Vrftjkry

I am trying to prove that $f:mathbb{R}^2rightarrowmathbb{R}^2$,
$$f(x)=x|x|$$
is differentiable as a part of a larger task.

I think there two ways to approach this:

1. By proving that for every $xinmathbb{R}^2$ there is a linear function $A$ for which
   $$f(x+h)=f(x)+Ah+|h|epsilon(h)$$
   where $epsilon(h)rightarrow0$ when $hrightarrow0$




2. By proving that all of the first order partial derivatives of $f$ exist and are continuous.

For the sake of my own understanding, I would like to know how to prove this with both of the ways. Here are my attempts so far:

1. $f(x+h)=(x_{1}sqrt{x_{1}^2+x_{2}^2}+h_{1},x_{2}sqrt{x_{1}^2+x_{2}^2}+h_{2})=(x_{1}sqrt{x_{1}^2+x_{2}^2},x_{2}sqrt{x_{1}^2+x_{2}^2})+(h_{1},h_{2})=f(x)+(h_{1},h_{2})$

I don't know how to go on with this since I'm not sure how I'm supposed to choose $A$. If $f$ is differentiable, $A$ should be $Df(x)$ but I don't know how should I manipulate the expression to achieve that.

2. For the second way, I'm not sure how to write out the general form of the first order partial derivatives for $f$.

Part 1.
Here is a derivation (no pun intended) of $f'(x)(h)$ from first
principles. It is valid not just in $mathbb{R}^2$, but in any real inner
product space $E$, not necessarily even finite-dimensional. ($E$ is
not even assumed to be complete; but if it isn't, then I don't think
one is allowed to speak of $f$ being "differentiable" at $x$.)

First, we make some estimates:

(i) By the Triangle Inequality,
$lvertlVert x + h rVert - lVert x rVertrvert leqslant
lVert h rVert$.

(ii) If $x ne 0$, then by (i), as $h to 0$,
$$
leftlvertfrac{2}{lVert x + h rVert + lVert x rVert}
- frac{1}{lVert x rVert}rightrvert =
frac{lvertlVert x rVert - lVert x + h rVertrvert}
{(lVert x + h rVert + lVert x rVert)lVert x rVert}
leqslant frac{lVert h rVert}{lVert x rVert^2} =
O(lVert h rVert).
$$

(iii) The Cauchy-Schwarz inequality
$lvertleftlangle x, h rightranglervert leqslant
lVert x rVert lVert h rVert$ gives
$leftlangle x, h rightrangle = O(lVert h rVert)$.

Now, for all $x, h in E$,
begin{align*}
f(x + h) - f(x) & = lVert x + h rVert(x + h) - lVert x rVert x
\ & = lVert x rVert h +
(lVert x + h rVert - lVert x rVert)(x + h)
\ & = lVert x rVert h +
(lVert x + h rVert - lVert x rVert)x + O(lVert h rVert^2),
&& text{by (i).}
end{align*}
This proves that $f'(x)(h) = 0$ for all $h$ when $x = 0$. From now
on, we assume that $x ne 0$.
begin{gather*}
f(x + h) - f(x) - lVert x rVert h =
frac{lVert x + h rVert^2 - lVert x rVert^2}
{lVert x + h rVert + lVert x rVert}x +
O(lVert h rVert^2)
\ =
frac{2leftlangle x, h rightrangle + lVert h rVert^2}
{lVert x + h rVert + lVert x rVert}x +
O(lVert h rVert^2)
=
frac{2}{lVert x + h rVert + lVert x rVert}
leftlangle x, h rightrangle x +
O(lVert h rVert^2)
\ =
frac{leftlangle x, h rightrangle}{lVert x rVert}x +
left(frac{2}{lVert x + h rVert + lVert x rVert} -
frac{1}{lVert x rVert}right)
leftlangle x, h rightrangle x +
O(lVert h rVert^2).
end{gather*}
Therefore, by (ii) and (iii),
$$
f(x + h) = f(x) + lVert x rVert h +
frac{leftlangle x, h rightrangle}{lVert x rVert}x +
O(lVert h rVert^2).
$$

^{This agrees with the formula for $f'(x)(h)$ in my earlier brief comment.
(The main result used there - apart from the Chain Rule, and the formula for
the derivative of the square root function on $mathbb{R}_{>0}$ - is
Frechet derivative for bilinear map.)}

Part 2.
For simplicity [but at some risk of confusion with the earlier use of the symbol
'$x$'!], I'll use the notation $(x, y)$, instead of $(x_1, x_2)$, and write
$$
(u, v) = f(x, y) = r(x, y) = (rx, ry),
text{ where } r = sqrt{x^2 + y^2}.
$$

The case $(x, y) = (0, 0)$ was dealt with in an earlier answer, but as that
answer has now been deleted, I'll go over the same ground here.

We have $f(h, 0) = (h|h|, 0)$, $f(0, k) = (0, k|k|)$, and so
begin{align*}
|u(h, 0)| & = h^2, v(h, 0) = 0, \
|v(0, k)| & = k^2, u(0, k) = 0,
end{align*}
showing that the partial derivatives
$D_1u(0, 0)$, $D_1v(0, 0), D_2v(0, 0)$, $D_2u(0, 0)$ exist, and are all zero.

Assume now that $(x, y) ne (0, 0)$. Then $r > 0$, and
$partial r/partial x = x/r$, $partial r/partial y = y/r$, whence
$$
begin{pmatrix}
frac{partial u}{partial x} & frac{partial u}{partial y} \
frac{partial v}{partial x} & frac{partial v}{partial y}
end{pmatrix}
begin{pmatrix} h \ k end{pmatrix} =
begin{pmatrix}
frac{x^2}{r} + r & frac{xy}{r} \
frac{xy}{r} & frac{y^2}{r} + r
end{pmatrix}
begin{pmatrix} h \ k end{pmatrix} =
r begin{pmatrix} h \ k end{pmatrix} +
frac{xh + yk}{r} begin{pmatrix} x \ y end{pmatrix},
$$
in agreement with the previous result.

In a convenient but admittedly loose notation, simply denoting the
separate convergence of all four matrix entries,
$$
lim_{(x, y) to (0, 0)}
begin{pmatrix}
frac{partial u}{partial x} & frac{partial u}{partial y} \
frac{partial v}{partial x} & frac{partial v}{partial y}
end{pmatrix} =
lim_{(x, y) to (0, 0)}
begin{pmatrix}
frac{x^2}{r} + r & frac{xy}{r} \
frac{xy}{r} & frac{y^2}{r} + r
end{pmatrix} = begin{pmatrix} 0 & 0 \ 0 & 0 end{pmatrix},
$$
showing that all four partial derivatives are continuous everywhere. $square$