I want to show that the sum of two Hölder continuous functions is Hölder continuous. Definition: $f:[0, 1] \to \mathbb R$ is Hölder continuous if there exist $\alpha,M>0$, s.t. $\forall x,y\in[0,1]$ $|f(x)-f(y)| \le M|x-y|^\alpha$.

Suppose $f,g$ are Hölder continuous. Thus there exist $M,N,\alpha,\beta>0$ s.t. $\forall x,y \in [0,1]$:

$|f(x)-f(y)| \le M|x-y|^\alpha$, and $|g(x) – g(y)| \le N|x-y|^\beta$, thus:

\begin{align}

|f(x) +g(x) -f(y)-g(y)| &\le |f(x)-f(y)|+|g(x)-g(y)| \\

&\le M|x-y|^\alpha+ N|x-y|^\beta=*.

\end{align}

Now let $A = \max(M,N)$ and $\sigma=\max(\alpha, \beta)$. Then we get:

$* \le A|x-y|^\sigma$, hence $f+g$ is Hölder continuous.

My intuition says I did something wrong can someone tell me what went wrong or if it went good?

## Best Answer

It would be better to take $\sigma=\min(\alpha,\beta)$ because you would get $N\lvert x-y\rvert^\alpha=N\lvert x-y\rvert^{\alpha-\sigma+\sigma}\leqslant N\lvert x-y\rvert^\sigma$ (as $\alpha-\sigma\geqslant 0$) and similarly, $M\lvert x-y\rvert^\beta\leqslant M\lvert x-y\rvert^\sigma$. Taking $A=2\max(M,N)$ gives the inequality you want.

Intuitively, the closer the Hölder exponent to one is, the more regularity you have and for a sum, in general (unless for example $f=g$), you cannot hope for a better exponent than the minimum of the exponents of the two functions.