从一个小定理出发，用定义理解狄拉克函数

证明一个小的定理，或者说一个结论：

δ [f (t)] = \sum_{i = 1}^{N} \frac{1}{f (t)^{'}} δ (t - t_{i}) f (t_{i}) = 0, i = 1, 2, 3...

$\delta[f(t)] = \sum_{i=1}^N \frac{1}{f(t)'} \delta(t-t_i)\\ f(t_i)=0,i=1,2,3...$

由于有： $\delta(t)= \lim\limits_{\Delta t \to 0} \frac{1}{\Delta t},t \leq \frac{1}{2} |\Delta t|$ ，则将t替换成f(t)后显然有如下推导过程：

δ [f (t)] = lim_{Δ f (t) \to 0} \frac{1}{Δ f (t)} = lim_{Δ f (t) \to 0} \frac{1}{f (t)^{'} Δ t} = lim_{t \to t_{i}} \frac{1}{f (t)^{'} Δ t} = lim_{Δ (t - t_{i}) \to 0} \frac{1}{f (t_{i})^{'} Δ t} = \frac{1}{f (t_{i})^{'}} δ (t - t_{i}), f (t_{i}) = 0

$\delta[f(t)]=\lim\limits_{\Delta f(t) \to 0} \frac{1}{\Delta f(t)}=\lim\limits_{\Delta f(t) \to 0} \frac{1}{f(t)' \Delta t} \\ =\lim\limits_{ t \to t_i} \frac{1}{f(t)' \Delta t}=\lim\limits_{\Delta(t-t_i) \to 0} \frac{1}{f(t_i)'\Delta t} \\ =\frac{1}{f(t_i)'} \delta(t-t_i),f(t_i)=0$

做如下说明：

由于 $\Delta f(t)=f(t)'\Delta t$ ，故可等效为 $\Delta f(t) \to 0 \Leftrightarrow f(t)' \to 0, \Delta t \to 0$