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

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?

• Did you truly mean to want your norms to tend to $\infty$? Commented Mar 17 at 11:54
• @BrunoB indeed, I have corrected the norms to tend to $0$. Commented Mar 17 at 12:04

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 $$L^{p}$$].

• Could you elaborate on the steps you used? Along the line of what I was trying to do with Vitali's convergence. Commented Mar 17 at 12:05
• @TheLimitDoesNotExist The answer is just applying Vitali's convergence theorem to $g_{n}$ which is uniformly integrable (as it's $L^{q}$ norm is bounded for some $q>p$) and it converges to $0$ a.e. Thus $g_{n}\xrightarrow{L^{1}}0$ which is the same as $f_{n}-f$ converges in $L^{p}$ to $0$. Commented Mar 17 at 12:10
• @TheLimitDoesNotExist Is the proof clear now? Commented Mar 17 at 12:16
• Not sure why this answer would be downvoted. Commented Mar 17 at 13:05

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:

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

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

• 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.