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What is the kernel of such a linear map

  1. Aug 10, 2007 #1
    1. The problem statement, all variables and given/known data

    This is a problem related to linear map over vector spaces of functions and finding kernels.

    Let V be the vector space of functions which have derivatives of all orders, and let D:V->V be the derivative. Problem1: What is the kernal of D?

    Problem2: Let L=D-I,where I is the identity map of V. What is the kernel of L?

    2. Relevant equations
    problem1: let f be an unknown element of the kernel, then D(f)=0
    problem2: let g be an unknown element of its kernel, then L(g)=0

    Note:The above 0 is a 0 function.

    3. The attempt at a solution

    My Solving problem1: D(f)=0, but the unknown is a function. I don't have a clue to solve such "equations" (with the unknowns of functions). So I can only guess the answer.

    My Solving problem2: L(g)=0 => (D-I)(g)=0 =>D(g)-I(g)=D(g)-g=0 noting D-I must also be a linear map. So I need to solve D(g)=g. Since g is a function, again I can only guess the answer.

    There is a problem with my "guessed" answer, because I can only prove my "guessed" answer (is a set) meet the equation, but I have no idea whether there exists a solution which does not lie in my guessed set.

  2. jcsd
  3. Aug 10, 2007 #2

    matt grime

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    I suspect that you're supposed to have met differentiation before - have you taken any courses in differential equations, say? Because the fact that f' = 0 implies f is constant is one of the first things you'll meet. The second part is writing out the solutions to the differential equation f'=f, which, again is about the first thing you'll meet in a differential equations course.

    The answers can also be found by some variation on the intermediate value theorem.
  4. Aug 10, 2007 #3

    The problem is simple but you still responded to it.Thank you!

    Yes,I have knowledge of differential equations in calculus course. But that procedure is about the solution of unknowns about reals, not about functions. My original "guess" can now be proved to be rigorous as follows.

    f'(x)=0 => f(x)=constant ... (a)
    f'=0 <=> f'(x)=0 for all x ... (b)

    With the experience in (a) with real x, I can now only guess at the moment (since x in (b) can be others things than reals) the solution of (b) is f(x)=constant (the solution is actually the corresponding map) for any x. If there exists another solution other than f(x)=constant for (b), then contradiction occurs , i.e. the solution in (a) is not a "complete" solution. This completes the proof (my original guess is proved to be rigorous, although it is so much obvious maybe).

    Ok. My problems, I think, have been worked out completely.

    The next is my other imagination, please give some advice if possible.

    1 Originally, I also wrongly thought about the stuff about f'=df/dsth. dsth here seems to be sth in functional space stuff. If sth in here (f'=df/dsth) is related to the vector space, then maybe it will belong to some advanced subjects in the 2 ?? (in my original problem f'=df/dx and x has nothing much to do with the vector space mentioned)

    2. I heard about variational principal and functional analysis, where the argument or unknowns are function (or functional, I am not sure) itself. It seems so much related to the problem f'=0, where the unknown is function. And such a solution can be worked out in those subjects, or depending on what f' means maybe ?

    3. "some variation on the intermediate value theorem" belongs to the variational principal or functional analysis?

    best regards
    Last edited: Aug 10, 2007
  5. Aug 10, 2007 #4

    matt grime

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    I did not understand any of that. You say you have done a course in differential equations. So what is the issue? The solution to the D.E. df/dx=0 are the constants, and what about df/dx=f? I have no idea what you mean to do by invoking the variational principal, or things along those lines. If I give you the differential equation df/dx=f, you can write down the solutions, right?
  6. Aug 10, 2007 #5

    Please forget about my above explanation stuff together variational principal etc. Let me explain why I was afraid of function as an element of a vector space below.

    The special thing about the vector space of functions is a little different from that of a usual one. After learning calculus only, I have no knowledge of any operation of function (function is a mapping, right?). What does f+g means is beyond a student with only calculus knowledge, right?, unless it is defined.

    Whether a set of functions forms a vector space actually requires a new definition , i.e. the operations of functions, for example f+g is defined such that (f+g)(x)=f(x)+g(x). Such a kind of operation is new to me (maybe obvious to you). There are also, (cf)(x)=cf(x), zero function etc. These new definitions are made in order that the vector space axioms are met.Based on these new definitions, we can then say the vector space of functions.

    Now, the function is an element of the vector space defined. The element is a function such as f ,g etc, NOT THE FUNCTION VALUE and the value of f is f(x),which can be taken an a real.

