# Real Analysis - Convergence to Essential Supremum

• MHB
• joypav
In summary, we are trying to prove that the limit of the ratio of integrals of $|f|$ to the power of $n+1$ and $n$ is equal to the infinity norm of $f$. We first use the definition of convergence and then use Hölder's inequality to show that the inferior limit of the ratio is greater than or equal to the infinity norm of $f$. By taking arbitrary values of $\epsilon$, we can show that the liminf of the ratio is also greater than or equal to the infinity norm of $f$. Therefore, the limit of the ratio is equal to the infinity norm of $f$.
joypav
Problem:
Let $\left(X, M, \mu\right)$ be a probability space. Suppose $f \in L^\infty\left(\mu\right)$ and $\left| \left| f \right| \right|_\infty > 0$. Prove that
$lim_{n \rightarrow \infty} \frac{\int_{X}^{}\left| f \right|^{n+1} \,d\mu}{\int_{X}^{}\left| f \right|^{n} \,d\mu} = \left| \left| f \right| \right|_\infty$

Proof:
So... I'm not really sure how to approach this problem.
$\left(X, M, \mu\right)$ is a probability space $\implies \mu\left(X\right) = 1$
Which then, $\left| \left| f \right| \right|_p \leq \left| \left| f \right| \right|_\infty$ for all $p$.

I thought maybe to just use our usual definition for convergence? Meaning,
$\forall \epsilon>0, \exists \delta>0, n>\delta \implies \left|\frac{\int_{X}^{}\left| f \right|^{n+1} \,d\mu}{\int_{X}^{}\left| f \right|^{n} \,d\mu} - \left| \left| f \right| \right|_\infty \right| \leq \epsilon$

But again, I'm not seeing what the next step would be... if someone could give some help I would really appreciate it!

Hello joypav,

If $\alpha_n = \int_X \lvert f\rvert^n\, d\mu$, then using the fact $\lvert f(x)\rvert \le \|f\|_\infty$ a.e., one finds $\alpha_{n+1} \le \|f\|_\infty \alpha_n$, forcing $\limsup_{n\to \infty} \alpha_{n+1}/\alpha_n \le \|f\|_\infty$. On the other hand, if $\varepsilon \in (0, \|f\|_\infty)$, the set $A := \{x\in X : |f(x)| > \|f\|_\infty - \varepsilon\}$ has finite positive measure. Using Hölder's inequality, show that $$\left(\frac{1}{\mu(A)} \int_A \lvert f\vert^{n+1}\, d\mu\right)^{1/(n+1)} \ge \left(\frac{1}{\mu(A)}\int_A \lvert f\vert^n\, d\mu\right)^{1/n}$$ and consequently $$\alpha_{n+1} \ge \int_A \lvert f\rvert^{n+1}\, d\mu \ge \mu(A)^{-1/n}(\|f\|_\infty - \varepsilon)^{n+1}$$ Since $\alpha_n \le \|f\|_\infty^n$, then $\alpha_{n+1}/\alpha_n \ge \mu(A)^{-1/n}(\|f\|_\infty - \varepsilon)$. Taking inferior limits as $n \to \infty$ results in $$\liminf_{n\to \infty} \frac{\alpha_{n+1}}{\alpha_n} \ge \|f\|_\infty-\varepsilon$$As $\epsilon$ was arbitrary, $\liminf_{n\to \infty} \alpha_{n+1}/\alpha_n \ge \|f\|_\infty$. Hence $\alpha_{n+1}/\alpha_n$ converges to $\|f\|_\infty$.

Hi there! I think you're on the right track with using the definition of convergence. Here's how I would approach this problem:

We know that $\left| \left| f \right| \right|_\infty > 0$, which means that there exists some $x_0 \in X$ such that $\left| f\left(x_0\right) \right| > 0$. Now, let's consider the sequence of functions $g_n = \left| f \right|^n$. We have:
$$\lim_{n \rightarrow \infty} g_n\left(x_0\right) = \lim_{n \rightarrow \infty} \left| f\left(x_0\right) \right|^n = 0$$
since $\left| f\left(x_0\right) \right| < 1$. This means that $g_n$ converges pointwise to the function $g = 0$.

Now, we can use the dominated convergence theorem to show that the sequence $\frac{\int_{X}^{} g_{n+1} \,d\mu}{\int_{X}^{} g_{n} \,d\mu}$ converges to $\frac{\int_{X}^{} g \,d\mu}{\int_{X}^{} g \,d\mu} = 0$. Since $\left| \left| f \right| \right|_\infty > 0$, we know that $\left| f \right|^n$ is dominated by $g_n$ for all $n$, so we can use the theorem.

This means that for any $\epsilon > 0$, there exists some $N$ such that for all $n > N$, we have:
$$\left| \frac{\int_{X}^{} \left| f \right|^{n+1} \,d\mu}{\int_{X}^{} \left| f \right|^{n} \,d\mu} - 0 \right| < \epsilon$$
which is equivalent to:

\left| \frac{\int_{X}^{} \left| f \right|^{n+1} \,d\mu}{\int_{X}^{} \left| f \right|^{n} \,

## 1. What is the definition of convergence to essential supremum?

Convergence to essential supremum is a concept in real analysis that describes the behavior of a sequence of functions. It means that as the index of the sequence increases, the values of the functions approach the supremum (or least upper bound) of the set of their essential values.

## 2. How is convergence to essential supremum different from pointwise convergence?

Pointwise convergence only requires that the values of a sequence of functions approach a given point for each value of the independent variable. In contrast, convergence to essential supremum requires that the values approach the supremum of the essential values, which may be different from the pointwise limit.

## 3. What is the importance of convergence to essential supremum in real analysis?

Convergence to essential supremum is important in real analysis because it allows us to study the behavior of a sequence of functions in a more general and flexible way. It also allows us to prove theorems and make conclusions about the behavior of functions that would not be possible with pointwise convergence alone.

## 4. How is convergence to essential supremum related to the concept of measurability?

Convergence to essential supremum is closely related to the concept of measurability in real analysis. A function is said to be measurable if its essential supremum and essential infimum are finite and equal. Measurability is important because it allows us to apply tools from measure theory to study the behavior of functions.

## 5. Can a sequence of functions converge to essential supremum but not converge pointwise?

Yes, it is possible for a sequence of functions to converge to essential supremum but not converge pointwise. This can happen when the pointwise limit does not exist or is different from the supremum of the essential values. In this case, the sequence of functions may still have a well-defined behavior in terms of convergence to essential supremum.

Replies
4
Views
386
Replies
1
Views
662
Replies
4
Views
2K
Replies
2
Views
604
Replies
2
Views
887
Replies
2
Views
1K
Replies
2
Views
1K
Replies
21
Views
2K
Replies
4
Views
873
Replies
2
Views
1K