# The sum of Hölder continuous functions is Hölder continuous

holder-spacesreal-analysissolution-verification

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?

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.