Suppose that $f,g$ are uniformly bounded by $C$ on some set $X$ with finite measure $\mu$. Show that $$\|f-g\|_p\leq C^{p-2}\|f-g\|_2.$$
By Holder inequality $$\int_X|f-g|^pd\mu\leq \|f-g\|_2\|f-g\|_\infty^{p-1}\mu(X)^{1/2},$$ but this does not give me what I need.
On
The good inequality is indeed $$\tag{*} \|f-g\|_p^p\leq (2C)^{p-2}\|f-g\|_2^2$$ [the inequality $$\|f-g\|_p\leq C^{p-2}\|f-g\|_2 $$ has a problem in term of homogeneity, that is, if we replace $f$ by $a \cdot f$ and $g$ by $a\cdot g$ for some positive $a$, then $a\cdot f$ and $a\cdot g$ are bounded by $a\cdot C$ and we should have $a\|f-g\|_p\leq a^pC^{p-2}\|f-g\|_2$, which cannot hold for all $a$. ]
In order to prove $(*)$, observe that $$\left\lvert f(x)-g(x)\right\rvert^p=\left\lvert f(x)-g(x)\right\rvert^2 \left\lvert f(x)-g(x)\right\rvert^{p-2} $$ and $$\left\lvert f(x)-g(x)\right\rvert^{p-2} \leqslant\left(2C\right)^{p-2} $$ hence $$\left\lvert f(x)-g(x)\right\rvert^p\leqslant \left\lvert f(x)-g(x)\right\rvert^2 \left(2C\right)^{p-2} $$ and the result follows by integration.
It is worth noticing that finiteness of the measure space is not needed.
I'm posting this as an answer, since it doesn't fit in a comment, and seems to be pretty darn close to what is required.
I get the following bound: \begin{align} \|f-g\|_p^p &= \| |f-g|^p \|_1 \\ &\le \| |f-g| |f-g|^{p-1} \| \\ &= \| f-g \|_2 \| |f-g|^{p-1} \|_2 \tag{by Hölder} \\ &= \|f-g\|_2 \left( \int |f-g|^{2p-2} \right)^{\frac{1}{2}} \\ &\le \|f-g\|_2 \left( \int (|f|-|g|)^{2p-2} \right)^{\frac{1}{2}} \tag{triangle inequality} \\ &\le \|f-g\|_2 \left( \int (2C)^{2p-2} \right)^{\frac{1}{2}} \\ &= (2C)^{p-1} \sqrt{\mu(X)} \|f-g\|_2 \end{align} This differs from the desired bound in three ways: (1) the exponent on $C$ is $p-1$, not $p-2$ (this could be a clerical error on my part), (2) the extra 2 in the constant, and (3) it appears to me that the inequality also depends on the measure of the space.