Does uniformly bounded sequence in Lp which converges almost everywhere converge in norm?

Question

I'm trying to solve problem $17$ from chapter $6$ of Royden's Real Analysis. The problem is -

Let $(f_n)_{n=1}^\infty$ be a sequence of functions in $L^p([0,1]), p\in(1,\infty)$, which converge almost everywhere to some $f\in L^p$. Suppose that there is a constant $M$ s.t $||f_n||_p\leq M$ for all $n$. Then for each $g\in L^q$ (where $\frac{1}{q}+\frac{1}{p}=1$) we have -

$\int fg=lim_{n\rightarrow \infty}\int f_ng$

I want to solve this problem using the Riesz representation theorem for $L^p$ spaces, which states that any bounded linear functional in $(L^p)^*=L^q$ is of the form - $\phi _g (h)=\int hg$, where $g\in L^q$. Since $g$ corresponds to a bounded (and thus continuous) linear functional, the problem is basically solved, if I prove that $f_n\rightarrow f$ in norm. But I'm not sure that this is the case.

I know that generally, $f_n\rightarrow f$ almost everywhere does not imply convergence in norm. But since $f_n$ are uniformly bounded in norm by $M$, I'm hoping that I can somehow conclude that this is indeed the case (I know that if moreover $||f_n||_p\rightarrow ||f||_p$ then there actually is convergence in norm, maybe I can somehow use this fact?).

Does anyone know if this it true and how to prove this? If not, can anyone suggest a different approach?

Thanks in advance.

Answer 1

It need not be the case that $f_n \to f$ in $L^p$.For example, consider $f_n = n^{1/p} \chi_{[0,\frac{1}{n}]}$ as functions on $[0,1]$ with the Lebesgue measure. Then $f_n \to 0$ a.e. but $\|f_n\|_p = 1$ so that $f_n$ is bounded in $L^p([0,1])$ but does not converge to $0$ in $L^p([0,1])$.

In fact, in Royden's book the Riesz representation theorem for $L^p$ is presented after this exercise so it's good to guess that we shouldn't use it.

Instead we will make use of Egoroff's theorem. First, since $g \in L^q$, for any $\varepsilon > 0$ there is a $\delta > 0$ such that $m(E) < \delta$ implies that $$\int_E |g|^q dm < \bigg(\frac{\varepsilon}{4M}\bigg)^q.$$

Then by Egoroff's theorem there is measurable $E \subseteq [0,1]$ such that $m(E) < \delta$ and $f_n \to f$ uniformly on $F := [0,1] \setminus E$. As a result, there is an $N$ such that $n \geq N$ and $x \in F$ implies that $$|f_n(x) - f(x)| \leq \frac{\varepsilon}{2m(F)^{1/p}\|g\|_q}.$$

Finally we get that for $n \geq N$ \begin{align} \bigg|\int_0^1 (f_n - f)g dm\bigg| \leq& \int_E |f_n-f||g| dm + \int_F |f_n - f||g|dm \\ \leq& \|f_n - f\|_p \frac{\varepsilon}{4M} + \bigg(\int_F |f_n - f|^pdm\bigg)^{1/p} \|g\|_q \\ \leq& \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{align} where to get the second line I applied Holder's inequality and substituted the previous bound on the integral of $|g|^q$ over $E$. This is exactly the desired convergence result.