Vtyjkyuk

question

I am trying to understand the concept of continuity better and for that I wonder, if the following function

$$
f(x) =
begincases
0 &textif x geq 0\
x^2 &textif x < 0
endcases
$$

continuous and its derivatives continuous in two levels or more such that it is a function of class $C^2$?

context

Someone reasoned that $f(x^2)=0 Rightarrow f(t)=0$ to eliminate a term in an expression, but it seems to me that it isn't necessarily true if $t < 0$ because of cases like the function I am asking about. The function has to be of class $C^2$, however, so the function $f$ might not be contradicting the reasoning if $f$ isn't of class $C^2$.

It's certainly continuously differentiable (i.e. $C^1$), but it isn't $C^2$. To see this first compute the derivative. For $x_0 > 0$, the function is constant around $x_0$, and hence has a derivative of $0$. For $x_0 < 0$, the function is equal to $x^2$ around $x_0$, so the derivative is $2x_0$. The only case left to check is $x = 0$.

We have
beginalign*
f'(0) &= lim_h to 0 fracf(h) - f(0)h = lim_h to 0 fracf(h)h \
&= lim_hto 0 begincases 0 & textif h ge 0 \ h & textif h < 0 endcases.
endalign*
Thus clearly we have $lim_hto 0^+ fracf(h)h = lim_hto 0^- fracf(h)h = 0$, so $f'(0) = 0$. Hence,
$$f'(x) = begincases 0 & textif x ge 0 \ 2x & textif x < 0 endcases.$$
Such a function could only be discontinuous at $0$, but the left and right limits both agree with the function value, so it's continuous.

It is not, however, differentiable at $0$. Note that
beginalign*
f''(0) &= lim_h to 0 fracf'(h) - f'(0)h = lim_h to 0 fracf'(h)h \
&= lim_hto 0 begincases 0 & textif h ge 0 \ 2 & textif h < 0 endcases.
endalign*
Clearly, in this case, the left and right limits are $2$ and $0$ respectively. Hence the limit (and the second derivative at $0$) doesn't exist.