I am struggling with this question:
Show that if $f(x)$ is convex then the function $ yf(x/y)$ is convex on $\{(x, y): y>0\}$.
I have tried starting from the standard definition of convexity but it just leads to a lot of algebra that doesn't go anywhere.
Define $g(x,y)=yf(x/y)$. Then, $$g(\alpha x_1+(1-\alpha)x_2,\alpha y_1+(1-\alpha)y_2)=(\alpha y_1+(1-\alpha)y_2)f\left(\frac{\alpha x_1+(1-\alpha)x_2}{\alpha y_1+(1-\alpha)y_2}\right)\\ \stackrel{\le}{\tiny{\mbox{by convexity of $f$}}} (\alpha y_1+(1-\alpha)y_2)\left[\frac{\alpha y_1}{(\alpha y_1+(1-\alpha)y_2)}f(x_1/y_1)+\frac{(1-\alpha) y_2}{(\alpha y_1+(1-\alpha)y_2)}f(x_2/y_2)\right]\\=\alpha g(x_1,y_1)+(1-\alpha)g(x_2,y_2)$$ Note that this where we require $y>0$.
Denote the set $\{(x, y): y > 0\}$ by $S$, and let $g: S \to \mathbb{R}^1$ be the function that sends $p = (x, y) \in S$ to $yf(x/y)$.
For every $\lambda \in (0, 1)$, and $p_1 = (x_1, y_1) \in S, p_2 = (x_2, y_2) \in S$, we have \begin{align*} & g(\lambda p_1 + (1 - \lambda)p_2) \\ = & g((\lambda x_1 + (1 - \lambda)x_2, \lambda y_1 + (1 - \lambda)y_2) \\ = & (\lambda y_1 + (1 - \lambda)y_2)f((\lambda x_1 + (1 - \lambda)x_2)/(\lambda y_1 + (1 - \lambda)y_2)) \\ = & (\lambda y_1 + (1 - \lambda)y_2)f((\lambda {\color{red}{y_1}} (x_1/{\color{red}{y_1}}) + (1 - \lambda){\color{red}{y_2}}(x_2/{\color{red}{y_2}}))/(\lambda y_1 + (1 - \lambda)y_2)) \\ \leq & (\lambda y_1 + (1 - \lambda)y_2)\left[\frac{\lambda y_1}{\lambda y_1 + (1 - \lambda)y_2}f(x_1/y_1) + \frac{(1 - \lambda) y_2}{\lambda y_1 + (1 - \lambda)y_2}f(x_2/y_2) \right]\quad \text{ by convexity of } f.\\ = & \lambda g(p_1) + (1 - \lambda) g(p_2) \end{align*} proves the assertion.