discussing the existence of a convex function If $g$ is a positive function on $[0,1]$ such that $g(x)$ tends to $\infty$ as $x$ tends to $0$, then there is a convex function $h$ on $[0,1]$ such that $ h \leq g$ and $h(x)$ tends to $\infty$ as $x$ tends to $0$.
Is that true or false and why?
 A: First observe that it does not really make sense for your function $g$ to be a real-valued function on $[0,1]$ (note that $0 \in [0,1]$) if you also assume $g(x) \to \infty$ for $x \downarrow 0$. Hence, I will in the following assume that we are actually working on the interval $(0,1]$.
In this case, it turns out that such a function always exists.
To see this, first observe that if $F$ is a (nonempty) set of convex functions on an interval $\emptyset \neq I \subset \Bbb{R}$, such that
$$
g(x) := \sup_{f \in F} f(x)
$$
is finite for all $x \in I$, then $g$ is a convex function.
To see this, let $\lambda \in [0,1]$ and $x,y \in I$. For any $f \in F$, we have
$$
f(\lambda x + (1-\lambda) y ) \leq \lambda f(x) + (1-\lambda) f(y) \leq \lambda g(x) + (1-\lambda) g(y).
$$
After taking the supremum, we arrive at $g(\lambda x + (1-\lambda) y) \leq \lambda g(x) + (1-\lambda) g(y)$.
Now, for $\alpha, \beta \in \Bbb{R}$, let $g_{\alpha, \beta} := (x \mapsto \alpha x + \beta)$, considered as a function on $(0,1]$. Observe that each $g_{\alpha, \beta}$ is affine-linear and hence convex.
Set
$$
F := \{g_{\alpha, \beta} \mid g_{\alpha, \beta}(x) \leq g(x) \,\forall x\in (0,1]\}.
$$
The above considerations imply that the function $h(x) := \sup_{\varphi \in F} \varphi(x)$ is a convex function on $(0,1]$ with $h(x) \leq g(x)$ for all $x$. This function is called the convex envelope of $g$. It suffices to show $h(x) \to \infty$ for $x \downarrow 0$.
To see this, let $M > 0$ be arbitrary. Because of $g(x) \to \infty$ for $x \downarrow 0$, there is $\delta > 0$ with $g(x) > 2M$ for all $x \in (0,\delta)$.
Let $\gamma \in (0,\delta)$ be arbitrary and let $\alpha, \beta \in \Bbb{R}$, so that
$$
g_{\alpha, \beta} (x) = -\frac{2M}{\gamma} (x - \frac{\gamma}{2}) + M.
$$
Note that $g_{\alpha, \beta}$ is strictly decreasing.
For $x \in (0,1]$, there are now two cases:


*

*We have $x \in (0,\gamma]$. Then $g_{\alpha, \beta}(x) \leq g_{\alpha, \beta}(0) = 2M \leq g(x)$ by choice of $\delta, \gamma$.

*We have $x \in (\gamma, 1]$. Then $g_{\alpha, \beta}(x) \leq g_{\alpha, \beta}(\gamma) = 0 \leq g(x)$, because $g$ is non-negative.
All in all, we see $g_{\alpha, \beta}(x) \leq g(x)$ for all $x$ and hence
$$
h(\gamma/2) \geq g_{\alpha, \beta}(\gamma/2) = M. 
$$
This proves $h(x) \geq M$ for all $x \in (0,\delta/2)$ and hence $h(x) \to \infty$ for $x \downarrow 0$, because $M>0$ was arbitrary.
