Second derivative of $f(x)=\max(0,x-c)$

Question

I am unable to prove the claim that $f''(x)=1$ if $x=c$, and $f''(x)=0$ otherwise.

Let $f(x)=\max(0,x-c)$ for some $c\in\mathbb R$ given. By definition, $f''(x)=\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}$. If $x<c$, each term in the numerator is $0$ for small $h$. If $x=c$, the numerator is $h$, so the limit is $0$. If $x>c$, the numerator is $0$.

Clearly I'm missing something. Any guidance is greatly appreciated. Thanks in advance.

EDIT: Context of the question is the exposition of Dupire's formula in Bergomi's Stochastic Volatility Modeling, page 27:

Answer 1

$$\begin{align}f(x)~&:=~\max(0,x-c)~=~(x-c) ^+,\cr f^{\prime}(x)~&=~\theta(x-c),\cr f^{\prime\prime}(x)~&=~\delta(x-c).\end{align} $$

Answer 2

The second derivative is the Dirac delta 'function', which is not actually a function - that explains why you are stuck! When calculating the Dupire local volatility the only property of the Dirac delta function that you need to know is that

$ \mathbb{E} \left[ \frac{1}{2} \delta(S_T - K) \sigma_T^2 S_T^2 \right] = \frac{1}{2} K^2 \mathbb{E} \left[ \delta(S_T-K) \sigma_T^2 \right]$

You may also find it helpful to note that

$ \sigma^2_{\text{LV}}(t,s) := \mathbb{E} \left[ \sigma_t^2 | S_t = s \right ] = \frac{\mathbb{E} \left[ \sigma_t^2 \delta(S_t-s) \right ]}{\mathbb{E} \left[ \delta(S_t-s \right ] }$