The Lp(E) Normed Linear Space

# The Lp(E) Normed Linear Space

Definition: Let $E$ be a Lebesgue measurable set and let $1 \leq p < \infty$. Then the $L^p(E)$ Space is the set $L^p(E) = \{ f \: \mathrm{measurable} : \int_E |f|^p < \infty$ with the norm $\| \cdot \|_p : L^p(E) \to [0, \infty)$ defined for each $f \in L^p(E)$ by $\displaystyle{\| f \|_p = \left ( \int_E |f|^p \right )^{1/p}}$. |

Observe that $L^p(E)$ is the set of measurable functions such that $|f|^p$ is Lebesgue integrable on $E$, and for every $f \in L^p(E)$, the $p$-norm of $f$ is simply $p^{\mathrm{th}}$ root of the integral of $|f|^p$. Sometimes, a measurable function $f$ is said to be $p$-integrable if $|f|^p$ is integrable, and hence $L^p(E)$ is the set of $p$-integrable functions on $E$.

Proposition 1: $(L^p(E), \| \cdot \|_p)$ is a normed space. |

**Partial Proof:**Again, since $L^p(E)$ is a subset of the set of measurable functions (which is a linear space), all we need to show is that $L^p(E)$ is closed under addition, closed under scalar multiplication, and contains the zero function $0$ to show that it is a linear space.

- Let $f, g \in L^p(E)$. Then $f$ and $g$ are measurable and so $f + g$ is measurable. Hence:

\begin{align} \: \: \: \int_E |f + g|^p \leq \int_E [|f| + |g|]^p \leq \int_E [2 \max \{ |f|, |g| \}]^p = \int_E 2^p \max \{ |f|, |g| \}^p = 2^p \int_E \max \{ |f|^p, |g|^p \} \leq 2^p \int_E (|f|^p + |g|^p) = 2^p \int_E |f|^p + 2^p \int_E |g|^p < \infty \end{align}

- Therefore $(f+g) \in L^p(E)$.

- Let $\alpha \in \mathbb{R}$ and let $f \in L^p(E)$. Then $f$ is measurable and so therefore $\alpha f$ is measurable, and:

\begin{align} \quad \int_E |\alpha f|^p = \int_E |\alpha|^p |f|^p = |\alpha|^p \int_E |f|^p < \infty \end{align}

- Therefore $\alpha f \in L^p(E)$.

- Lastly, $0 \in L^p(E)$ since $\int_E 0^p = \int_E 0 = 0 < \infty$. Therefore, $L^p(E)$ is a linear space.

- All that remains to show is that $\| \cdot \|_p$ is a norm on $L^p(E)$.

**Showing that $\| f \|_p = 0$ if and only if $f = 0$ a.e. on $E$:**Suppose that $\| f \|_p = 0$. Then $\left ( \int_E |f|^p \right )^{1/p} = 0$. So $|f|^p = 0$ a.e. on $E$ which implies that $f = 0$ a.e. on $E$. Conversely, suppose that $f = 0$ a.e. on $E$. Then $|f|^p = 0$ a.e. on $E$ so $\| f \|_p = \left ( \int_E |f|^p \right )^{1/p} = 0^{1/p} = 0$.

**Showing that $\| \alpha f \|_p = |\alpha| \| f \|_p$:**Let $\alpha \in \mathbb{R}$ and let $f \in L^p(E)$. Then:

\begin{align} \quad \| \alpha f \|_p = \left ( \int_E |\alpha f|^p \right )^{1/p} = \left ( \int_E |\alpha|^p |f|^p \right )^{1/p} = \left ( |\alpha|^p \int_E |f|^p \right )^{1/p} = |\alpha| \left ( \int_E |f|^p \right )^{1/p} = |\alpha| \| f \|_p \end{align}

**Showing that $\| f + g \|_p \leq \| f \|_p + \| g \|_p$:**Unfortunately it is at this final step in the proof that we come to a problem. The proof of the triangle inequality for $\| \cdot \|_p$ is known as**Minkowski's Inequality**and requires a lot of preparation. We will postpone the proof until later. $\blacksquare$