$f\in \mathcal{L}^{1}_{loc}(\Bbb{R}) $ and $\text{supp}(f) \subset [0, 1]$ and $\int_{0}^{1}f=1$
Consider a sequence $(f_k)_{k\in\Bbb{N}}$ where $f_k: \Bbb{R}\to \Bbb{R}$ defined by $$f_k(x) =kf(kx) $$
Calim :
- $(f_k)\to \textbf{0} $ pointwise on $\Bbb{R}\setminus\{0\}$.
- $(T_k) \to \delta $ in the space of distribution $\mathcal{D'}(\mathbb{R}) $.
Fix $x\in \Bbb{R}$. For $x<0$ , $\forall k\in\Bbb{N}$ we have $f_k(x) =0$.
For $x>0$ , $\exists N\in \Bbb{N}$ such that $Nx>1$ and hence $\forall k\in\Bbb{N} , k\ge N$ we have $f_k(x) =0$
Hence $f_k(x) \to 0$ on $ \Bbb{R}\setminus\{0\}$
$$T_k(\varphi(x)) =\int_{\Bbb{R} }f_k(x) \varphi(x) dx$$
where $\varphi\in \mathcal{D}(\Bbb{R}) =C_{c}^{\infty}(\Bbb{R}) $
To show : $T_k\to \delta$ in $\mathcal{D}'(\Bbb{R}) $ i.e $T_k(\varphi(x)) \to \delta(\varphi(x))=\varphi(0) $ on $\Bbb{R}$ for all $\varphi\in \mathcal{D}'(\Bbb{R}) $
$|T_k(\varphi(x))-\varphi(0) |\\=|\int_{0}^{1} f(t) \varphi(\frac{t}{k})dt-\int_{0}^{1} \varphi(0)f(t) dt |\\\le |\varphi(\frac{t}{k})-\varphi(0)|\int_{0}^{1}|f(t)| dt \\\le M |\varphi(\frac{t}{k})-\varphi(0)|<\epsilon$
By uniform continuity of $\varphi $ on $[0, 1]$.
Please help me to prove. Thanks in advance.