Measure Theory/L^p spaces

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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[edit]

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


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)+\beta\left(f(x)-\int_Xfd\mu\right)\geq 0, which completes the proof.


  1. Putting \phi(x)=e^x,
    \displaystyle e^{(\int_Xfd\mu)}\leq\int_Xe^fd\mu
  1. If X is finite, \mu is a counting measure, and if f(x_i)=p_i, then
    \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)

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

Holder's inequality[edit]

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



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


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


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


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


The L^p space with the \|\cdot\|_p norm is a normed linear space, that is,

  1. \|f\|_p\geq0 for every f\in L^p, further, \|f\|_p=0\iff f=0
  2. \|\lambda\|_p=|\lambda|\|f\|_p
  3. \|f+g\|_p\leq\|f\|_p+\|g\|_p . . . (Minkowski's inequality)


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:



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\}