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Note: $K_\varepsilon(x_0):=(x_0 - \varepsilon, x_0 + \varepsilon )$

let $r > 0$ and $f:K_r(0) \rightarrow \mathbb R$ with $\exists C>0 \forall x \in K_r(0) : |f(x)| \le Cx^2$

Show, that $f$ is differentiable in $0$.

My proof:

$Cx^2$ is divergent for $\infty$, because there is for every $M>0$ a $R \in K_r(0)$ with $f(x)>M$ for all $x>R$.

$$\lim_{x \to 0} \quad f(x) = \infty$$

there must be $|Cx^2 - |f(x)| | < \varepsilon$ for all $x \in K_r(0)$

for $f(x)$ there is $M>0$ for ever $R \in K_r(0)$ so $f(x)>M$ and $Cx^2 > f(x)$. So $Cx^2 > f(x) > M$. $\rightarrow f(x)$ is divergent for $\infty$ and so it is for $K_r(0)\setminus \{0\}$ differentiable.

It is known that Cx^2 is differentiable for 0. Because of $|Cx^2|- |f(x)||$ $f(x)$ must be a Polynom with an even degree and so it is differentiable for $x=0$.

Question: Is that proof correct? If not, what is missing, or what could be wrong?

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    I can't understand your proof at all. This is simply a question of using the definition of the derivative at $0.$ The graph of $f$ lies between $y=-Cx^2$  and $y = Cx^2.$ Doesn't that suggest that the tangent line is the $x$-axis? – 2017-01-03
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    @zhw:I second that. To the OP: sorry, but this text doesn't make much sense. In any proof, each claim has to have a reason. Can you say that about what you wrote here? Can you explain (not to us, to yourself) each phrase in your "proof": what you said there and why you said that? – 2017-01-03

1 Answers 1

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Since $ 0 \leq |f(0)| \leq C \cdot0^2$, we can conclude that $ f(0)=0 $
Looking at the difference quotient, there holds $$ 0 \leq|1/t(f(0+t)-f(0))|=|1/t| \cdot |f(0+t)| \leq | 1/t| \cdot Ct²=C|t| \to 0 $$
as $ t \to 0 $. Hence f is differentiable in 0.

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    He's in $\mathbb R,$ not $\mathbb R^n.$ – 2017-01-03
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    Thanks for pointing it out, I modified my response. – 2017-01-03