I am trying to show that the function $H(f,t):\operatorname{Diff}^r (\mathbb{R^n})\times [0,1]\rightarrow \operatorname{Diff}^r(\mathbb{R^n})$ that sends $(f,t)\rightarrow H(f,t)$, where $H(f,t)(x)=\frac{f(tx)-f(0)}{t}$ if $t\neq 0$ and $H(f,t)(x)=Df(0)x$, if $t=0$ will be continuous if we give $\operatorname{Diff}(\mathbb{R^n})$ the weak topology and won't be continuous if we give it the strong topology.
Now for the weak topology proving continuity is where I am having some trouble. So any help here is aprecciated.
Now for the strong topology I believe the following works. Let's suppose that $H(f,t)$ is continuous so that when we have a sequence $f_n\rightarrow g$ and $t_n\rightarrow t$ then $H(f_n,t_n)\rightarrow H(g,t)$ , both in the strong topology. Now take $t_n=t=1, $ and $f_n\rightarrow g$ such that $f_n(0)\neq g(0)$. Since we know that $f_n\rightarrow g$ in the strong topology therer exists a compact set $K'\in \mathbb{R^n}$ and $m'\in \mathbb{N}$ such that for all $n>m'$ and $x\in \mathbb{R^n}-K'$ we have that $f_n(x)=g(x)$., this is an exercise in Hirsch's Differential topology. Now for the same reason there exists a compact set $K$ and a natural number $m$ such that when $n>m$ and $x\in \mathbb{R^n}-K$ we have that $H(f_n(x),1)=H(g(x),1)$ and so $f_n(x)-f(0)=g(x)-g(0)$. Now consider $x\in \mathbb{R}^n-K-K'$ and $M=\max\{m,m'\}$ we have that for $n>m$ this will give us that $f_n(0)=g(0)$, giving a contradiction with our choice of sequence. Now I think this works but would appreciate some input.
Any hints regarding the weak topology are appreciated. Thanks in advance.