Is $(\mathbb{E}[[X-\mathbb{E}[X]]_+^p])^{1/p}$ convex?

We say that a risk measure $\rho(\cdot)$ is convex if $$ \rho(tX+(1-t)Y) \leq t\rho(X) + (1-t)\rho(Y) $$ for all $X,Y \in L^p$ be random varaibles and $t\in [0,1]$.

With the definition above, I wonder if the following random function is convex or not: Let $X \in L^p$, consider $$ \rho(X):=\left( \mathbb{E}\left[ \left( X-\mathbb{E}[X]\right)_+^p \right] \right)^{1/p} $$ where $(a)_+:=\max\{a,0\}$ and $p \in [1,\infty)$ .

My attempt: If $X \leq \mathbb{E}[X]$, then $\left( X-\mathbb{E}[X]\right)_+ = 0$. Hence, $\rho(X)$ is convex trivially. On the other hand, if $X > \mathbb{E}[X]$, then $\left( X-\mathbb{E}[X]\right)_+ = X-\mathbb{E}[X]$. Thus, in this case, we have $$ \rho(X) = \mathbb{E}[(X - \mathbb{E}[X])^p]^{1/p}. $$ If $X,Y$ are given and $t \in [0,1]$, then $$\begin{align*} \rho(tX+(1-t)Y) &=\mathbb{E}\left[\left( (tX+(1-t)Y) - \mathbb{E}[tX +(1-t)Y] \right)^p \right]^{1/p}\\ &\leq \|(tX+(1-t)Y) - \mathbb{E}[tX + (1-t)Y]\|_p \\ &= \|t(X - \mathbb{E}[X]) + (1-t) (Y - \mathbb{E}[Y])\|_p\\ & \leq t\|X-\mathbb{E}[X]\|_p + (1-t) \|Y-\mathbb{E}[Y]\|_p \end{align*} $$ Then I get stuck on how to achive $t\rho(X) + (1-t) \rho(Y)$ on the right hand side. Any help is appreciated.

Define $ a_- = | \min\{ a, 0 \} | $, so that for any real number $ a $, $ a = a_+ - a_- $. Define $ f(X) = (X - \mathbb{E}[X] )_+ $ and $ g(Y) = (\mathbb{E}[| Y |^p])^{1/p} $. $ f(X) $ is convex:

$\begin{align*} & f(t X + (1-t) Y) \\ &= ( t (X - \mathbb{E}[X] ) + (1-t) (Y - \mathbb{E}[Y] ) )_+ \\ &= ( t (X - \mathbb{E}[X] )_+ - t (X - \mathbb{E}[X] )_- + (1-t) (Y - \mathbb{E}[Y] ) )_+ - (1-t) (Y - \mathbb{E}[Y] ) )_- )_+ \\ &\le t (X - \mathbb{E}[X] )_+ + (1-t) (Y - \mathbb{E}[Y] ) )_+ \\ &= t f(X) + (1-t) f(Y) , \end{align*}$

where $ 0 \le t \le 1 $. Now, $ g(Y) $ is the $ L_p $ norm of $ Y $, and hence $ g(Y) $ is convex for $ p \ge 1 $. Further, if $ 0 \le Y_1 \le Y_2 $, then $ | Y_1 |^p \le | Y_2 |^p $, and hence $ g(Y_1) \le g(Y_2) $.

Note that $ \rho(X) = g(f(X)) $. Therefore, $\begin{align*} & \rho(t X + (1-t) Y) \\ &= g(f(t X + (1-t) Y)) \\ &\le g(t f(X) + (1-t) f(Y)) \;\; \text{because } 0 \le f(t X + (1-t) Y) \le t f(X) + (1-t) f(Y) \\ &\le t g(f(X)) + (1-t) g(f(Y)) = t \rho(X) + (1-t) \rho(Y) , \end{align*}$

which proves the convexity of $ \rho(\cdot) $.