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The answer is in the negative.

Let $f$ and $g$ be two upper densities (in the sense of the OP), and let $\alpha \in [0,1]$ and $q \in [1,\infty[$. Then the function $$h: \mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto (\alpha (f(X))^q + (1-\alpha) (g(X))^q)^{\frac{1}{q}}$$ is an upper density too (in particular, condition (F3) follows from Minkowski's inequality, which is why we need $q \ge 1$).

Next, fix a set $X \subseteq 2\cdot\mathbf N^+$, and suppose to a contradiction that $h$ is ``weakly additive'' (that is, $f(A \cup B) = f(A) + f(B)$ for all disjoint $A, B \subseteq \mathbf N^+$ such that $B$ is an (infinite) arithmetic progression), regardless of the actual values of the parameters $\alpha$ and $q$. Then, also $f$ and $g$ are weakly additive, and using that $f(2\cdot\mathbf N^+ + 1) = g(2\cdot\mathbf N^+ + 1)=\frac{1}{2}$, we obtain $$ (\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}} + 1, $$ where $x := 2f(X)$ and $y := 2g(X)$.

On the other hand, an appropriate choice of $f$, $g$ and $X$ makes it possible to have $x$ equal to zero while $y$ takes any value in the interval $[0,1]$: This can be achieved, for instance, by letting $f$ be the upper asymptotic density (on $\mathbf N^+$), $g$ the upper Banach density, and $X$ a suitable subset of the intersection, $S$, of $\bigcup_{n \ge 1} [\![2^n, 2^n + n]\!]$ and $2 \cdot\mathbf N^+$, and by considering that (i) the upper Banach density of $S$ is $\frac{1}{2}$, (ii) the upper asymptotic and upper Banach densities are upper densities, and (iii) upper densities have the strong, and hence the weak, Darboux property (by the main theorem here).

Accordingly, we should have $$(\alpha + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (1-\alpha)^{\frac{1}{q}}y + 1$$ for all $\alpha, y \in [0,1]$ and $q \in [1,\infty[$, which, however, is blatantly false. []

Added later. If you assume $\alpha = \frac{1}{2}$ and $q = 2$ in the last displayed equation, you don't even need to know that the upper Banach density has the weak Darboux property, since then you end up with the equation $$\sqrt{1 + (1+y)^2} = y + \sqrt{2},$$ which has a unique solution for $y \in \bf R$ (namely, $y = 0$).

The answer is in the negative.

Let $f$ and $g$ be two upper densities (in the sense of the OP), and let $\alpha \in [0,1]$ and $q \in [1,\infty[$. Then the function $$h := (\alpha f^q + (1-\alpha) g^q)^{\frac{1}{q}}$$ is an upper density too (in particular, condition (F3) follows from Minkowski's inequality, which is why we need $q \ge 1$).

Next, fix a set $X \subseteq 2\cdot\mathbf N^+$, let $x := 2f(X)$, $y := 2g(X)$ and $Y := 2 \cdot \mathbf N^+ + 1$, and suppose to a contradiction that $h$ is ``weakly additive'' (that is, $h(A \cup B) = h(A) + h(B)$ for all disjoint $A, B \subseteq \mathbf N^+$ such that $B$ is an (infinite) arithmetic progression), regardless of the actual values of the parameters $\alpha$ and $q$. Then, also $f$ and $g$ are weakly additive, and using that $f(Y) = g(Y)=\frac{1}{2}$, we obtain $$ \begin{split} 2h(X \cup Y) & = 2(\alpha (f(X \cup Y))^q + (1-\alpha) (g(X \cup Y))^q)^{\frac{1}{q}} \\ & = 2(\alpha (f(X) + f(Y))^q + (1-\alpha) (g(X) + g(Y))^q)^{\frac{1}{q}} \\ & = (\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}} \end{split} $$ and $$ \begin{split} h(X) + h(Y) & = (\alpha (f(X))^q + (1-\alpha) (g(X))^q)^{\frac{1}{q}} + \frac{1}{2} \\ & = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}}+ \frac{1}{2} \end{split} $$ which, together with $h(X \cup Y) = h(X) + h(Y)$, yields $$ (\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}} + 1. $$ On the other hand, an appropriate choice of $f$, $g$ and $X$ makes it possible to have $x$ equal to zero while $y$ takes any prescribed value in the interval $[0,1]$: This can be achieved, for instance, by letting $f$ be the upper asymptotic density (on $\mathbf N^+$), $g$ the upper Banach density, and $X$ a suitable subset of the intersection, $S$, of $\bigcup_{n \ge 1} [\![2^n, 2^n + n]\!]$ and $2 \cdot\mathbf N^+$, and by considering that (i) the upper asymptotic density of $S$ is $0$, (ii) the upper Banach density of $S$ is $\frac{1}{2}$, (iii) the upper asymptotic and upper Banach densities are upper densities, and (iv) upper densities have the strong, and hence the weak, Darboux property (by the main theorem here).

