank you!
– ShellRox
Nov 28 '18 at 11:07
add a comment |
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1 Answer
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I suppose that it is implicit here that $V$ is the space of all fonctions $fcolonmathbb{R}longrightarrowmathbb R$ which have derivatives of all orders. Then$$ker D=left{fin V,middle|,f'=0right}.$$But the functions $f$ such that $f'$ is the null function are the constant functions and only those functions.
answered Nov 28 '18 at 10:42
José Carlos Santos
151k22123224
Oh, I should've utilized the definition of kernel more appropriately... Hence every point of the function's derivative is stationary, not just some specific one. Thank you!
– ShellRox
Nov 28 '18 at 11:07
add a comment |
I suppose that it is implicit here that $V$ is the space of all fonctions $fcolonmathbb{R}longrightarrowmathbb R$ which have derivatives of all orders. Then$$ker D=left{fin V,middle|,f'=0right}.$$But the functions $f$ such that $f'$ is the null function are the constant functions and only those functions.
answered Nov 28 '18 at 10:42
José Carlos Santos
151k22123224
Oh, I should've utilized the definition of kernel more appropriately... Hence every point of the function's derivative is stationary, not just some specific one. Thank you!
– ShellRox
Nov 28 '18 at 11:07
add a comment |
I suppose that it is implicit here that $V$ is the space of all fonctions $fcolonmathbb{R}longrightarrowmathbb R$ which have derivatives of all orders. Then$$ker D=left{fin V,middle|,f'=0right}.$$But the functions $f$ such that $f'$ is the null function are the constant functions and only those functions.
answered Nov 28 '18 at 10:42
José Carlos Santos
151k22123224
I suppose that it is implicit here that $V$ is the space of all fonctions $fcolonmathbb{R}longrightarrowmathbb R$ which have derivatives of all orders. Then$$ker D=left{fin V,middle|,f'=0right}.$$But the functions $f$ such that $f'$ is the null function are the constant functions and only those functions.
answered Nov 28 '18 at 10:42
José Carlos Santos
151k22123224
answered Nov 28 '18 at 10:42
José Carlos Santos
151k22123224
answered Nov 28 '18 at 10:42
José Carlos Santos
151k22123224
answered Nov 28 '18 at 10:42
José Carlos Santos
151k22123224
151k22123224
Oh, I should've utilized the definition of kernel more appropriately... Hence every point of the function's derivative is stationary, not just some specific one. Thank you!
– ShellRox
Nov 28 '18 at 11:07
add a comment |
Oh, I should've utilized the definition of kernel more appropriately... Hence every point of the function's derivative is stationary, not just some specific one. Thank you!
– ShellRox
Nov 28 '18 at 11:07
Oh, I should've utilized the definition of kernel more appropriately... Hence every point of the function's derivative is stationary, not just some specific one. Thank you!
– ShellRox
Nov 28 '18 at 11:07
Oh, I should've utilized the definition of kernel more appropriately... Hence every point of the function's derivative is stationary, not just some specific one. Thank you!
– ShellRox
Nov 28 '18 at 11:07
add a comment |
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@DavidC.Ullrich Thanks, it was supposed to say "a function in the kernel has the zero function as derivative", of course. I was trying to make a different point, but that incorrect formulation certainly didn't help... :-)
– StackTD
Nov 28 '18 at 14:49