Consider a continuous function $f\colon \mathbb R \to \mathbb R$ and $C^\alpha(A)$, the space of Hölder-continuous functions on a bounded set $A \subseteq \mathbb R$, and the induced function $F \colon C^\alpha(A) \to C^\alpha(A)$, $F(g) := f \circ g$.
Question: Can I, under any conditions or modifications which do not force $f$ to be an affine function, guarantee that $F$ will be continuous?
First, I directly tried to prove that $F$ is Lipschitz: $$ \Vert F(g) - F(h) \Vert_{C^\alpha(A)} \le L\Vert g - h\Vert_{C^\alpha(A)}, $$ which would, in particular, imply $$ \vert f(g(x)) - f(h(x)) - f(g(y)) + f(h(y))\vert \\ = \vert (F(g) - F(h))(x) - (F(g) - F(h))(y)\vert \\ \le L \Vert g - h\Vert_{C^\alpha(A)} \vert x - y\vert^\alpha,$$ but this seems to be not provable that because it seems impossible to split the sum into a product of $g - h$ and $x - y$.
This superposition operator, also known as (autonomous) Nemytskii operator, is continuous on Hölder spaces if and only if $f$ is $C^1$-smooth. This is Theorem 2 in
If you find this pre-TeX paper hard to read, there's a more general form of the same result in
Finally, there is a painfully detailed further study of this problem in section 7.4 of the book