$\lim_{n\to \infty}(\int_0^1 f(x)^{2n}g(x)^n h(x)~dx)^{1/n}$ where $f,g,h$ are positive continuous functions on $[0,1]$

Question

I want to find $\lim_{n\to \infty}(\int_0^1 f(x)^{2n}g(x)^n h(x)~dx)^{1/n}$ where $f,g,h$ are positive continuous functions on $[0,1]$. By Holder's inequality, this limit is greater than or equal to $\lim_{n\to \infty} \int_0^1 f(x)^2g(x)h(x)^{1/n}~dx$ which is equal to $\int_0^1 f(x)^2g(x)~dx$ by the DCT. But I can't find an upper bound. I tried Jensen's inequality, but it does not seems to work I think because $x\mapsto x^{1/n}$ is not convex. Can I get a hint?

Answer 1

Let $k=f^{2}g$, $M_{k}=\max_{x\in[0,1]}k(x)$, and $M_{h}=\max_{x\in[0,1]}h(x)$. Firstly, observe that \begin{eqnarray*} & & \left\{ \int_{0}^{1}k^{n}(x)h(x)dx\right\} ^{\frac{1}{n}}\\ & \leq & M_{k}M_{h}^{\frac{1}{n}}. \end{eqnarray*} Therefore $\limsup_{n}\left\{ \int_{0}^{1}k^{n}(x)h(x)dx\right\} ^{\frac{1}{n}}\leq M_{k}$.

Choose $x_{0}\in[0,1]$ such that $k(x_{0})=M_{k}$. Let $\alpha\in(0,1)$ be arbitrary. Since $k(x_{0})>\alpha M_{k}$, by continuity of $k$ at $x_{0}$, there exists $\delta>0$ such that $k(x)>\alpha M_{k}$ whenever $x\in[x_{0}-\delta,x_{0}+\delta]$ (if $x_{0}=0$ or $x_{0}=1$, adjust the interval accordingly). Let $b=\min_{x\in[0,1]}h(x)>0$. We have that \begin{eqnarray*} & & \left\{ \int_{0}^{1}k^{n}(x)h(x)dx\right\} ^{\frac{1}{n}}\\ & \geq & \left\{ \int_{x_{0}-\delta}^{x_{0}+\delta}k^{n}(x)h(x)dx\right\} ^{\frac{1}{n}}\\ & \geq & \alpha M_{k}b^{\frac{1}{n}}(2\delta)^{\frac{1}{n}}. \end{eqnarray*} Therefore, $\liminf_{n}\left\{ \int_{0}^{1}k^{n}(x)h(x)dx\right\} ^{\frac{1}{n}}\geq\alpha M_{k}$. Further letting $\alpha\rightarrow1$, we have $\liminf_{n}\left\{ \int_{0}^{1}k^{n}(x)h(x)dx\right\} ^{\frac{1}{n}}\geq M_{k}$. Hence, $\lim_{n}\left\{ \int_{0}^{1}k^{n}(x)h(x)dx\right\} ^{\frac{1}{n}}$ exists and equals to $M_{k}=\max_{x\in[0,1]}f^{2}(x)g(x)$.

Answer 2

Like I said in my comment to your question. Let $m_h=\min h$ and $M_h=\max h$ ($\min$ and $\max$ over the unit interval). These are finite and positive by your hypothesis. THen

$$m^{1/n}_h\|f^2g\|_n\leq \|f^2gh^{1/n}\|_n\leq \|f^2g\|_nM^{1/n}_h$$

Since $\|f^2g\|_n\xrightarrow{n\rightarrow\infty}\|f^2g\|_\infty$, passing to the limit you obtain

$$\|f^2gh^{1/n}\|_n\xrightarrow{n\rightarrow\infty}\|f^2g\|_\infty$$.

Here we have used a well known result: if $f$ is integrable with respect a finite measure, then $\|f\|_p$ converges to $\|f\|_\infty$ as $p\rightarrow\infty$.