Proving $\int f_n gd \mu \to \int f g d\mu$ for $f \in L^p, g \in L^q$ where $p$ and $q$ are Holder conjugate exponents.

convergence-divergencelp-spacesmeasure-theoryreal-analysis

I usually have a hard time with proofs about convergence, so I want to make sure if I did it right or not. The problem is:

If $f_n \to f$ in $L^p$, $g \in L^q$, and $\frac{1}{p} + \frac{1}{q} = 1$, then $$\int f_n g d\mu \to \int fg d\mu$$

By Hölder's inequality, for each $n$,

$$\int |f_n g| d\mu \leq ||f_n||_p ||g||_q < \infty$$

Since $f_n \to f$ in $L^p$, that means that $f$ is integrable and, therefore, measurable. Since $g \in L^q$, it is integrable.

Therefore, by the Dominated Convergence theorem

$$\lim_{n\to \infty}\int f_n g d\mu = \int\lim_{n\to \infty}(f_ng )d\mu = \int fgd\mu$$

Best Answer

You implicitly said that $f,g\in L^{p}$ implies that they are integrable, this is false in general. Being integrable means that $f$ (or $g$) belongs to $L^{1}$. We know that $L^{p}$ functions do not necessarily belong to $L^{1}$ unless the finite measure is in issue.

To use the Lebesgue Dominated Convergence Theorem, you need an integrable upper bound, that is, are you able to find an $L^{1}$ function $h$ such that $|f_{n}(x)g(x)|\leq h(x)$? As commented by @ABP there.

In this case, I cannot see any of such an integrable upper bound, so a simpler way to do it is to do it directly:

\begin{align*} \left|\int f_{n}g-\int fg\right|=\left|\int(f_{n}-f)g\right|\leq\int|f_{n}-f||g|\leq\|f_{n}-f\|_{L^{p}}\|g\|_{L^{q}}\rightarrow 0 \end{align*} as $n\rightarrow\infty$. Note that $\|g\|_{L^{q}}$ is just a constant.

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