Proving the Lp norm is a norm.

Question

I want to prove the $L_p$ norm on continuous functions is in fact a norm. I have proven definitiveness and homogeneity but am struggling with the triangle inequality. I am using the fact that if the closed unit ball is convex then the triangle inequality holds.

Attempt so far:

Let $f$,$g$ be two functions in the closed unit ball, let $h=af+(1-a)g$, for some $a \in [0,1]$.

Then the $L_p$ norm of $h$ is

$$\left(\int|af+(1-a)g|^p\right)^{1/p}$$

I think i want to try and decompose this into 2 components, one containing the integral of $f$, one containing the integral of $g$ but am unsure how to do this given the power of $p$. Thanks in advance Dan

Answer 1

I can't understand the intuition of taking some functions that belong in the closed unit ball, as $\mathcal{L}^p$ is generally the space :

$$\mathcal{L}^p = \left\{f:[a,b] \to \mathbb R, \; \text{measurable} : \left(\int_a^b |f(x) |^p\mathrm{d}x\right)^{1/p}< + \infty \right\}$$

Hint :

Let $f,g \in \mathcal{L}^P$. Then, it is :

\begin{align*} \|f+g \|_p & = \left(\int_a^b |f(x) + g(x)|^p\mathrm{d}x\right)^{1/p} \\ &\leq \left(\int_a^b \left(|f(x)| + |g(x)| \right)^p \mathrm{d}x \right)^{1/p} \end{align*}

Answer 2

There's a few ways to proceed. Probably the easiest is to not worry about the unit ball, and just prove the triangle inequality directly (that proof is available everywhere, for example on Wikipedia, if you want to look), but you asked for a version via convexity, so I'll give it a go:

First, note that $\|af+(1-a)g\|_{L_p} \leq 1$ if and only if $\|af+(1-a)g\|_{L_p}^p \leq 1$, so we'll deal with the latter, since it gets rid of that awkward power of $1/p$.

At this point, things are relatively easy: first, we note that $x \mapsto x^p$ is convex on the non-negative real half-line (feel free to insert a proof of this here if you don't have one already), so (combining this with the triangle inequality for $|\cdot|$) we have $|af(x)+(1-a)g(x)|^p \leq \left(a|f(x)| + (1-a)|g(x)| \right)^p\leq a|f(x)|^p + (1-a)|g(x)|^p$.

Thus, by standard properties of integrals, the definition of $\|\cdot\|_{L_p}$, and the fact that $f$ and $g$ lie in the closed unit ball, we have

\begin{align*}\|af+(1-a)g\|_{L_p}^p &= \int |af(x)+(1-a)g(x)|^p\mathrm{d}x \\&\leq \int a|f(x)|^p+(1-a)|g(x)|^p\mathrm{d}x \\&= a\int|f(x)|^p\mathrm{d}x + (1-a)\int|g(x)|^p\mathrm{d}x \\&= a\|f\|_{L_p}^p + (1-a)\|g\|_{L_p}^p \\&\leq a + (1 -a) \\&= 1\end{align*}

Thus, $\|af+(1-a)g\|_{L_p} \leq 1$ also, so our open unit ball is convex, hence we have the triangle inequality.