Let $X$ be a Banach space. Let $p > 1$ and, consider the functional $X \to \mathbb{C}$ given by:
$$x \mapsto \frac{1}{p}\|x\|^p$$
I would like the know if the above functional is convex. That is, for $x_1, x_2 \in X$ and $t \in (0,1)$, do we have $\frac{1}{p}\|tx_1 + (1-t)x_2\|^p \le t\frac{1}{p}\|x_1\|^p + (1-t) \frac{1}{p}\|x_2\|^p$?
I am reading through a paper that states $x \mapsto \frac{1}{p}\|x\|^p$ is a "convex functional," but I'm not sure if they mean the same convexity that I am describing above.
I have went ahead and tried to establish (my definition of) convexity on my own, using the well-known inequality:
$$|\alpha + \beta|^p \le 2^{p-1}(|\alpha|^p + |\beta|^p).$$
for $\alpha, \beta \in \mathbb{C}$. But I have gotten stuck:
$$\frac{1}{p}\|tx_1 + (1-t)x_2\|^p \le \frac{1}{p}(t\|x_1\| + (1-t)\|x_2\|)^p \le \frac{2^{p-1}}{p}(t^p\|x_1\|^p + (1-t)^p\|x_2\|^p),$$
and I'm not sure if I can do any good by proceeding from this last step.
Hints or solutions are greatly appreciated.
Let $f(s) = s^p$ for some $p>1$, then for every $s\geq 0$m we have $$f''(s) = \underbrace{p(p-1)}_{>0}\,\underbrace{s^{p-2}\vphantom{)}}_{\geq 0} \geq 0$$ and thus $f$ is convex on the non negative numbers. So for any $x_1,x_2 \in X$ and $t \in (0,1)$ we get $$\frac{1}{p}\|x_1t + (1-t)x_2 \|^p \overset{(1)}{\leq} \frac{1}{p}(t\| x_1\|+(1-t)\|x_2\|)^p \overset{(2)}{\leq} \frac{t}{p}\|x_1\|^p + \frac{1-t}{p}\|x_2\|^p,$$ where $(1)$ is obtained by the triangle inequality and homogeneity of the norm and $(2)$ by convexity of $f$.