Showing non-differentiability of a function at $0$

Question

Starting from a function $f:t\mapsto f(t)$ such that:

• $f(0)=0\tag{1}$
• $\liminf\limits_{t\to\infty} \sqrt{t}f\left(\dfrac{1}{t}\right)=-\infty\tag{2}$
• $\limsup\limits_{t\to\infty} \sqrt{t}f\left(\dfrac{1}{t}\right)=+\infty\tag{3}$

setting:

• $f_l(t)=\liminf\limits_{\delta\to0} \dfrac{f(t+\delta)-f(t)}{\delta}$;
• $f^u(t)=\limsup\limits_{\delta\to0} \dfrac{f(t+\delta)-f(t)}{\delta}$.
  

I have to show that:

1. $f^u(0)=+\infty$;
2. $f_l(0)=-\infty$.
   

I succeeded in showing 1. by the following steps: \begin{align} f^u(0)=\limsup\limits_{t\to\infty} \dfrac{f\left(\dfrac{1}{t}\right)-f(0)}{\dfrac{1}{t}}&\geq\limsup_{t\to\infty}\sqrt{t}f\left(\dfrac{1}{t}\right)\\&=+\infty \end{align} with the inequality coming from $(1)$ plus the fact that $\dfrac{f\left(\dfrac{1}{t}\right)}{\dfrac{1}{t}}=tf\left(\dfrac{1}{t}\right)\geq\sqrt{t}f\left(\dfrac{1}{t}\right)$ for $t\geq1$ and the last equality following from $(3)$.

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However, I cannot show 2. (I have tried to show it by the same token used for showing 1., but I did not succeed). Could you please give me some hint?

Answer 1

$\lim \inf_{t \to 0} \frac {f(t)} t=\lim \inf_{t \to \infty} t f(\frac 1 t)\leq \lim \inf_{t \to \infty} \sqrt t f(\frac 1 t)=-\infty$. [The inequality follows from the fact that $f(\frac 1 t)<0, t>1$ implies $tf(\frac 1 t) \leq \sqrt t f(\frac1 t)$].