1

We first use basic Hölder’s inequality to show Hölder’s inequality for two nonnegative simple functions \(\varphi\) and \(\psi\). Suppose we decompose these simple functions as \[ \varphi= \sum_{i=1}^n x_i \chi_{E_i} \text{ and } \psi = \sum_{i=1}^n y_j \chi_{E_i} \] with \(E_1,\dots,E_n\) pairwise disjoint measurable subsets, Then we have \[\begin{align*} \left\lVert\varphi\psi\right\rVert_1 &= \int \varphi\psi\ d\mu \\ &= \sum_{i=1}^n x_i\;y_i\;\mu(E_i) \\ &= \sum_{i=1}^n x_i\;\mu(E_i)^{1/p}\;y_i\;\mu(E_i)^{1/q} \\ &\leq \left(\sum_{i=1}^n {x_i}^p\; \mu(E_i)\right)^{1/p} \left(\sum_{i=1}^n {y_i}^q\;\mu(E_i)\right)^{1/q} \\ &= \left(\int \varphi^p\ d\mu\right)^{1/p} \left(\int \varphi^q\ d\mu\right)^{1/q} \\ &= \left\lVert\varphi\right\rVert_p \left\lVert\psi\right\rVert_q. \end{align*}\]

Following hint, without loss of generality assume \(f,g\) are nonnegative.

Let \(\varphi_n\) and \(\psi_n\) be two monotonically increasing sequences of nonnegative simple functions converging to \(f\) and \(g\) respectively. Basic Hölder inequality for simple functions gives us that \[ \left\lVert\varphi_n\psi_n\right\rVert_1 \leq \left\lVert\varphi_n\right\rVert_p\left\lVert\psi_n\right\rVert_q \] for each \(n\).

Now \(\varphi_n\psi_n \to fg\) and \(\varphi_1\psi_1\leq \varphi_2\psi_2\leq \dots\), so by monotone convergence (thrice) \[\begin{align*} \left\lVert fg\right\rVert_1 &= \lim_{n\to\infty} \left\lVert\varphi_n\psi_n\right\rVert \\ &= \lim_{n\to\infty} \left\lVert\varphi_n\right\rVert_p\left\lVert\psi_n\right\rVert_q \\ &= \left\lVert f\right\rVert_p\left\lVert g\right\rVert_q. \end{align*}\]

2

Similarly we use basic Minkowski inequality to show the case for nonnegative simple functions \(\varphi\) and \(\varphi\). Decompose them similarly as previous question with \(E_1,\dots,E_n\) measurable satisfying the same disjoint conditions. Then we have \[\begin{align*} \left\lVert\varphi+\psi\right\rVert_p &= \left( \int \left(\varphi+\psi\right)^p\ d\mu \right)^{1/p} \\ &= \left( \sum_{i=1}^n \left(x_i + y_i\right)^p\;\mu(E_i) \right)^{1/p} \\ &= \left( \sum_{i=1}^n \left(x_i\;\mu(E_i)^{1/p} + y_i\;\mu(E_i)^{1/p}\right)^p \right)^{1/p} \\ &\leq \left( \sum_{i=1}^n {x_i}^p\mu(E_i) \right)^{1/p} + \left( \sum_{i=1}^n {y_i}^p\mu(E_i) \right)^{1/p} \\ &= \left( \int \varphi^p\ d\mu \right)^{1/p} + \left( \int \psi^p\ d\mu \right)^{1/p} \\ &= \left\lVert\varphi\right\rVert_p + \left\lVert\psi\right\rVert_p. \end{align*}\]

Similarly applying monotone convergence we get the result for the general case.

3

By the definition of \(L^p\) we see that it is closed under scalar multiplication. From Minkowski inequality we can deduce that \(L^p\) is closed under addition.

We proceed to show that \(\left\lVert\cdot\right\rVert_p\) is a norm. Positivity and definiteness easily arise from the definitions, same for homogeneity. The triangle inequality for \(\left\lVert\cdot\right\rVert_p\) is just Minkowski’s inequality.

4

Let \((f_n)\) be a Cauchy sequence in \(L^p\). Without loss of generality we can assume that it contracts quickly enough in the sense that \(\left\lVert f_n - f_m\right\rVert_p < 2^{-N}\) whenever \(n,m>N\).

By the condition that \(\left\lVert f_n - f_{n+1}\right\rVert_p < 2^{1-n}\), it follows that the set \(E_n := \left\{x\in X: \left\lvert f_n- f_{n+1}>2^{-\frac{n}{2}}\right\rvert\right\}\) satisfies \(\mu(E_n) < 2^p\cdot 2^{1-\frac{n}{2}}\). Since \(\sum_{n=1}^\infty \mu(E_n)\) is finite, by Borel-Cantelli lemma \(N = \limsup E_n\) has measure zero. Define \(f\) to be the pointwise limit of \(f_n\) on \(X\setminus N\) and \(0\) everywhere else.

Next we show the \(\mu\)-a.e. pointwise limit \(f\) is also the limit in \(L^p\), we have \[ \left\lvert f(x)\right\rvert \leq g(x) := \begin{cases} \left\lvert f_1(x)\right\rvert + \sum_{i=1}^\infty \left\lvert f_{i+1}(x) - f_i(x)\right\rvert &\text{for } x\notin N, \\ +\infty &\text{for } x\in N. \end{cases} \] which means \(f\in L^p\) because \(\left\lVert f\right\rVert_p \leq \left\lVert g\right\rVert_p \leq \left\lVert f_1\right\rVert_p + 2\).

Now as \(\left\lvert f_n-f\right\rvert^p\) is dominated by \(g^p\), by dominated convergence, \[ \lim_{n\to\infty} {\left\lVert f_n-f\right\rVert_p}^p = \lim_{n\to\infty} \int \left\lvert f_n-f\right\rvert^p\ d\mu = 0 \] which shows that \(f\) is also the limit in \(\left\lVert\cdot\right\rVert_p\).