# Determining if the limit exists of the sequence

I am trying to determine whether the limit of the following sequence exists and if so, find the limit. $$f$$ is a positive continuous function of $$[a,b]$$.

\begin{align*} \lim_{n\to\infty} \Big[\int_{a}^{b}f(x)^{n}dx\Big]^{\frac{1}{n}} \end{align*}

My thoughts so far on the problem: By definition, I know \begin{align*} \lim_{n\to\infty} \Big[\int_{a}^{b}|f(x)|^{n}dx\Big]^{\frac{1}{n}} = \sup\{|f(x)|: x \in [a,b] \} \end{align*}

because $$f$$ is positive I believe I can say the limits of these two sequences will be equivalent. We know because $$f$$ is continuous and on an compact interval $$[a,b]$$ it has it achieves it sup. I feel like I'm missing something here.

If anyone can provide any hints or insights that would be great. Thanks.

• What part of that equality is "by definition"? – Thorgott Apr 18 '19 at 20:59
• @Thorgott I am under the impression this is the definition of the $L^{p}$ norm as $p\to\infty$. – Matt Apr 18 '19 at 21:01
• The LHS is the definition of the limit of the $L^p$ norm of $f$ as $p\rightarrow\infty$ (although you use the index $n$ which suggests that you're viewing it as a sequence). The RHS is the $L^{\infty}$ norm of $f$. The limit of something is already a notion with mathematical meaning, so you can't just define it to be whatever you want. The stated equality is true, but it certainly does not follow by definition; rather, it has to be demonstrated, which takes some potentially non-trivial effort. – Thorgott Apr 18 '19 at 21:50
• @Thorgott Yes, thank you. – Matt Apr 19 '19 at 21:11

$$\int_a^{b}[f(x)]^{n} dx \leq \int_a^{b} M^{n}=M^{n} (b-a)$$ where $$M=\sup \{f(x): a \leq x \leq b\}$$. Hence $$(\int_a^{b}[f(x)]^{n} dx)^{1/n} \leq M (b-a)^{1/n}$$. Letting $$n \to \infty$$ we get LHS $$\leq$$ RHS.
Now there is a point $$x_0$$ such that $$f(x_0)=M$$. Suppose $$a < x_0 . Let $$\epsilon >0$$. By continuity there exists $$\delta >0$$ such that $$f(x) >M-\epsilon$$ for $$x_0-\delta . Hence $$\int_a^{b}[f(x)]^{n} dx \geq \int_{x_0-\delta}^{x_0+\delta}[f(x)]^{n} dx \geq (M-\epsilon)^{n}(2\delta)$$. This gives $$(\int_a^{b}[f(x)]^{n} dx)^{1/n} \geq (M-\epsilon)(2\delta)^{1/n}$$. Letting $$n \to \infty$$ we get LHS $$\geq$$RHS. I will leave the cases $$x_0=a$$ and $$x_0=b$$ to you.
• Also, the cases $x_0 = a and b$ are analogous to the case you have shown above with the difference we would restrict our neighbourhood to the right side of a for $x_0 = a$ (or left side of b for $x_0 = b$). This is because continuity of $f$ is only within the interval $[a,b]$. Then perform similar calculations as above. – Matt Apr 19 '19 at 23:07