Show that $||f_n-f||_p \rightarrow 0$ as $n \rightarrow \infty$

Question

Given $(X, \mathcal{A}, \mu)$ be a finite measure space and $f_n \in L^p(X, \mu)$ where $f_n(x) \rightarrow f(x)$ almost everywhere as $n \rightarrow \infty$ and $1 \leq p < \infty$.

Now suppose $||f_n||_q \leq M < \infty \ \forall n \ $and $M \in \mathbb{R}^+$ is a constant, plus $q > p $.

Then show that $||f_n-f||_p \rightarrow 0$ as $n \rightarrow \infty$

My Attempt

We have been asked to use the hint to use Vitali's Convergence Theorem. Now firstly as $||f_n||_q \leq M < \infty \implies f_n \in L^q(X, \mu)$. Using $L^p$ and $L^q$ space inclusion I can say that $f_n \in L^1(X, \mu) \ \forall n $. This $\implies f_n$ is uniformly integrable, again using For every $\epsilon>0$ there exists $\delta>0$ such that $\int_A|f(x)|\mu(dx) < \epsilon$ whenever $\mu(A) < \delta$

These above conditions now let me use Vitali's convergence theorem. But that does not involve a $||.||_p$ norm anywhere. How do I proceed from here?

Answer 1

Here is a proof by contradiction which does not directly use Vitali or uniform integrability (but essentially proves these results in this setting).

Suppose that $||f_n - f||_p \not \to 0$. Then, $\exists \varepsilon >0$ s.t. (WLOG restricting to a subsequence) $||f_n - f||_p \ge \varepsilon$ for every $n$.

Let $\delta >0$ and consider $S_n = \{|f_n - f|^p \ge \delta\}$. Then, since convergence a.e. implies convergence in measure for finite measure spaces, we have that $\mu(S_n) \downarrow 0$ (remark: this also follows from continuity from above for finite measures).

But then, $$\varepsilon \le ||f_n - f||_p = \mu(|f_n - f|^p)= \mu(|f_n - f|^p[1(S_n) + 1(S_n^c)])$$

$$\varepsilon \le \mu(|f_n - f|^p1(S_n)) + \delta T$$

where $T = \mu(X)$, noting that $|f_n-f|^p$ is bounded by $\delta$ on $S_n^c$ by definition. Let $\delta$ be sufficiently small such that $\eta = \varepsilon - \delta T > 0$. Then, we have that:

$$\mu(|f_n - f|^p 1(S_n)) \ge \eta$$

for every $n$. If we now apply Hölder with $\frac{q}{p}$, we find that:

$$\mu(|f_n - f|^q) \mu(1(S_n)) = \mu(|f_n - f|^q) \mu(1(S_n)^{\frac{q}{q-p}}) \ge \eta$$

$$\implies \mu(|f_n - f|^q) \ge \frac{\eta}{\mu(1(S_n))} \to \infty$$

which is a contradiction.

Intuitively:

• If $f_n \not \to f$ in $L^p$, then $f_n$ certainly must differ from $f$ on sets of positive measure.
• However, $f_n \to f$ a.e. on a finite measure space implies that the sets on which they differ must have measure shrinking to zero.
• Therefore, $|f_n-f|$ must be getting larger and larger on these sets -- in fact, unbounded, since the $L^p$ norm is bounded away.
• The effect on the integral is amplified in $L^q$, resulting in divergence.

Answer 2

Let $g_n=|f_n-f|^{p}$. Then $g_n \to 0$ a.e. and $\int g_n^{q/p}d\mu\le 2^{q}[\int |f_n|^{q}d\mu+\int |f|^{q}d\mu]$. Note that the last term is finite by Fatou's Lemma. Since $q/p >1$ it follws that $(g_n)$ is uniformly integrable. Hence, $\int g_n d\mu \to 0$ ,as required.

If you want to apply Vitali's Theorem just apply it to $(g_n)$ and conclude that $g_n \to 0$ in $L^{1}(\mu)$. [Apply the theorem to $L^{1}$ space instead of $L^{p}$].