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.