f(x+y)=f(x)+f(y)+P(x;y).Prove that: f is a function of the class $C^2$ Let $P:\mathbb{R}^2 \rightarrow \mathbb{R}$ be a mapped of the class $C^2$.
Let $f:\mathbb{R} \rightarrow \mathbb{R} $ be a continuous function such that: $\forall (x;y)\in\mathbb{R}^2, f(x+y)=f(x)+f(y)+P(x;y)$
Prove that: f is a function of the class $C^2$
Because f is continuous in  $\mathbb{R}$,  $F:\mathbb{R} \rightarrow \mathbb{R},x \mapsto \int_{0}^{x}f$ is a mapped of the class $C^1$ in  $\mathbb{R}$ and $F'=f$.
We have $\forall (x;y)\in\mathbb{R}^2, \int_{0}^{y}f(t+x)dt=\int_{0}^{y}(f(t)+f(x)+P(t;x))dt=F(y)+yf(x)+\int_{0}^{y}P(t;x)dt$.
However: $\int_{0}^{y}f(t+x)dt=\int_{x}^{x+y}f(u)du=F(x+y)-F(x) $
So $\forall (x;y)\in\mathbb{R}^2,F(x+y)=F(x)+F(y)+yf(x)+\int_{0}^{y}P(t;x)dt$.
Instead $y=1$, we have: $\forall (x)\in\mathbb{R},f(x)=F(x+1)-F(x)-F(1)-\int_{0}^{1}P(t;x)dt$
Because P is a mapped of the class $C^2$ and $P,\frac{\partial P}{\partial y} \frac{\partial^2 P}{\partial y^2}$ exists and is continuous in [0;1]x$\mathbb{R}$, $x \mapsto \int_{0}^{1}P(t;x)dt$ is a mapped of the class $C^2$.
I need to prove that F is a mapped of the class $C^2$ .
Could you help me prove that? Thanks for helping
 A: To prove $f$ is of class $C^2$, I'll take a different approach.
I'll take $\lim_{t\to 0}\frac{f(x+t)-f(x)}{t} = \lim_{t\to 0}\frac{f(x)+f(t)+P(x,t)-f(x)}{t}=\lim_{t\to 0}\frac{f(t)+P(x,t)}{t}$
Then, let's caculate $f(x)=f(x)+f(0)+P(x,0) \Rightarrow f(0) = -P(x,0)$
I know that $P(x,y)$ is $C^2$, so I there is $\frac{\partial P(x,0)}{\partial x} = -f'(0)$
Now let's take the limit on $0 \Rightarrow f'(0) = \lim_{t\to 0}\frac{f(0+t)-f(0)}{t} = \lim_{t\to 0}\frac{f(t)+P(t,0)}{t}$
As $\lim_{t\to 0}\frac{P(t,0)}{t}$ exists and is finite and $f'(0)$ exists, we have $\lim_{t\to 0}\frac{f(t)}{t}$
The same way $\lim_{t\to 0}\frac{P(x,t)}{t}$ exists, and as $\lim_{t\to 0}\frac{f(t)}{t}$ also exists, finally I have $\lim_{t\to 0}\frac{f(t)+P(x,t)}{t}$ which is $f'(x).$
Now we redo everything for $f''(x)$.
First, $f'(x+y) = f'(x)+\frac{\partial P(x,y)}{\partial x}$
Let's now calculate $\frac{\partial^2P(0,y)}{\partial x^2} = \lim_{x\to 0}\frac{\frac{\partial P(x,y)}{\partial x}-\frac{\partial P(0,y)}{\partial x}}{x} = \lim_{x \to \infty}\frac{f'(x+y)-f'(x)-f'(y)+f'(0)}{x}=\lim_{x \to \infty}\frac{(f'(x+y)-f'(y))-(f'(x)-f'(0))}{x}$
We know this limit exist as $P$ is $C^2$.The first parenthesis is $f''(y)$, which is what we want, the second parenthesis is $f''(0)$. So, all is left is proving the existence of $f''(0)$.
we have $f'(y)=f'(0)+\frac{\partial P(0,y)}{\partial x}$ and $\frac{\partial P(0,0)}{\partial x}=0$
$\lim_{y\to 0}\frac{f'(y)-f'(0)}{y}=\lim_{y\to 0}\frac{\frac{\partial P(0,y)}{\partial x}}{y}=\lim_{y\to 0}\frac{\frac{\partial P(0,y)}{\partial x}-\frac{\partial P(0,0)}{\partial x}}{y} = \frac{\partial^2P(0,0)}{\partial y \partial x} = f''(0)$