Accordingly, we should have $$(\alpha + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (1-\alpha)^{\frac{1}{q}}y + 1$$ for all $\alpha, y \in [0,1]$ and $q \in [1,\infty[$, which, however, is blatantly false. []

Added later. If you assume $\alpha = \frac{1}{2}$ and $q = 2$ in the last displayed equation, you don't even need to know that the upper Banach density has the weak Darboux property, since then you end up with the equation $$\sqrt{1 + (y+1)^2} = y + \sqrt{2},$$ which has a unique solution for $y \in \bf R$ (namely, $y = 0$).

The answer is in the negative.

Let $f$ and $g$ be two upper densities (in the sense of the OP), and let $\alpha \in [0,1]$ and $q \in [1,\infty[$. Then the function $$h: \mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto (\alpha (f(X))^q + (1-\alpha) (g(X))^q)^{\frac{1}{q}}$$ is an upper density too (this is a special case of Proposition 11 here, where, however, there is a minor mistake with the range of the parameter $q$).

Next, fix a set $X \subseteq 2\cdot\mathbf N^+$, and suppose to a contradiction that $h$ is additive, regardless of the actual values of the parameters $f$, $g$, $\alpha$ and $q$.

  Then, using that $f(2\cdot\mathbf N^+ + 1) = g(2\cdot\mathbf N^+ + 1)=\frac{1}{2}$, we see that the additivity of $h$ implies $$ (\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}} + 1, $$ where $x := 2f(X)$ and $y := 2g(X)$.

But an appropriate choice of $f$, $g$ and $X$ makes it possible to have $x$ equal to zero while $y$ takes any value in the interval $[0,1]$: This can be achieved, for instance, by letting $f$ be the upper asymptotic density (on $\mathbf N^+$), $g$ the upper Banach density, and $X$ a suitable subset of the intersection, $S$, of $\bigcup_{n \ge 1} [\![2^n, 2^n + n]\!]$ and $2 \cdot\mathbf N^+$, and by considering that (i) the upper Banach density of $S$ is $\frac{1}{2}$, (ii) the upper asymptotic and upper Banach densities are upper densities, and (iii) upper densities have the strong, and hence the weak, Darboux property (by the main theorem here).

Accordingly, we should have $$(\alpha + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (1-\alpha)^{\frac{1}{q}}y + 1$$ for all $\alpha, y \in [0,1]$ and $q \in [1,\infty[$, which, however, is blatantly false. []

Added later. If you assume $\alpha = \frac{1}{2}$ and $q = 2$ in the last displayed equation, you don't even need to know that the upper Banach density has the weak Darboux property, since then you end up with the equation $$\sqrt{1 + (1+y)^2} = y + \sqrt{2},$$ which has a unique solution for $y \in \bf R$ (namely, $y = 0$).

The answer is in the negative.

Let $f$ and $g$ be two upper densities (in the sense of the OP), and let $\alpha \in [0,1]$ and $q \in [1,\infty[$. Then the function $$h: \mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto (\alpha (f(X))^q + (1-\alpha) (g(X))^q)^{\frac{1}{q}}$$ is an upper density too (in particular, condition (F3) follows from Minkowski's inequality, which is why we need $q \ge 1$).

Next, fix a set $X \subseteq 2\cdot\mathbf N^+$, and suppose to a contradiction that $h$ is ``weakly additive'' (that is, $f(A \cup B) = f(A) + f(B)$ for all disjoint $A, B \subseteq \mathbf N^+$ such that $B$ is an (infinite) arithmetic progression), regardless of the actual values of the parameters $\alpha$ and $q$. Then, also $f$ and $g$ are weakly additive, and using that $f(2\cdot\mathbf N^+ + 1) = g(2\cdot\mathbf N^+ + 1)=\frac{1}{2}$, we obtain $$ (\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}} + 1, $$ where $x := 2f(X)$ and $y := 2g(X)$.

On the other hand, an appropriate choice of $f$, $g$ and $X$ makes it possible to have $x$ equal to zero while $y$ takes any value in the interval $[0,1]$: This can be achieved, for instance, by letting $f$ be the upper asymptotic density (on $\mathbf N^+$), $g$ the upper Banach density, and $X$ a suitable subset of the intersection, $S$, of $\bigcup_{n \ge 1} [\![2^n, 2^n + n]\!]$ and $2 \cdot\mathbf N^+$, and by considering that (i) the upper Banach density of $S$ is $\frac{1}{2}$, (ii) the upper asymptotic and upper Banach densities are upper densities, and (iii) upper densities have the strong, and hence the weak, Darboux property (by the main theorem here).

