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The supremum you are looking for is $\frac{-\ln(3/4)}{\ln(4)}$. I leave my proof below, although as user479223 mentions, this is the same as showing Hölder continuity of the Weierstrass function, which is well known (see this MSE answer).

Note that for all $x,y\in\mathbb{R}$, $|g(x)-g(y)|\leq\min(|x-y|,1)$. Thus, for all $x,h\in\mathbb{R}$,

$$|f(x+h)-f(x)|=\sum_n\frac{3^ng(4^n(x+h))}{4^n}-\frac{3^ng(4^nx)}{4^n}\leq \sum_n\frac{3^n}{4^n}\min(|4^nh|,1)=$$ $$=\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right).$$

Note that when $h$ is close to $0$, the sequence $\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)$ first increases exponentially and then decreases exponentially as follows:

$$h,3h,9h,27h,\dots,3^{k_1}h,\left(\frac{3}{4}\right)^{k_2},\left(\frac{3}{4}\right)^{k_2+1},\left(\frac{3}{4}\right)^{k_2+2},\dots$$ where both $3^{k_1}h$ and $\left(\frac{3}{4}\right)^{k_2}$ are, within a factor of ten, equal to $h^{\frac{-\ln(3/4)}{\ln(4)}}$. Thus,

$$|f(x+h)-f(x)|\leq\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)\leq100h^{\frac{-\ln(3/4)}{\ln(4)}}.$$

This means that any constant $\alpha<\frac{-\ln(3/4)}{\ln(4)}$ should make the limit of the question be $0$ for all $x$. However, if $\alpha=\frac{-\ln(3/4)}{\ln(4)}$ and my computations below are correct, then the limit $\lim\limits_{h\to0}\frac{f(h)}{h^\alpha}$ need not exist. E.g. taking $h_m=1/4^m$; then we have

$$f(h_m)=\sum_{n=1}^m\frac{3^n}{4^n}g(4^{n-m}) =\sum_{n=1}^m\frac{3^n}{4^n}4^{n-m}=\sum_{n=1}^m\frac{3^n}{4^m}.$$ Thus, $$ \lim_m\frac{f(h_m)}{h_m^\alpha}=\lim_m\frac{\left(\frac{3^m\cdot\frac{3}{2}}{4^m}\right)}{(3/4)^m}=\frac{3}{2}. $$

However, $$f(2h_m)=\sum_{n=1}^{m-1}\frac{3^n}{4^n}g(2\cdot4^{n-m})=2\sum_{n=1}^{m-1}\frac{3^n}{4^m}.$$ Thus, $$ \lim_m\frac{f(2h_m)}{(2h_m)^\alpha}=\lim_m\frac{\left(\frac{3^{m-1}\cdot\frac{3}{2}}{4^m}\right)} {2^\alpha(3/4)^m}=\frac{1}{2\cdot2^\alpha}=\frac{\sqrt{3}}{4}. $$

The supremum you are looking for is $-\frac{\ln(3/4)}{\ln(4)}$. I leave my proof below, although as user479223 mentions, this is the same as showing Hölder continuity of the Weierstrass function, which is well known (see this MSE answer).

Note that for all $x,y\in\mathbb{R}$, $|g(x)-g(y)|\leq\min(|x-y|,1)$. Thus, for all $x,h\in\mathbb{R}$, $$ \begin{split} |f(x+h)-f(x)|&=\sum_n\frac{3^ng(4^n(x+h))}{4^n}-\frac{3^ng(4^nx)}{4^n} \\ &\leq \sum_n\frac{3^n}{4^n}\min(|4^nh|,1)\\ &=\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right). \end{split} $$ Note that when $h$ is close to $0$, the sequence $\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)$ first increases exponentially and then decreases exponentially as follows:

$$h,3h,9h,27h,\dots,3^{k_1}h,\left(\frac{3}{4}\right)^{k_2},\left(\frac{3}{4}\right)^{k_2+1},\left(\frac{3}{4}\right)^{k_2+2},\dots$$ where both $3^{k_1}h$ and $\left(\frac{3}{4}\right)^{k_2}$ are, within a factor of ten, equal to $h^{\frac{-\ln(3/4)}{\ln(4)}}$. Thus,

$$|f(x+h)-f(x)|\leq\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)\leq100h^{\frac{-\ln(3/4)}{\ln(4)}}.$$

