I have a smooth function $f:\mathbb{R}^2\to\mathbb{R}$ (smooth means have derivatives of all orders).
Is it true that $f(x+y)=f(x)+f(y)$ ? (or can someone please give a counter example)
I have a smooth function $f:\mathbb{R}^2\to\mathbb{R}$ (smooth means have derivatives of all orders).
Is it true that $f(x+y)=f(x)+f(y)$ ? (or can someone please give a counter example)
Really your fact is not true. There are many counterexamples.
Nevertheless, i can tell you a fact some similar, but different.
If you have a function $f: \mathbb{R}\rightarrow \mathbb{R}$ and these function is continue and concave in [a,b] (for example). The Jansen's Theorem says that for $x,\ y \in (a,b)$ $ f\left ( \frac{x+y}{2}\right )\geq \frac{f(x)+f(y)}{2} $ So, is very interesting, can you say the same?.
And more general and useful for proofs inequalities, you have $ f\left (\sum_{i=1} ^n \frac{x_i}{n} \right )\geq \frac{1}{n} \sum_{i=1}^n f(x_i) $
Regards,
No. Try $f(p)=|p|^2$. Then $f(1,0) = 1$, but $f(2,0) = 4 \neq 2 = f(1,0) + f(1,0)$.
The condition $f(x+y) = f(x)+f(y)$ is additivity. It is absolutely not a dense condition, let alone a generic condition, among smooth functions.
Edit: Changed "linearity" $\longrightarrow$ "additivity." Thanks for the catch.
Consider $F:\mathbb{R}^2 \to\mathbb{R}$ given by $F(x,y)=y^3$. This $F$ clearly does not satisfy what you ask.