## Riesz-Thorin interpolation theorem

I had, a while ago, the great pleasure of going through the proof of the Riesz-Thorin interpolation theorem. I believe I understand the general strategy of the proof, though for sure, I glossed over some details. It is my hope that in writing this, I can fill in the holes for myself at the more microscopic level.

Let us begin with a statement of the theorem.

Riesz-Thorin Interpolation Theorem. Suppose that $(X,\mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$ are measure spaces and $p_0, p_1, q_0, q_1 \in [1, \infty]$. If $q_0 = q_1 = \infty$, suppose also that $\mu$ is semifinite. For $0 < t < 1$, define $p_t$ and $q_t$ by

$\frac{1}{p_t} = \frac{1-t}{p_0} + \frac{t}{p_1}, \qquad \frac{1}{q_t} = \frac{1-t}{q_0} + \frac{t}{q_1}$.

If $T$ is a linear map from $L^{p_0}(\mu) + L^{p_1}(\mu)$ into $L^{q_0}(\nu) + L^{q_1}(\nu)$ such that $\left\|Tf\right\|_{q_0} \leq M_0 \left\|f\right\|_{p_0}$ for $f \in L^{p_0}(\mu)$ and $\left\|Tf\right\|_{q_1} \leq M_1 \left\|f\right\|_{p_1}$ for $f \in L^{p_1}(\mu)$, then $\left\|Tf\right\|_{q_t} \leq M_0^{1-t}M_1^t \left\|f\right\|_{p_t}$ for $f \in L^{p_t}(\mu)$, $0 < t < 1$.

We begin by noticing that in the special case where $p = p_0 = p_1$,

$\left\|Tf\right\|_{q_t} \leq \left\|Tf\right\|_{q_0}^{1-t} \left\|Tf\right\|_{q_1}^t \leq M_0^{1-t}M_1^t \left\|f\right\|_p$,

wherein the first inequality is a consequence of Holder’s inequality. Thus we may assume that $p_0 \neq p_1$ and in particular that $p_t < \infty$.

Observe that the space of all simple functions on $X$ that vanish outside sets of finite measure has in its completion $L_p(\mu)$ for $p < \infty$ and the analogous holds for $Y$. To show this, take any $f \in L^p(\mu)$ and any sequence of simple $f_n$ that converges to $f$ almost everywhere, which must be such that $f_n \in L^p(\mu)$, from which follows that they are non-zero on a finite measure. Denote the respective spaces of such simple functions with $\Sigma_X$ and $\Sigma_Y$.

To show that $\left\|Tf\right\|_{q_t} \leq M_0^{1-t}M_1^t \left\|f\right\|_{p_t}$ for all $f \in \Sigma_X$, we use the fact that

$\left\|Tf\right\|_{q_t} = \sup \left\{\left|\int (Tf)g d\nu \right| : g \in \Sigma_Y, \left\|g\right\|_{q_t'} = 1\right\}$,

where $q_t'$ is the conjugate exponent to $q_t$. We can rescale $f$ such that $\left\|f\right\|_{p_t} = 1$.

From this it suffices to show that across all $f \in \Sigma_X, g \in \Sigma_Y$ with $\left\|f\right\|_{p_t} = 1$ and $\left\|g\right\|_{q_t'} = 1$, $|\int (Tf)g d\nu| \leq M_0^{1-t}M_1^t$.

For this, we use the three lines lemma, the inequality of which has the same value on its RHS.

Three Lines Lemma. Let $\phi$ be a bounded continuous function on the strip $0 \leq \mathrm{Re} z \leq 1$ that is holomorphic on the interior of the strip. If $|\phi(z)| \leq M_0$ for $\mathrm{Re} z = 0$ and $|\phi(z)| \leq M_1$ for $\mathrm{Re} z = 1$, then $|\phi(z)| \leq M_0^{1-t} M_1^t$ for $\mathrm{Re} z = t$, $0 < t < 1$.

This is proven via application of the maximum modulus principle on $\phi_{\epsilon}(z) = \phi(z)M_0^{z-1} M_1^{-z} \mathrm{exp}^{\epsilon z(z-1)}$ for $\epsilon > 0$. The $\mathrm{exp}^{\epsilon z(z-1)}$ serves of function of $|\phi_{\epsilon}(z)| \to 0$ as $|\mathrm{Im} z| \to \infty$ for any $\epsilon > 0$.

We observe that if we construct $f_z$ such that $f_t = f$ for some $0 < \mathrm{Re} t < 1$. To do this, we can express for convenience $f = \sum_1^m |c_j|e^{i\theta_j} \chi_{E_j}$ and $g = \sum_1^n |d_k|e^{i\theta_k} \chi_{F_k}$ where the $c_j$‘s and $d_k$‘s are nonzero and the $E_j$‘s and $F_k$‘s are disjoint in $X$ and $Y$ and take each $|c_j|$ to $\alpha(z) / \alpha(t)$ power for such a fixed $t$ for some $\alpha$ with $\alpha(t) > 0$. We let $t \in (0, 1)$ be the value corresponding to the interpolated $p_t$. With this, we have

$f_z = \displaystyle\sum_1^m |c_j|^{\alpha(z)/\alpha(t)}e^{i\theta_j}\chi_{E_j}$.

