I am able to prove the following generalization of Schur's test using the Riesz-Thorin interpolation theorem, however I have been stuck for days now trying to prove it using Young's inequality:

Let the integral operator $T$ from functions $f: X \to \Bbb C$ to functions $Tf: Y \to \Bbb C$ be defined via the kernel $K: X \times Y \to \Bbb C$ , which is some measurable function. Moreover, let

$$ \| K(x,\cdot) \| _{L^{q_0}(Y) } \leq 1 $$

$$\| K(\cdot,y)\|_{L^{p_1'}(X) } \leq 1$$

for all $x\in X$ and all $y\in Y$.

Then for every $0<\theta<1$ and all $f\in L^{p_\theta}$

$$\| Tf \| _{L^{q_\theta}(Y) } \leq \| f\|_{L^{p_\theta}(X) } $$


$${1\over p_\theta} = {1-\theta\over p_0} + {\theta\over p_1} $$

$${1\over q_\theta} = {1-\theta\over q_0} + {\theta\over q_1} $$

$${1} = {1\over p_1} + {1\over p_1'} $$

and $p_0=1$ and $q_1=+\infty$.

I am able to prove the special case $p_1'=q_0=1$ via Hölder's as well as Young inequalities. However, I am making zero progress trying to prove the general case. I am struggling with this now for almost a week and I would greatly appreciate help! From online sources I know that a proof based on Young's inequality is possible. Young's inequality for non-negative reals $x,y$ is $$ xy \leq x^r/r + y^s/s$$ for dual exponents satisfying $1/r+1/s=1$ for $1<r<\infty$. Thanks in advance.

  • $\begingroup$ I think I finally got it! I had been using Young's inequality only with two exponents, i.e. dual exponents. But there is a version of Young's inequality for more then 2 exponents. For example $wxyz\leq w^a/a+x^b/b+ y^c/c+z^d/d$ for positive finite exponents satisfying $1/a+1/b+1/c+1/d=1$. Once I realized this trick, I was able to make some progress. When I finish, I hope to write it up and post it here. But if anybody has a different approach, I would be grateful for any additional insight. $\endgroup$ – Gandhi Viswanathan May 15 '13 at 15:03
  • $\begingroup$ The 3-way Young's or Hölder's inequality approach does not work. I have now hit a wall, having tried everything I know. $\endgroup$ – Gandhi Viswanathan Apr 30 '14 at 20:08

We want to show that $$\int_Y \int_X |K(x, y)| |f(x)| |h(y)| \, dx \, dy \le 1$$ for $f \in L^{p_\theta}(X)$ with $||f||_{L^{p_\theta}(X)} = 1$ and $h \in L^{q_\theta'}(Y)$ with $||h||_{L^{q_\theta'}} = 1$.

To show this, we use Young's inequality for three variables. That is, for any $x, y, z \ge 0$ and $1 < r_1, r_2, r_3 < \infty$ such that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = 1$, we have $$xyz \le \frac{1}{r_1}x^{r_1} + \frac{1}{r_2}y^{r_2} + \frac{1}{r_3}z^{r_3}.$$ Pick $r_1, r_2, r_3$ by $$\frac{1}{r_1} = \frac{1}{q_\theta}, \hspace{0.2in} \frac{1}{r_2} = \frac{1}{p_\theta'}, \hspace{0.2in} \frac{1}{r_3} = \frac{1}{p_\theta} - \frac{1}{q_\theta} = \frac{1}{q_\theta'} - \frac{1}{p_\theta'}.$$ Note $$\frac{1}{r_3} = 1 - \frac{1 - \theta}{q_0} - \frac{\theta}{p_1'} \ge 1 - (1 - \theta) - \theta = 0.$$ If $\frac{1}{r_3} = 0$, then we proceed as below but using Young's inequality for the two variables $r_1$ and $r_2$ and ignoring all terms with $r_3$.

Also let $$r = \frac{q_0}{r_1}, \hspace{0.2in} s = \frac{p_1'}{r_2}, \hspace{0.2in} t = \frac{p_\theta}{r_1},$$ $$u = \frac{p_\theta}{r_3}, \hspace{0.2in} v = \frac{q_\theta'}{r_2}, \hspace{0.2in} w = \frac{q_\theta'}{r_3}.$$ By construction, these satisfy $$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = 1,$$ $$r + s = 1, \hspace{0.2in} t + u = 1, \hspace{0.2in} v + w = 1,$$ $$rr_1 = q_0, \hspace{0.1in} sr_2 = p_1', \hspace{0.1in} tr_1 = p_\theta, \hspace{0.1in} vr_2 = q_\theta', \hspace{0.1in} ur_3 = p_\theta, \hspace{0.1in} wr_3 = q_\theta'.$$ Thus by Young's inequality for three variables (or two variables if $r_3$ is infinite), $$\left|K(x, y) f(x) h(y)\right| = (|K(x, y)|^{r}|f(x)|^t)(|K(x, y)|^s |h(y)|^v)(|f(x)|^u |h(y)|^w) \le \frac{1}{r_1} |K(x, y)|^{rr_1}|f(x)|^{tr_1} + \frac{1}{r_2} |K(x, y)|^{sr_2} |h(y)|^{vr_2} + \frac{1}{r_3} |f(x)|^{ur_3} |h(x)|^{wr_3}$$ $$\le \frac{1}{r_1} |K(x, y)|^{q_0} |f(x)|^{p_\theta} + \frac{1}{r_2} |K(x, y)|^{p_1'} |h(y)|^{q_\theta'} + \frac{1}{r_3}|f(x)|^{p_\theta} |h(y)|^{q_\theta'} \hspace{0.2in} (1).$$ Using Fubini-Tonelli and the assumption that $||K(x, \cdot)||_{L^{q_0}(Y)} \le 1$ for almost every $x \in X$ \begin{eqnarray*} \int_{Y} \int_X |K(x, y)|^{q_0} |h(y)|^{q_\theta'} \, dx \, dy & = & \int_X \int_Y |K(x, y)|^{q_0} |h(y)|^{q_\theta'} \, dy \, dx\\ & \le & \int_X ||K(x, \cdot)||_{L^{q_0}(Y)}^{q_0} \, dx \int_Y |h(y)|^{q_\theta'} \, dy \le ||h||_{L^{q_\theta'}(Y)}^{q_\theta'} \le 1. \end{eqnarray*} Likewise, using the assumption that $||K(\cdot, y)||_{L^{p_1'}(X)} \le 1$ for almost every $y \in Y$, $$\int_Y \int_X |K(x, y)|^{p_1'} |f(x)|^{p_\theta} \, dx \, dy \le 1.$$ Finally, $$\int_Y \int_X |f(x)|^{p_\theta} |h(y)|^{q_\theta'} \, dx \, dy \le ||f||_{L^{p_\theta}(X)}^{p_\theta}||h||_{L^{q_\theta'}(Y)}^{q_\theta'} \le 1.$$ Integrating Inequality (1) and substituting these three estimates, $$\int_Y \int_X |K(x, y)| |f(x)| |h(y)| \, dx \, dy \le \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = 1.$$