    My orginal question is something like f'=0, actually means f'= zero function NOT NUMBER 0. The zero function is an element of the vector space of functions (and number 0 does not lie in my vector space of functions really). Therefore, f'(x)=0 learned in differential equations (where 0 is a real number only) CANNOT be directly used in my original case.

    In differential equations, I learned f'(x)=0 (where x and 0 are both real numbers), with the solution f(x)=constant. Now my case is f'=0 , which is somewhat different, isn't it?

    Also for my original problem2, you may say the kernal is {ce^x} since we have f'=f. But in fact the kernal should be the set of the functions NOT THE FUNCTION VALUEs. Therefore, the real kernal should be a set like this (my mathematical language is not standard, but hope you know what I mean)
    {f:f(x)=ce^x, f belongs to V, c a real number}.(YOU ONLY NEED TO CONFIRM THIS paragraph PLEASE if you think it is hard to catch what I mean, I have only a first degree of engineering and do not have the standard mathematical language.)

    Also, the set of FUNCTION VALUE f(x) is not the image of linear map D. The image of D is a set of functions. The FUNCTION here has not much to do with the linear map, it is only an element.

    Thank you
    best regards
  7. Aug 10, 2007 #6

    matt grime

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    Sorry, but you appear to have misunderstood what f'(x)=0 means. It means precisely that f'(x) is the zero function, i.e. that f'(x)=0 for all x. It meant that in your differential equation class, it means it now.

    not in the slightest. In fact you did not learn what you claim you learned, or at least you ought not to have. That 0 is still the 0 function, i.e. the function that returns the real number 0 for any input.

    In what way is the set of functions of the form c*exp(x) not a set of functions? I see no mention at all of function values by anyone but you.
  8. Aug 11, 2007 #7
    You said you did not mention the function value, but I think you did.

    You used c*exp(x) to describe the function of x, but I think it is the function value of x.

    f(x) is the value of f at x, or the image of x under f. The symbol f(x) are read
    "f of u". (Bishop and Goldenburg, Serge Lang, also write f(x) as fx). Also see wiki

    A specific input in a function is called an argument of the function. For each argument value x, the corresponding unique y in the codomain is called the function value at x, or the image of x under f. The image of x may be written as f(x) or as y

    Again, f'(x)=0 can be written g(x)=0. So I still think zero here is a real number.

    Maybe, if you define f(x) as the function (the action, not the value) of x, then I also agree with you.

    best regards
  9. Aug 11, 2007 #8


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    You are making this much too difficult! Yes, if "f" is a function, then f(x) represents a "value of the function", but it is standard notation to write "the function f(x)" with the understanding that you are talking about the function having those values for all x.

    Earlier, you say:
    Yet, you say you have taken differential equations? Differential equations ARE functional equations and you solve them for functions! D(f)= 0, since you say that D represents differentiation, is f '(0)= 0 and has solution f(x)= C (some arbitrary constant) for all x. The kernel of D is the set of constant functions.

    (D-I)(g)= 0 is precisely the differential equation g'- g= 0. That is, g'= g and so g has the property that it is its own derivative. One of the first things you should have learned in differential equations is that the general solution to that is g(x= Cex for all x (again, C is an arbitrary constant). The kernel of L= D- I is the set of all functions of the form Cex.

    You seem to have got the notion that "functional analysis" is involved here. Yes, functional analysis studies "function spaces" but looks at much deeper properties than are needed here. In particular, you should have say "What does f+g means is beyond a student with only calculus knowledge, right?" No, the definition of f+ g for f and g functions is taught in basic algebra classes, well before calculus.
  10. Aug 11, 2007 #9

    matt grime

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    f(x) is a function, not a value. f(3) would be a value, if 3 were in the domain of f. f(x) evaluated at x=3 would produce a value. f(x) is just a function.
  11. Aug 11, 2007 #10
    Thanks very much, matt_grime and Hallsofivy.

    Both of your detailed explanations (including differential equations stuff)
    have made sense to me now.

    best regards
  12. Aug 11, 2007 #11


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    You can get heated arguments about that in any Mathematics department!

    I am of the school that holds that "x" represents a number, and so, if "f" is a function, then f(x) is the value of f at that represented number! If you want to talk about a function, you say "f". If you say "f(x)" then you are talking about a (general) value of that function. However, I do concede that cleaving to that leads to very comples notation! I recall a
    mathematician who wrote a text in which he followed that concept rigorously: instead of "f(x)= x2", he would write "f( )= ( )2". I don't think that's the wave of the future!
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