This means that any constant $\alpha<\frac{-\ln(3/4)}{\ln(4)}$ should make the limit of the question be $0$ for all $x$. However, if $\alpha=\frac{-\ln(3/4)}{\ln(4)}$ and my computations below are correct, then the limit $\lim\limits_{h\to0}\frac{f(h)}{h^\alpha}$ need not exist. E.g. taking $h_m=1/4^m$; then we have

$$f(h_m)=\sum_{n=1}^m\frac{3^n}{4^n}g(4^{n-m}) =\sum_{n=1}^m\frac{3^n}{4^n}4^{n-m}=\sum_{n=1}^m\frac{3^n}{4^m}.$$ Thus, $$ \lim_m\frac{f(h_m)}{h_m^\alpha}=\lim_m\frac{\left(\frac{3^m\cdot\frac{3}{2}}{4^m}\right)}{(3/4)^m}=\frac{3}{2}. $$

However, $$f(2h_m)=\sum_{n=1}^{m-1}\frac{3^n}{4^n}g(2\cdot4^{n-m})=2\sum_{n=1}^{m-1}\frac{3^n}{4^m}.$$ Thus, $$ \lim_m\frac{f(2h_m)}{(2h_m)^\alpha}=\lim_m\frac{\left(\frac{3^{m-1}\cdot\frac{3}{2}}{4^m}\right)} {2^\alpha(3/4)^m}=\frac{1}{2\cdot2^\alpha}=\frac{\sqrt{3}}{4}. $$

The supremum you are looking for is $\frac{-\ln(3/4)}{\ln(4)}$.

Note that for all $x,y\in\mathbb{R}$, $|g(x)-g(y)|\leq\min(|x-y|,1)$. Thus, for all $x,h\in\mathbb{R}$,

$$|f(x+h)-f(x)|=\sum_n\frac{3^ng(4^n(x+h))}{4^n}-\frac{3^ng(4^nx)}{4^n}\leq \sum_n\frac{3^n}{4^n}\min(|4^nh|,1)=$$ $$=\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right).$$

Note that when $h$ is close to $0$, the sequence $\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)$ first increases exponentially and then decreases exponentially as follows:

$$h,3h,9h,27h,\dots,3^{k_1}h,\left(\frac{3}{4}\right)^{k_2},\left(\frac{3}{4}\right)^{k_2+1},\left(\frac{3}{4}\right)^{k_2+2},\dots$$ where both $3^{k_1}h$ and $\left(\frac{3}{4}\right)^{k_2}$ are, within a factor of ten, equal to $h^{\frac{-\ln(3/4)}{\ln(4)}}$. Thus,

$$|f(x+h)-f(x)|\leq\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)\leq100h^{\frac{-\ln(3/4)}{\ln(4)}}.$$

This means that any constant $\alpha<\frac{-\ln(3/4)}{\ln(4)}$ should make the limit of the question be $0$ for all $x$. However, if $\alpha=\frac{-\ln(3/4)}{\ln(4)}$ and my computations below are correct, then the limit $\lim\limits_{h\to0}\frac{f(h)}{h^\alpha}$ need not exist. E.g. taking $h_m=1/4^m$; then we have

$$f(h_m)=\sum_{n=1}^m\frac{3^n}{4^n}g(4^{n-m}) =\sum_{n=1}^m\frac{3^n}{4^n}4^{n-m}=\sum_{n=1}^m\frac{3^n}{4^m}.$$ Thus, $$ \lim_m\frac{f(h_m)}{h_m^\alpha}=\lim_m\frac{\left(\frac{3^m\cdot\frac{3}{2}}{4^m}\right)}{(3/4)^m}=\frac{3}{2}. $$

