Prove $\pi_{i}:\mathbb{R}^n \to \mathbb{R}$ such $\pi (\bar{x})=x_{i}$ for every $\bar{x}=(x_{1},...x_{n}) \in \mathbb{R}^n$ is differentiable.

Let $\pi_{i}:\mathbb{R}^n \to \mathbb{R}$ such $\pi (\bar{x})=x_{i}$ for every $\bar{x}=(x_{1},...x_{n}) \in \mathbb{R}^n$. I want to prove that for every $i \in \lbrace 1,2,....n \rbrace$ the $i$-th projection, $\pi_{i}$ is differentiable in every $\bar{x} \in \mathbb{R}^n$ and calculate its differential.

Im wondering if the particular case I did for $\mathbb{R}^2$ is right? Any help proving the general case or alternate proofs will be apreciated.

Here is how I attacked this problem:

Trying to prove this for $\pi_{1,2}: \mathbb{R}^2 \to \mathbb{R}$ to get the idea and then generalize to $\mathbb{R}^n$. Following the Marsden-Tromba calculus book definition I should prove for an abritrary $(x_{1},x_{2}) \in \mathbb{R}^2$ first that $\frac{\partial \pi_{i} }{\partial x}$ and $\frac{\partial \pi_{i} }{\partial y}$ exist at the arbitrary $(x_{1},x_{2})$. Without lost of generality, lets consider the $1$-th projection. So

$$\frac{\partial \pi_{1} }{\partial x}(x_{1},x_{2})=\lim_{h \to 0}\frac{\pi_{1}(x_{1}+h,x_{2})- \pi_{1}(x_{1},x_{2})}{h}=lim_{h \to 0} \frac{x_{1}+h-x_{1}}{h}=1.$$

And,

$$\frac{\partial \pi_{1} }{\partial y}(x_{1},x_{2})=\lim_{h \to 0}\frac{\pi_{1}(x_{1},x_{2}+h)- \pi_{1}(x_{1},x_{2})}{h}=lim_{h \to 0} \frac{x_{1}-x_{1}}{h}=0.$$

Proving the required partial derivatives exist at the arbitrary point.

So then , I should prove that

$$lim_{(x,y) \to (x_{1},x_{2})} \frac{\pi_{1}(x,y)-\pi_{1}(x_{1},x_{2})-\frac{\partial \pi_{1} }{\partial x}(x_{1},x_{2})(x-x_{1})-\frac{\partial \pi_{1} }{\partial y}(x_{1},x_{2})(y-x_{2})}{||(x-x_{0},y-y_{0})||}=0$$

Developing the numerator part of the last limit I got $$x-x_{1}-1(x-x_{1})-0(y-x_{2})=-2x_{1}=0.$$

For the denominator I got;

$$\sqrt{(x-x_{1})^2+(y-x_{2})^2}$$.

So,

$$lim_{(x,y) \to (x_{1},x_{2})} \frac{0}{\sqrt{(x-x_{1})^2+(y-x_{2})^2}}=0$$.

Proving the case for $\mathbb{R}^2$.

Solution 1:

Recall the following definition of differentiability:

Definition: Let $(V, \lVert \cdot \rVert_V)$ and $(W, \lVert \cdot \rVert_W)$ be normed vector spaces (not assumed finite-dimensional), and $A$ an open subset of $V$. Let $f: A \to W$ be a given function, and let $a \in A$. We say $f$ is differentiable at $a$ if there is a continuous linear transformation $T:V \to W$ such that \begin{equation} \lim_{h \to 0} \dfrac{\lVert f(a+h) - f(a) - T(h)\rVert_W}{\lVert h \rVert_V} = 0. \end{equation} In this case one can show $T$ is unique; this allows us to unambiguously denote $T$ by $df_a$ or $Df(a)$ or any other convenient symbol.

Now, we have the following theorem:

Theorem: With $V$ and $W$ as in the definition above, let $f:V \to W$ be a continuous linear transformation. Then for every $a \in V$, $f$ is differentiable at $a$ and $df_a = f$ (this is an equality of linear transformations from $V$ to $W$)

The proof of this is very easy. What we are claiming is if in the definition above, we put $T = f$ (which by assumption is continuous), then the limit will be $0$. To prove this, notice that for $h \neq 0$, the difference quotient is \begin{align} \dfrac{\lVert f(a+h) - f(a) - f(h) \rVert_W}{\lVert h \rVert_V} = 0 \end{align} It equals $0$ by linearity of $f$, so in the limit $h \to 0$, we still get $0$. This is exactly what it means for $f$ to be differentiable at $a$, and to have $df_a = f$.

I assume you're only interested in the finite dimensional case, for instance $V = \mathbb{R^n}$ and $W =\mathbb{R^m}$. In this case, every linear transformation between finite dimensional vector spaces is automatically continuous, so in the definition above, you do not have to worry about checking for continuity of $T$.

Now, all the projections $\pi_i$ you are considering are clearly linear maps between finite-dimensional spaces, hence the above theorem applies; so they are all differentiable everywhere, and for every $a = (a_1, \dots a_n) \in \mathbb{R^n},$ we have $d(\pi_i)_a (\cdot)= \pi_i (\cdot)$.