Measure Theory/L^p spaces

Recall that an $$\mathcal{L}^p$$ space is defined as $$\mathcal{L}^p(X)=\{f:X\to\mathbb{C}:f\text{ is measurable,}\int_X|f|^pd\mu<\infty\}$$

Jensen's inequality
Let $$(X,\Sigma,\mu)$$ be a probability measure space.

Let $$f:X\to\mathbb{R}$$, $$f\in\mathcal{L}^1$$ be such that there exist $$a,b\in\mathbb{R}$$ with $$a<f(x)<b$$

If $$\phi$$ is a convex function on $$(a,b)$$ then,

$$\displaystyle\phi\left(\int_Xfd\mu\right)\leq\int_X\phi\circ fd\mu$$

Proof

Let $$t=\displaystyle\int_Xfd\mu$$. As $$\mu$$ is a probability measure, $$a<t<b$$

Let $$\beta=\sup\{\frac{\phi(t)-\phi(s)}{t-s}:a<s<t<b\}$$

Let $$t<u<b$$; then $$\beta\leq\displaystyle\frac{\phi(u)-\phi(t)}{u-t}$$

Thus, $$\displaystyle\frac{\phi(t)-\phi(s)}{t-s}\leq\frac{\phi(u)-\phi(t)}{u-t}$$, that is $$\phi(t)-\phi(s)\leq\beta(t-s)$$

Put $$s=f(x)$$

$$\phi\left(\int_Xfd\mu\right)-\phi f(x) \leq \beta\left(f(x)-\int_Xfd\mu\right)$$, which completes the proof.

Corollary
$$\displaystyle e^{(\int_Xfd\mu)}\leq\int_Xe^fd\mu$$ $$\displaystyle e^{\left(\frac{p_1+\ldots+p_n}{n}\right)}\leq\frac{1}{n}\left(e^{p_1}+e^{p_2}+\ldots+e^{p_n}\right)$$
 * 1) Putting $$\phi(x)=e^x$$,
 * 1) If $$X$$ is finite, $$\mu$$ is a counting measure, and if $$f(x_i)=p_i$$, then

For every $$f\in\mathcal{L}^p$$, define $$\|f\|_p=\left(\int_X|f|^pd\mu\right)^{\frac{1}{p}}$$

Holder's inequality
Let $$1<p,q<\infty$$ such that $$\displaystyle\frac{1}{p}+\frac{1}{q}=1$$. Let $$f\in\mathcal{L}^p$$ and $$g\in\mathcal{L}^q$$.

Then, $$fg\in\mathcal{L}^1$$ and

$$\|fg\|\leq\|f\|_p\|g\|_q$$ Proof

We know that $$\log$$ is a concave function

Let $$0\leq t\leq 1$$, $$0<a<b$$. Then $$t\log a+(1-t)\log b\leq \log(at+b(1-t))$$

That is, $$a^tb^{1-t}\leq ta+(1-t)b$$

Let $$t=\frac{1}{p}$$, $$a=\left(\frac{|f|}{\|f\|_p}\right)^p$$, $$b=\left(\frac{|f|}{\|f\|_q}\right)^q$$

$$\displaystyle\frac{|f|}{\|f\|_p}\frac{|g|}{\|g\|_q}\leq\frac{1}{p}\frac{|f|^p}{\|f\|^p_p}+\frac{1}{q}\frac{|g|^q}{\|g\|^q_q}$$

Then, $$\displaystyle\frac{1}{\|f\|_p\|g\|_q}\int_X|f||g|d\mu\leq\frac{1}{p\|f\|_p^p}\int_X|f|^pd\mu+\frac{1}{p\|g\|_q^q}\int_X|g|^qd\mu=1$$,

which proves the result

Corollary
If $$\mu(X)<\infty$$, $$1<s<r<\infty$$ then $$\mathcal{L}^r\subset \mathcal{L}^s$$

Proof

Let $$\phi\in\mathcal{L}^s$$, $$p=\frac{r}{s}\geq 1$$, $$g\equiv 1$$

Then, $$f=|\phi|^s\in\mathcal{L}^1$$, and hence $$\displaystyle\int_X|\phi|^sd\mu\leq\left(\int_X\left(|\phi|^s\right)^{\frac{r}{s}}d\mu\right)^{\frac{s}{r}}\mu(X)^{1-\frac{s}{r}}$$

We say that if $$f,g:X\to\mathbb{C}$$, $$f=g$$ almost everywhere on $$X$$ if $$\mu(\{x|f(x)\neq g(x)\})=0$$. Observe that this is an equivalence relation on $$\mathcal{L}^p$$

If $$(X,\Sigma,\mu)$$ is a measure space, define the space $$L^p$$ to be the set of all equivalence classes of functions in $$\mathcal{L}^p$$

Theorem
The $$L^p$$ space with the $$\|\cdot\|_p$$ norm is a normed linear space, that is,
 * $$\|f\|_p\geq0$$ for every $$f\in L^p$$, further, $$\|f\|_p=0\iff f=0$$
 * $$\|\lambda\|_p=|\lambda|\|f\|_p$$
 * $$\|f+g\|_p\leq\|f\|_p+\|g\|_p$$ . . . (Minkowski's inequality)

Proof

1. and 2. are clear, so we prove only 3. The cases $$p=1$$ and $$p=\infty$$ (see below) are obvious, so assume that $$0<p<\infty$$ and let $$f,g\in L^p$$ be given. Hölder's inequality yields the following, where $$q$$ is chosen such that $$1/q+1/p = 1$$ so that $$p/q = p-1$$:

$$\displaystyle\int_X|f+g|^pd\mu=\int_X|f+g|^{p-1}|f+g|d\mu\leq\int_X|f+g|^{p-1}(|f|+|g|)d\mu$$

$$\leq\displaystyle\left(\int_X|f+g|^{(p-1)q}d\mu\right)^\frac{1}{q}\|f\|_p+\left(\int_X|f+g|^{(p-1)q}d\mu\right)^{\frac{1}{q}}\|g\|_p=\|f+g\|_p^{\frac{p}{q}}\|f\|_p+\|f+g\|_p^{\frac{p}{q}}\|g\|_p.$$

Moreover, as $$t\mapsto t^p$$ is convex for $$p>1$$,

$$\displaystyle \frac{|f+g|^p}{2^p} = \left|\frac{f}{2}+\frac{g}{2}\right|^p\leq \left(\frac{|f|}{2}+\frac{|g|}{2}\right)^p \leq \frac{1}{2}|f|^p + \frac{1}{2}|g|^p.$$

This shows that $$\|f+g\|_p<\infty$$ so that we may divide by it in the previous calculation to obtain $$\|f+g\|_p\leq\|f\|_p+\|g\|_p$$.

Define the space $$L^{\infty}=\{f|X\to\mathbb{C},f\text{ is bounded almost everywhere}\}$$. Further, for $$f\in L^{\infty}$$ define $$\|f\|_{\infty}=\sup\{|f(x)|:x\notin E\}$$