However, $$f(2h_m)=\sum_{n=1}^{m-1}\frac{3^n}{4^n}g(2\cdot4^{n-m})=2\sum_{n=1}^{m-1}\frac{3^n}{4^m}.$$ Thus, $$ \lim_m\frac{f(2h_m)}{(2h_m)^\alpha}=\lim_m\frac{\left(\frac{3^{m-1}\cdot\frac{3}{2}}{4^m}\right)} {2^\alpha(3/4)^m}=\frac{1}{2\cdot2^\alpha}=\frac{\sqrt{3}}{4}. $$

The supremum you are looking for is $\frac{-\ln(3/4)}{\ln(4)}$. I leave my proof below, although as user479223 mentions, this is the same as showing Hölder continuity of the Weierstrass function, which is well known (see this MSE answer).

Note that for all $x,y\in\mathbb{R}$, $|g(x)-g(y)|\leq\min(|x-y|,1)$. Thus, for all $x,h\in\mathbb{R}$,

$$|f(x+h)-f(x)|=\sum_n\frac{3^ng(4^n(x+h))}{4^n}-\frac{3^ng(4^nx)}{4^n}\leq \sum_n\frac{3^n}{4^n}\min(|4^nh|,1)=$$ $$=\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right).$$

Note that when $h$ is close to $0$, the sequence $\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)$ first increases exponentially and then decreases exponentially as follows:

$$h,3h,9h,27h,\dots,3^{k_1}h,\left(\frac{3}{4}\right)^{k_2},\left(\frac{3}{4}\right)^{k_2+1},\left(\frac{3}{4}\right)^{k_2+2},\dots$$ where both $3^{k_1}h$ and $\left(\frac{3}{4}\right)^{k_2}$ are, within a factor of ten, equal to $h^{\frac{-\ln(3/4)}{\ln(4)}}$. Thus,

$$|f(x+h)-f(x)|\leq\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)\leq100h^{\frac{-\ln(3/4)}{\ln(4)}}.$$

This means that any constant $\alpha<\frac{-\ln(3/4)}{\ln(4)}$ should make the limit of the question be $0$ for all $x$. However, if $\alpha=\frac{-\ln(3/4)}{\ln(4)}$ and my computations below are correct, then the limit $\lim\limits_{h\to0}\frac{f(h)}{h^\alpha}$ need not exist. E.g. taking $h_m=1/4^m$; then we have

$$f(h_m)=\sum_{n=1}^m\frac{3^n}{4^n}g(4^{n-m}) =\sum_{n=1}^m\frac{3^n}{4^n}4^{n-m}=\sum_{n=1}^m\frac{3^n}{4^m}.$$ Thus, $$ \lim_m\frac{f(h_m)}{h_m^\alpha}=\lim_m\frac{\left(\frac{3^m\cdot\frac{3}{2}}{4^m}\right)}{(3/4)^m}=\frac{3}{2}. $$

However, $$f(2h_m)=\sum_{n=1}^{m-1}\frac{3^n}{4^n}g(2\cdot4^{n-m})=2\sum_{n=1}^{m-1}\frac{3^n}{4^m}.$$ Thus, $$ \lim_m\frac{f(2h_m)}{(2h_m)^\alpha}=\lim_m\frac{\left(\frac{3^{m-1}\cdot\frac{3}{2}}{4^m}\right)} {2^\alpha(3/4)^m}=\frac{1}{2\cdot2^\alpha}=\frac{\sqrt{3}}{4}. $$

The supremum you are looking for is $\frac{-\ln(3/4)}{\ln(4)}$.

