Let we have the following function $$ψ:R^{*+} \to R$$ $$x\to ψ(x)=x^3$$ How can I prove that $ψ$ is a convex function by using the definition ?
I meant that I have to prove that $$ψ(λx+(1-λ)y) \le λψ(x)+ (1-λ)ψ(y)$$ where $λ \in [0,1] $ $x$ and $y$ belong to $R^{*+}$
Take $a,b$ positive reals. Then for $a \ge b$, we have $a^2+ab \ge 2b^2$, that is, $a^2 + ab+b^2 \ge 3b^2$. Multiplying by $(a-b)$, we get $a^3 - b^3 \ge 3b^2(a-b)$.
For $a\le b$, we have $a^2+ab \le 2b^2$, that is, $a^2+ab+b^2 \le 3b^2$. Here $(a-b) \le 0$, so multiplying by $(a-b)$, we have $a^3-b^3 \ge 3b^2(a-b)$.
Thus for any positive reals $a,b$, we have $a^3-b^3 \ge 3b^2(a-b)$, that is, $a^3 \ge b^3 + 3b^2(a-b). \tag{1}$
Taking $a = x, b = \lambda x + (1- \lambda)y$ in the above inequality, we get $$ x^3 \ge (\lambda x + (1-\lambda)y)^3 + 3b^2(1-\lambda)(x-y) \tag{2}$$
Taking $a = y, b = \lambda x + (1- \lambda)y$ in $(1)$, we get $$ y^3 \ge (\lambda x + (1-\lambda)y)^3 + 3b^2(-\lambda)(x-y) \tag{3}$$
Now, $\lambda \times (2) + (1-\lambda) \times (3)$ gives $$\lambda x^3 + (1-\lambda)y^3 \ge (\lambda x + (1-\lambda)y)^3 $$
That proves convexity the way you want.
Note though that everything we did above follows from Taylor's theorem and non-negativity of the second derivative and is in fact a proof in disguise of the fact that a non-negative second derivative implies convexity. You are better off just using that.
Note that, if $f(x) = x^3$, then $f''(x) =6x $.
Therefore $f(x)$ is concave (down) for $x < 0$ and convex (up) for $x > 0$.
At $x = 0$ it is neither.
Applying the linear definition is harder.
Hint: since $\psi$ is continuous, it is sufficient to prove that it is midpoint convex since Midpoint-Convex and Continuous Implies Convex.
Midpoint convexity on $\mathbb{R}^{+}$ amounts to proving that for $\forall a,b \ge 0\,$:
$$ \begin{align} \left(\frac{a+b}{2}\right)^3 \le \frac{a^3+b^3}{2} \quad & \iff \quad a^3+3a^2b+3ab^2+b^3 \le 4a^3+4b^3 \\ & \iff \quad 0 \le 3(a^3-a^2b-ab^2+b^3) \\ & \iff \quad 0 \le 3(a+b)(a-b)^2 \end{align} $$
