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Equality of Two Linear Mappings

  1. Mar 17, 2013 #1
    Let ##L_1## and ##L_2## be two linear mappings from ##R \rightarrow R^n## satisfying the following theorem:
    "The mapping ##f: R \rightarrow R^n## is differentiable at ##a \in R## if and only if there exists a linear mapping ##L: R \rightarrow R^n## such that
    [tex]\lim_{h \to 0}\frac{f(a + h) - f(a) - L(h)}{h} = {\bf{0}}[/tex]
    in which case L is defined by ##L(h) = df_a(h) = hf'(a)##." (From Edwards 'Advanced Calculus').
    I am to prove that ##L_1 = L_2##. The text hints that I should first show that:
    [tex]lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = 0[/tex]
    My first idea was to use the definition of ##L## from the theorem so that
    [tex]L_1(h) - L_2(h) = h\lim_{h \to 0}\Big(\frac{f(a + h) - f(a)}{h}\Big) - h\lim_{h \to 0}\Big(\frac{f(a + h) - f(a}{h}\Big) = h\lim_{h \to 0}\Big(\frac{f(a + h) - f(a) - f(a + h) + f(a)}{h}\Big) = h\lim_{h \to 0}(0) = 0[/tex]
    And so [tex]lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = \lim_{h \to 0}\frac{0}{h} = 0[/tex]
    But looking back, I think that whole process rested on the assumption that ##L_1 = L_2## in the first place. I am not quite sure what to do.
     
  2. jcsd
  3. Mar 17, 2013 #2

    jbunniii

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    Start by assuming
    $$\lim_{h \to 0}\frac{f(a + h) - f(a) - L_1(h)}{h} = 0$$
    and
    $$\lim_{h \to 0}\frac{f(a + h) - f(a) - L_2(h)}{h} = 0$$
    Now subtract one equation from the other. What do you end up with?
     
  4. Mar 17, 2013 #3
    [tex]\lim_{h \to 0}\frac{f(a + h) - f(a) - L_1(h)}{h} - \lim_{h \to 0}\frac{f(a + h) - f(a) - L_2(h)}{h} = 0 - 0 = 0[/tex]
    So [tex]\lim_{h \to 0}\frac{f(a + h) - f(a) - f(a +h) + f(a)}{h} - \lim_{h \to 0}\frac{L_2(h) - L_1(h)}{h} = \lim_{h \to 0}\frac{0}{h} + \lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = 0[/tex]
    Thus [tex]\lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = 0[/tex]
    Does the next step require a delta-epsilon proof to establish the validity of this limit? Seems to be pointing in that direction.
     
  5. Mar 17, 2013 #4

    jbunniii

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    Well what is the next step? Your goal is to show that ##L_1(x) = L_2(x)## for all ##x \in \mathbb{R}##.

    Hint: This would be a good time to take advantage of the linearity of ##L_1## and ##L_2##.
     
  6. Mar 17, 2013 #5
    Hm, I'm not yet making the connection from a mapping ##L(h)## to ##L({\bf{x}})##, but my first idea is since the above theorem is satisfied, there exist linear mappings ##L_1## and ##L_2## such that ##L_1(h{\bf{x}})## and ##L_2(h{\bf{x}})## = ##h{\bf{x}}_1## and ##h{\bf{x}}_2##, respectively for all ##{\bf{x}}_1,{\bf{x}}_2 \in R##. If I was allowed to say that ##L(h{\bf{x}}) = {\bf{x}}L(h)## I could see how using what I just showed above might be useful, but that doesn't seem right.
     
  7. Mar 17, 2013 #6

    jbunniii

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    But that's the nice thing about linear mappings: you CAN say that. Remember that a linear mapping from ##\mathbb{R}## to ##\mathbb{R}^n## satisfies
    $$L(x+y) = L(x) + L(y)$$
    for any ##x,y \in \mathbb{R}##, and
    $$L(ax) = aL(x)$$
    for any ##a,x \in \mathbb{R}##.

    Try using the second fact to simplify
    $$\frac{L_1(h)}{h}$$
    and
    $$\frac{L_2(h)}{h}$$
     
  8. Mar 17, 2013 #7
    [tex]\lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = 0[/tex]
    Since our theorem is satisfied, there exist linear mappings ##L_1## and ##L_2## such that ##L_1(h{\bf{x}})## and ##L_2(h{\bf{x}})## = ##h{\bf{x}}_1## and ##h{\bf{x}}_2##, respectively for all ##{\bf{x}}_1,{\bf{x}}_2 \in R##. By linearity, we have that ##L_1({\bf{x}}_1h) = {\bf{x}}_1L_1(h)## and ##L_2({\bf{x}}_2h) = {\bf{x}}_2L_2(h)##. Then,
    [tex] = \lim_{h \to 0}\frac{L_1(h{\bf{x}}_1) - L_2(h{\bf{x}}_2)}{h} = 0[/tex]
    And, by linearity:
    [tex] = \lim_{h \to 0}\frac{hL_1({\bf{x}}_1) - hL_2({\bf{x}}_2)}{h} = \lim_{h \to 0}\frac{h}{h}L_1({\bf{x}}_1) - \lim_{h \to 0}\frac{h}{h}L_2({\bf{x}}_2) =0 \rightarrow L_1({\bf{x}}_1) = L_2({\bf{x}}_2)[/tex]
    If I didn't need to make a distinction between ##{\bf{x}}## then perhaps this is plausible. ><
     
  9. Mar 17, 2013 #8

    jbunniii

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    Wait, isn't that what you are trying to prove? I don't see how you can say that at this point.

    What you have is
    $$\lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = 0$$
    We can rewrite this as follows:
    $$\lim_{h \to 0}\left(\frac{1}{h}L_1(h) - \frac{1}{h}L_2(h)\right) = 0$$
    and by linearity,
    $$\lim_{h \to 0}\left(L_1\left(\frac{1}{h}h\right) - L_2\left(\frac{1}{h}h\right)\right) = 0$$
    which is simply
    $$\lim_{h \to 0}\left(L_1\left(1\right) - L_2\left(1\right)\right) = 0$$
    What does this imply?
     
  10. Mar 17, 2013 #9
    Implying that:
    [tex]\lim_{h \to 0}L_1(1) = \lim_{h \to 0}L_2(1)[/tex]
    But as ##h \rightarrow 0##, doesn't this imply that ##L_1 = L_2##?
     
  11. Mar 17, 2013 #10

    jbunniii

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    No, there is no dependency on ##h## in ##L_1(1)## or ##L_2(1)##. So after taking limits the equation simply becomes
    $$L_1(1) = L_2(1)$$
    Now, can you see how this actually implies that ##L_1(x) = L_2(x)## for all ##x \in \mathbb{R}##? Use the linearity of ##L_1## and ##L_2## to show this.
     
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