Needless to say, we can do similarly for $g$, with $\beta(t) < 1$,

$g_z = \displaystyle\sum_1^n |d_k|^{(1-\beta(z))/(1-\beta(t))}e^{i\psi_k}\chi_{F_k}$.

Together these turn the LHS of the inequality we desire to prove to a complex function that is

$\phi(z) = \int (Tf_z)g_z d\nu$.

To use the three lines lemma, we must satisfy

$|\phi(is)| \leq \left\|Tf_{is}\right\|_{q_0}\left\|g_{is}\right\|_{q_0'} \leq M_0 \left\|f_{is}\right\|_{p_0}\left\|g_{is}\right\|_{q_0'} \leq M_0 \left\|f\right\|_{p_t}\left\|g\right\|_{q_t'} = M_0$.

It is not hard to make it such that $\left\|f_{is}\right\|_{p_0} = 1 = \left\|g_{is}\right\|_{q_0'}$. A sufficient condition for that would be integrands associated with norms are equal to $|f|^{p_t/p_0}$ and $|g|^{q_t'/q_0'}$ respectively, which equates to $\mathrm{Re} \alpha(is) = 1 / p_0$ and $\mathrm{Re} (1-\beta(is)) = 1 / q_0'$. Similarly, we find that $\mathrm{Re} \alpha(1+is) = 1 / p_1$ and $\mathrm{Re} (1-\beta(1+is)) = 1 / q_1'$. From this, we can solve that

$\alpha(z) = (1-z)p_0^{-1}, \qquad \beta(z) = (1-z)q_0^{-1} + zq_1^{-1}$.

With these functions inducing a $\phi(z)$ that satisfies the hypothesis of the three lines lemma, our interpolation theorem is shown for such simple functions, from which extend our result to all $f \in L^{p_t}(\mu)$.

To extend this to all of $L^p$, it suffices that $Tf_n \to Tf$ a.e. for some sequence of measurable simple functions $f_n$ with $|f_n| \leq |f|$ and $f_n \to f$ pointwise. Why? With this, we can invoke Fatou’s lemma (and also that $\left\|f_n\right\|_p \to \left\|f\right\|_p$ by dominated convergence theorem) to obtained the desired result, which is

$\left\|Tf\right\|_q \leq \lim\inf \left\|Tf_n\right\|_q \leq \lim\inf M_0^{1-t} M_1^t\left\|Tf_n\right\|_p \leq M_0^{1-t} M_1^t \left\|f\right\|_p$.

Recall that convergence in measure is a means to derive a subsequence that converges a.e. So it is enough to show that $\displaystyle\lim_{n \to \infty} \mu(\left\|Tf_n - Tf\right\| > \epsilon) = 0$ for all $\epsilon > 0$. This can be done by upper bounding with something that goes to zero. By Chebyshev’s inequality, we have

$\mu(\left\|Tf_n - Tf\right\| > \epsilon) \leq \frac{\left\|Tf_n - Tf\right\|_p^p}{\epsilon^p}$.

However, recall that in our hypotheses we have constant upper bounds on $T$ in the $p_0$ and $p_1$ norms respectively assuming that $f$ is in $L^{p_0}$ and $L^{p_1}$, which we can make use of.  So apply Chebyshev on any one of $q_0$ (let’s use this) and $q_1$, upper bound its upper bound with $M_0$ or $M_1$ times $\left\|f_n - f\right\|_{p_0}$, which must go to zero by pointwise convergence.

## Convergence in measure

Let $f, f_n (n \in \mathbb{N}) : X \to \mathbb{R}$ be measurable functions on measure space $(X, \Sigma, \mu)$. $f_n$ converges to $f$ globally in measure if for every $\epsilon > 0$,

$\displaystyle\lim_{n \to \infty} \mu(\{x \in X : |f_n(x) - f(x)| \geq \epsilon\}) = 0$.

To see that this means the existence of a subsequence with pointwise convergence almost everywhere, let $n_k$ be such that for $n > n_k$, $\mu(\{x \in X : |f_{n_k}(x) - f(x)| \geq \frac{1}{k}\}) < \frac{1}{k}$, with $n_k$ increasing. (We invoke the definition of limit here.) If we do not have pointwise convergence almost everywhere, there must be some $\epsilon$ such that there are infinitely many $n_k$ such that $\mu(\{x \in X : |f_{n_k}(x) - f(x)| \geq \epsilon\}) \geq \epsilon$. There is no such $\epsilon$ for the subsequence $\{n_k\}$ as $\frac{1}{k} \to 0$.

This naturally extends to every subsequence’s having a subsequence with pointwise convergence almost everywhere (limit of subsequence is same as limit of sequence, provided limit exists). To prove the converse, suppose by contradiction, that the set of $x \in X$, for which there are infinitely many $n$ such that $|f_n(x) - f(x)| \geq \epsilon$ for some $\epsilon > 0$ has positive measure. Then, there must be infinitely many $n$ such that $|f_n(x) - f(x)| \geq \epsilon$ is satisfied by a positive measure set. (If not, we would have a countable set in $\mathbb{N} \times X$ for bad points, whereas there are uncountably many with infinitely bad points.) From this, we have a subsequence without a pointwise convergent subsequence.