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Let $V , W_1$, and $W_2$ be finite-dimensional vector spaces, and let $W = W_1 \times W_2$. Given an open set $U \subseteq V$ and functions $f_1 :U \rightarrow W_1$ and $f_2 :U \rightarrow W_2$, the rule $f(x)=(f_1(x),f_2(x))$ defines a mapping from $U$ into $W$. Let $a \in U$.

Prove that $f$ is differentiable at $a$ if and only if $f_1$ and $f_2$ are differentiable at $a$.

$(\Rightarrow)$ Suppose $f$ is differentiable at $a$. Let $T:V\rightarrow W$ be a linear mapping. Then there exists $T_1:V\rightarrow W_1$ such that $f_1=T_1\circ f_1$. Then $f_1=T_1\circ f_1\implies$

$f_1(a)=(T_1\circ f_1)(a)\implies$

$(Df_1)(a)=D(T_1\circ f_1)(a) = (DT_1\circ Df_1)(a) = (T_1\circ Df)(a)$. Therefore $Df_1$ exits. The same argument is made for $f_2$.

$(\Leftarrow)$ Suppose $f_1$ and $f_2$ are differentiable at $a$. Let $T_1:V\rightarrow W_1$ and $T_2:V\rightarrow W_2$ be a linear mappings. Then there exists $T:V\rightarrow W$ such that $T(v)=(T_1(v), T_2(v))$ that is linear.

Is the forward part of my proof correct? I am trying to figure out the converse, but I want to be sure the first part is correct before spending too much time on it.

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Actually, the forward part of your proof doesn't make much sense.

You introduce $T$ for no reason, as you never reference it again. You claim the existence of some $T_1$ from $V$ into $W_1$, that somehow satisfies $f_1 = T_1\circ f_1$ even though $f_1$ has codomain $W_1$ and so cannot be composed with $T_1$, whose domain is $V$. From the rest, I am sure you meant $f_1 = T_1 \circ f$, but this is still not right, as the codomain of $f$ is $W$, not $V$.

In fact, it is appears to me that $T_1$ is actually supposed to be $\pi_1$, the projection from $W = W_1 \times W_2 \to W_1$. $\pi_1$ is a specific known function. It maps $(w_1, w_2) \in W$ to $w_1$. So you don't introduce it by saying something like "there exists $\pi_1$ ...". Instead you say "let $\pi_1$ be the projection ...". You are referring to a specific function, not some arbitrary one. It is true that $f_1 = \pi_1 \circ f$.

However if you already have the theorem that if $r, s$ are differentiable, the so is $r\circ s$ and the derivative is given by the chain rule: $D(r\circ s) = Dr\circ Ds$, then you can use this theorem to say that $f_1$ is differentiable and has derivative $Df_1 = D\pi_1 \circ Df$. (No further simplification is needed.)

And for the reverse implication, you are already introducing unnecessary arbitrary functions again, when you need to be referring to specific ones:

If $f_1, f_2$ are differentiable at $a$, then define a map $T\ :\ V \to W \ :\ v \mapsto (D_af_1(v), D_af_2(v))$. Then $T$ is linear.

Next, use the definition of $D_af$, and show that $T$ fits it. I.e., if the definition of the derivative that your course is using is:

$D_af$ is the unique linear map $T \ :\ V \to W$ such that $$\lim_{v\to 0}\frac{\|f(a + v) - f(a) - T(v)\|}{\|v\|} = 0$$

Then you need to show that the particular $T$ defined by $T(v) = (D_af_1(v), D_af_2(v))$ satifies this definition.

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