Note that for all $x,y\in\mathbb{R}$, $|g(x)-g(y)|\leq\min(|x-y|,1)$. Thus, for all $x,h\in\mathbb{R}$,

$$|f(x+h)-f(x)|=\sum_n\frac{3^ng(4^n(x+h))}{4^n}-\frac{3^ng(4^nx)}{4^n}\leq \sum_n\frac{3^n}{4^n}\min(|4^nh|,1)=$$ $$=\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right).$$

Note that when $h$ is close to $0$, the sequence $\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)$ first increases exponentially and then decreases exponentially as follows:

$$h,3h,9h,27h,\dots,3^{k_1}h,\left(\frac{3}{4}\right)^{k_2},\left(\frac{3}{4}\right)^{k_2+1},\left(\frac{3}{4}\right)^{k_2+2},\dots$$ where both $3^{k_1}h$ and $\left(\frac{3}{4}\right)^{k_2}$ are, within a factor of ten, equal to $h^{\frac{-\ln(3/4)}{\ln(4)}}$. Thus,

$$|f(x+h)-f(x)|\leq\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)\leq100h^{\frac{-\ln(3/4)}{\ln(4)}}.$$

This means that any constant $\alpha<\frac{-\ln(3/4)}{\ln(4)}$ should make the limit of the question be $0$ for all $x$. However, if $\alpha\geq\frac{-\ln(3/4)}{\ln(4)}$, then we can choose a sequence $(h_m)_m\to0$ such that $\frac{|f(h_m)|}{h_m^\alpha}\not\to0$; e.g. take $h_m=1/4^m$; then we have

$$f(h_m)=\sum_{n=1}^m\frac{3^n}{4^n}g(4^{n-m})\geq\frac{3^m}{4^m}g(1)=\frac{3^m}{4^m}=h_m^{\frac{-\ln(3/4)}{\ln(4)}}.$$

The supremum you are looking for is $\frac{-\ln(3/4)}{\ln(4)}$.

Note that for all $x,y\in\mathbb{R}$, $|g(x)-g(y)|\leq\min(|x-y|,1)$. Thus, for all $x,h\in\mathbb{R}$,

$$|f(x+h)-f(x)|=\sum_n\frac{3^ng(4^n(x+h))}{4^n}-\frac{3^ng(4^nx)}{4^n}\leq \sum_n\frac{3^n}{4^n}\min(|4^nh|,1)=$$ $$=\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right).$$

Note that when $h$ is close to $0$, the sequence $\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)$ first increases exponentially and then decreases exponentially as follows:

$$h,3h,9h,27h,\dots,3^{k_1}h,\left(\frac{3}{4}\right)^{k_2},\left(\frac{3}{4}\right)^{k_2+1},\left(\frac{3}{4}\right)^{k_2+2},\dots$$ where both $3^{k_1}h$ and $\left(\frac{3}{4}\right)^{k_2}$ are, within a factor of ten, equal to $h^{\frac{-\ln(3/4)}{\ln(4)}}$. Thus,

$$|f(x+h)-f(x)|\leq\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)\leq100h^{\frac{-\ln(3/4)}{\ln(4)}}.$$

This means that any constant $\alpha<\frac{-\ln(3/4)}{\ln(4)}$ should make the limit of the question be $0$ for all $x$. However, if $\alpha=\frac{-\ln(3/4)}{\ln(4)}$ and my computations below are correct, then the limit $\lim\limits_{h\to0}\frac{f(h)}{h^\alpha}$ need not exist. E.g. taking $h_m=1/4^m$; then we have

$$f(h_m)=\sum_{n=1}^m\frac{3^n}{4^n}g(4^{n-m}) =\sum_{n=1}^m\frac{3^n}{4^n}4^{n-m}=\sum_{n=1}^m\frac{3^n}{4^m}.$$ Thus, $$ \lim_m\frac{f(h_m)}{h_m^\alpha}=\lim_m\frac{\left(\frac{3^m\cdot\frac{3}{2}}{4^m}\right)}{(3/4)^m}=\frac{3}{2}. $$

However, $$f(2h_m)=\sum_{n=1}^{m-1}\frac{3^n}{4^n}g(2\cdot4^{n-m})=2\sum_{n=1}^{m-1}\frac{3^n}{4^m}.$$ Thus, $$ \lim_m\frac{f(2h_m)}{(2h_m)^\alpha}=\lim_m\frac{\left(\frac{3^{m-1}\cdot\frac{3}{2}}{4^m}\right)} {2^\alpha(3/4)^m}=\frac{1}{2\cdot2^\alpha}=\frac{\sqrt{3}}{4}. $$