Example a function $f$ such that non differentiable at $x=a$

Question

Example a function $f$ such that non differentiable at $x=a$ and $f(a)=0$ and $|f|$ differentiable at $x=a$

Answer 1

There are actually no such examples : if $|f|$ is differentiable at $a$ and $f(a)=0$, then $f$ is differentiable at $a$.

Indeed, $|f|$ is nonnegative and vanishes at $a$, so we must have $|f|'(a)=0$. This means that $$\lim_{x\to a}\frac{|f(x)|}{x-a}=0.$$ But then it is also true that $$\lim_{x\to a}\left|\frac{f(x)-f(a)}{x-a}\right|=\lim_{x\to a}\left|\frac{f(x)}{x-a}\right|=0;$$ and as a consequence $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=0.$$ So $f$ is differentiable at $a$ with $f'(a)=0$.

Answer 2

We prove that this is impossible. In other words, we wanna prove that if |f| is differentiable in $x=a$ and $f(a)=0$ (and so $|f|(a)=0$) then f itself should also be differentiable in $x=a$. First, we prove that $|f|'(a)=0$ (This makes sense and is compatible with intuition since if a nonnegative function gets back to zero in some point $x=a$ an it can't do it sharply then it must land smoothly on x-axis). Differentiability of $|f|$ in $x=a$ means that:$$\forall \epsilon>0 \quad \exists \delta\qquad |x-a|<\delta\to |\frac{|f(x)|}{x-a}-l|<\epsilon$$Take $g(x)=|f(x)|$. An alternative definition for differentiability in $x=a$ is:$$\forall \epsilon>0 \quad \exists \delta>0\qquad ,\quad |\frac{g(a+\delta)-g(a-\delta)}{2\delta}-l|<\epsilon$$for some $l$. This leads us to:$$|g(a+\delta)-g(a-\delta)-2l\delta|<2\epsilon\delta\\ g(a-\delta)+2\delta(l-\epsilon)<g(a+\delta)<g(a-\delta)+2\delta(l+\epsilon)$$we prove that $l$ can't be positive. Since $g(a)=0$ we can write it using $\epsilon$-$\delta$ approach:$$\forall \epsilon>0 \quad \exists \delta>0\quad,\quad g(a+\delta)<\epsilon$$from $g(a-\delta)+2\delta(l-\epsilon)<g(a+\delta)$ we conclude that $g(a-\delta)<g(a+\delta)-2\delta(l-\epsilon)$. Choose $\epsilon=\frac{l}{2}$ therefore $\epsilon<l-\epsilon$ and we have:$$\forall \epsilon>0(and \ \ l-\epsilon>0)\quad \exists\delta>0\quad,\quad g(a+\delta)<\epsilon<l-\epsilon\\and\\g(a-\delta)<g(a+\delta)-2\delta(l-\epsilon)<(l-\epsilon)(1-2\delta)<0$$for sufficiently small $\delta$ which leads us to contradiction since $g(a-\delta)>0$. To the same argument we can imply that $l$ is not negative(check it!) therefore $l=0$. So: $$\forall \epsilon>0 \quad \exists \delta\qquad |x-a|<\delta\to |\frac{|f(x)|}{x-a}-l|=|\frac{|f(x)|}{x-a}|=|\frac{f(x)}{x-a}|<\epsilon$$ which implies that $f$ is differentiable in $x=a$ ant its derivation is $0$. That's what we wanted to prove.