Prove a function is convex and relevant limits. Let $f:[0,1] \to \mathbb{R}$ be a strictly increasing continuous function with $f(0)>0$. Set $c=f(1)$ and $g(x)=\int _0 ^1 (f(t))^x dt$ for all $x \in \mathbb{R}$. Prove the following: 
A) $g(x)$ is convex on $\mathbb{R}$
B) $$\lim _{n \to \infty}g(n)^{1/n}=c$$
C) $$\lim _{n \to \infty} \frac {g(n)}{c^n}=0$$
D) $$\lim _{n \to \infty} \frac {g(n+1)}{g(n)}=c$$
For A, I know that I can compute the second derivative and show that it is positive, and thus $g(x)$ is convex. For parts B through D, I have no clue where to start. Any help is greatly appreciated, and a full solution to B in particular would be welcomed. 
 A: For Part $B)$, we proceed as follows.  First, write $g^{1/n}(n)$ as 
$$g^{1/n}(n)=\left(\int_0^1 f^n(t)\,dt\right)^{1/n}=c\left(\int_0^1 \left(\frac{f(t)}{c}\right)^n\,dt\right)^{1/n}\tag 1$$
Let $h(t)=f(t)/c$.  Then, $h(t)$ is continuous and increasing on $[0,1]$.  Furthermore $0<h(t)\le 1$.  Note that we can write for any $0<\epsilon<1$
$$\int_0^1 h^n(t)\,dt=\int_0^{1-\epsilon} h^n(t)\,dt+\int_{1-\epsilon}^1 h^n(t)\,dt$$
from which we can deduce the estimates
$$\epsilon^{1/n} h(1-\epsilon)\le \left(\int_0^1h^n(t)\,dt\right)^{1/n}\le 1$$
Letting $n\to \infty$, we find that for all $\epsilon\in (0,1)$ we have
$$h(1-\epsilon)\le \lim_{n\to \infty}\left(\int_0^1h^n(t)\,dt\right)^{1/n}\le 1 \tag 2$$
Since $(2)$ is true for all $\epsilon\in (0,1)$, then given that $h(t)$ is continuous and increasing with $h(1)=1$, we find that
$$\lim_{n\to \infty}\left(\int_0^1h^n(t)\,dt\right)^{1/n}=1 \tag 3$$
Putting $(3)$ together with $(1)$ yields the coveted limit

$$\lim_{n\to \infty}\left(\int_0^1 f^n(t)\,dt\right)^{1/n}=c$$

A: (This answers B.) The expression 
$$M(p)= \left( \int _0 ^1 [f(x)]^p dx \right)^{1/p}$$
is the generalised mean (or power mean with exponent $p$) of the function $f(x)$. The generalised power inequality (https://en.wikipedia.org/wiki/Generalized_mean) states that $M_p \le M_q$ for $p \le q$ and moreover
$$\lim_{p \uparrow \infty} M_p = M_{\infty} = \max f(x)$$
In your problem, since $f(x)$ is increasing, $\max f(x) = f(1) = c$, yielding
$$\lim _{n \to \infty} g(n)^{1/n} = \lim _{n \to \infty} M (n)=c$$