Accordingly, we should have $$(\alpha + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (1-\alpha)^{\frac{1}{q}}y + 1$$ for all $\alpha, y \in [0,1]$ and $q \in [1,\infty[$, which, however, is blatantly false. []

Added later. If you assume $\alpha = \frac{1}{2}$ and $q = 2$ in the last displayed equation, you don't even need to know that the upper Banach density has the weak Darboux property, since then you end up with the equation $$\sqrt{1 + (1+y)^2} = y + \sqrt{2},$$ which has a unique solution for $y \in \bf R$ (namely, $y = 0$).

The answer is in the negative.

Let $f$ and $g$ be two upper densities (in the sense of the OP), and let $\alpha \in [0,1]$ and $q \in {]0,1]}$. Then the function $$h: \mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto (\alpha (f(X))^q + (1-\alpha) (g(X))^q)^{\frac{1}{q}}$$ is an upper density too (as elementary as it may be, this is a special case of Proposition 11 here).

Next, fix a set $X \subseteq 2\cdot\mathbf N^+$, and suppose to a contradiction that $h$ is additive, regardless of the actual values of the parameters $f$, $g$, $\alpha$ and $q$.

Then, using that $f(2\cdot\mathbf N^+ + 1) = g(2\cdot\mathbf N^+ + 1)=\frac{1}{2}$, we see that the additivity of $h$ implies $$ (\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}} + 1, $$ where $x := 2f(X)$ and $y := 2g(X)$.

But an appropriate choice of $f$, $g$ and $X$ makes it possible to have $x$ equal to zero while $y$ takes any value in the interval $[0,1]$: This can be achieved, for instance, by letting $f$ be the upper asymptotic density (on $\mathbf N^+$), $g$ the upper Banach density, and $X$ a suitable subset of the intersection, $S$, of $\bigcup_{n \ge 1} [\![2^n, 2^n + n]\!]$ and $2 \cdot\mathbf N^+$, and by considering that (i) the upper Banach density of $S$ is $\frac{1}{2}$, (ii) the upper asymptotic and upper Banach densities are upper densities, and (iii) upper densities have the strong, and hence the weak, Darboux property (by the main theorem here).

Accordingly, we should have $$(\alpha + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (1-\alpha)^{\frac{1}{q}}y + 1$$ for all $\alpha, y \in [0,1]$ and $q \in {]0,1]}$, which, however, is blatantly false. []

The answer is in the negative.

Let $f$ and $g$ be two upper densities (in the sense of the OP), and let $\alpha \in [0,1]$ and $q \in [1,\infty[$. Then the function $$h: \mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto (\alpha (f(X))^q + (1-\alpha) (g(X))^q)^{\frac{1}{q}}$$ is an upper density too (this is a special case of Proposition 11 here, where, however, there is a minor mistake with the range of the parameter $q$).

Next, fix a set $X \subseteq 2\cdot\mathbf N^+$, and suppose to a contradiction that $h$ is additive, regardless of the actual values of the parameters $f$, $g$, $\alpha$ and $q$.

Then, using that $f(2\cdot\mathbf N^+ + 1) = g(2\cdot\mathbf N^+ + 1)=\frac{1}{2}$, we see that the additivity of $h$ implies $$ (\alpha (x+1)^q + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (\alpha x^q + (1-\alpha)y^q)^{\frac{1}{q}} + 1, $$ where $x := 2f(X)$ and $y := 2g(X)$.

But an appropriate choice of $f$, $g$ and $X$ makes it possible to have $x$ equal to zero while $y$ takes any value in the interval $[0,1]$: This can be achieved, for instance, by letting $f$ be the upper asymptotic density (on $\mathbf N^+$), $g$ the upper Banach density, and $X$ a suitable subset of the intersection, $S$, of $\bigcup_{n \ge 1} [\![2^n, 2^n + n]\!]$ and $2 \cdot\mathbf N^+$, and by considering that (i) the upper Banach density of $S$ is $\frac{1}{2}$, (ii) the upper asymptotic and upper Banach densities are upper densities, and (iii) upper densities have the strong, and hence the weak, Darboux property (by the main theorem here).

Accordingly, we should have $$(\alpha + (1-\alpha)(y+1)^q)^{\frac{1}{q}} = (1-\alpha)^{\frac{1}{q}}y + 1$$ for all $\alpha, y \in [0,1]$ and $q \in [1,\infty[$, which, however, is blatantly false. []

Added later. If you assume $\alpha = \frac{1}{2}$ and $q = 2$ in the last displayed equation, you don't even need to know that the upper Banach density has the weak Darboux property, since then you end up with the equation $$\sqrt{1 + (1+y)^2} = y + \sqrt{2},$$ which has a unique solution for $y \in \bf R$ (namely, $y = 0$).