Equality of Two Linear Mappings

[Parmenides]

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
\lim_{h \to 0}\frac{f(a + h) - f(a) - L(h)}{h} = {\bf{0}}
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:
lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = 0
My first idea was to use the definition of $L$ from the theorem so that
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
And so lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = \lim_{h \to 0}\frac{0}{h} = 0
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.

 

[jbunniii]

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?

 

[Parmenides]

\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
So \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
Thus \lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = 0
Does the next step require a delta-epsilon proof to establish the validity of this limit? Seems to be pointing in that direction.

 

[jbunniii]

Thus \lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = 0
Does the next step require a delta-epsilon proof to establish the validity of this limit? Seems to be pointing in that direction.

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 $L_2$.

 

[Parmenides]

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.

 

[jbunniii]

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.

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
$$\frac{L_2(h)}{h}$$

 

[Parmenides]

\lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = 0
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,
= \lim_{h \to 0}\frac{L_1(h{\bf{x}}_1) - L_2(h{\bf{x}}_2)}{h} = 0
And, by linearity:
= \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)
If I didn't need to make a distinction between ${\bf{x}}$ then perhaps this is plausible. ><

 

[jbunniii]

\lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = 0
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 ${\bf{x}}_1,{\bf{x}}_2 \in R$.

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?

 

[Parmenides]

Implying that:
\lim_{h \to 0}L_1(1) = \lim_{h \to 0}L_2(1)
But as $h \rightarrow 0$, doesn't this imply that $L_1 = L_2$?

 

[jbunniii]

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.