How Do I Check Convexity Using The Actual Definition? Suppose $f:\mathbb{R}^{+}\rightarrow\mathbb{R}$ is defined as follows:
\begin{eqnarray}
f(u)=\text{sgn}(\rho)\left(u^{\rho+1}-1\right),~u\geq 0,
\end{eqnarray}
where $\rho\in (-1,\infty)$, $\rho\neq 0$, is a fixed parameter. I need to show that $f$, as defined above, is convex on its domain. It is true that
\begin{eqnarray}
f''(u)=\rho(\rho+1)\text{sgn}(\rho)u^{\rho}\geq 0
\end{eqnarray}
for all values of $u\geq 0$ since $\rho>-1$ and $\rho(\text{sgn}(\rho))\geq 0$. This could serve as a means to show that $f$ is convex, but I would like to show the same from first principles, i.e., I would like to show:
Given any $u,v\in \mathbb{R}^{+}$ and any $\alpha \in [0,1]$, we have
\begin{eqnarray}
f(\alpha u+(1-\alpha)v)\leq \alpha f(u)+(1-\alpha)f(v).
\end{eqnarray}
In other words, I am looking for a proof of convexity of the function $g(x)=x^{\rho},~x\geq 0, ~\rho\geq 1$, using the above definition.
Can someone please guide me on how to do this? Any useful inequalities that I may need?
 A: I happened to find the solution to the question I asked during a discussion with my professor yesterday. Turns out, the inequality that I am after is a special case of the Holder's inequality. I am posting the solution here for the benefit of the community at large.
The following is the proof of convexity of the function $f(x)=x^{\rho},~\rho>1,~x\geq 0$. Suppose $\Omega=\{1,2\}$. Let $X$ and $Y$ be two real-valued random variables defined on $\Omega$ as follows:
\begin{eqnarray}
X(1)=u,~X(2)=v,~ u,v\in \mathbb{R}_{+}\\
Y(1)=1,~Y(2)=1.
\end{eqnarray}
Let $\mathbb{P}$ be a probability measure defined on the $\sigma$-algebra of the subsets of $\Omega$ such that $\mathbb{P}(1)=\alpha$ and $\mathbb{P}(2)=1-\alpha$, for some $0\leq \alpha\leq 1$. Then, applying Holder's inequality to $L^{\rho}(\mathbb{P})$, we have
\begin{eqnarray}
\int_{\Omega}XYd\mathbb{P} \leq \left(\int_{\Omega} X^{\rho}d\mathbb{P}\right)^{1/\rho} \left(\int_{\Omega} Y^{\frac{\rho}{\rho-1}}d\mathbb{P}\right)^{\frac{\rho-1}{\rho}}\\
\Rightarrow \left(\alpha u+(1-\alpha) v\right) \leq \left(\alpha u^{\rho}+(1-\alpha) v^{\rho}\right)^{1/\rho}\\
\Rightarrow \left(\alpha u+(1-\alpha) v\right)^{\rho} \leq \left(\alpha u^{\rho}+(1-\alpha) v^{\rho}\right),
\end{eqnarray}
which was the inequality desired.