There appears to be enough symmetry in the system of ten equations for the nine variables $r_1, r_2, r_3, r, s, t, u, v, w$ that one equation is redundant and we obtain a unique solution.

  • $\begingroup$ I have not checked this answer, but it basically tries to correct my old 3-way Young ineq. approach. But you still get an over-determined system: 10 eqs for 9 vars. This is the same situation as in my old 3-way Young approach. $\endgroup$ – Gandhi Viswanathan May 3 '14 at 19:17
  • $\begingroup$ See my comment at the bottom. The solution does work because of the slightly more general approach. $\endgroup$ – bwoodhouse729 May 4 '14 at 3:30
  • $\begingroup$ I need to understand better. Let me check your calculations and I'll get back to you within 24h. $\endgroup$ – Gandhi Viswanathan May 5 '14 at 13:50
  • $\begingroup$ I checked and it works! Thanks! $\endgroup$ – Gandhi Viswanathan May 8 '14 at 18:00
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    $\begingroup$ I saw that your approach didn't quite work, so I tried to solve a slightly more general system, expecting the approach to work on principle. The new system only took about 20 minutes to solve by hand. $\endgroup$ – bwoodhouse729 May 9 '14 at 15:12

[EDIT: This answer, in fact, is wrong. See comments.]

I was finally able to use a three-way Young's inequality to prove the theorem. However, I was able to see a very slightly more direct proof using Hölder's inequality. Both theorems exploit convexity, whereas Riesz-Thorin relies on complex analyticity.

Here is the sketch of the proof using Hölder's inequality:

Below, $\mu$ and $\nu$ are the measures on $X$ and $Y$ respectively.

The idea is to prove the theorem for simple functions $f$ of finite measure support and then use monotone convergence.

$$ \|~T f~\|_{L^{q_\theta} } \leq \sup_{\|{h}\|_{L^{q_\theta'}(Y) }\leq 1} | \int_Y \int_X |K(x,y)| |f(x)| ~d\mu(x) ~h(y) ~d\nu(y) ~| $$

so that it suffices to show that

$$ | \int_Y \int_X |K(x,y)f(x) h(y)| ~d\mu(X) ~d\nu(y) ~| \leq \|{f}\|_{L^{p_\theta}(X) } \|{h}\| _{L^{q_\theta'}(Y) } ~. $$

We can exploit homogenization symmetry in the claim. Let us normalize $\|{f}\|_{L^{p_\theta}(X) }=\|{h}\| _{L^{q_\theta'}(Y) }=1$.

Denoting $L^p(X\times Y)$ as $L^p$ for convenience, we can use Hölder's inequality for multiple exponents as follows:

\begin{align} \|{ K f h }\|_{L^1} &\leq \|{ K^{p_1'/r_1} f^{p_\theta/r_1} } \|_{L^{r_1}} \|{ K^{q_0/r_2} h^{q_\theta'/r_2} }\|_{L^{r_2}} \|{ f^{p_\theta/r_3} h^{q_\theta'/r_3} } \|_{L^{r_3}} \\ 1&= {1\over r_1} +{1\over r_2} +{1\over r_3} \\ 1&= {p_1'\over r_1} + {q_0\over r_2}\\ 1&= {p_\theta\over r_1} + {p_\theta\over r_3} \end{align}

which has solution the solution

\begin{align} {1\over r_1} &={1\over q_0}\\ {1\over r_2} &= {q_\theta-q_0 \over q_\theta p_1' }\\ {1\over r_3} &= {1\over q_0 } - {1\over p_\theta} ~. \end{align}

Since $q_\theta >q_0$ exponents $r_1,~r_2$ are finite. If $1/r_3$ becomes zero then the three-way Hölder's inequality becomes the standard Hölder's inequality.

Each of the $L^r$ norms can be evaluated by the Fubini-Tonelli theorem. Since each of 3 the norms on the RHS equals 1, we finally get, after putting everything together, $$ \| Tf \|_{q_\theta} \leq 1$$ and the claim follows.


I don't have enough points to comment, but the above solution doesn't seem to work, for example, one can just take $\theta = 1/7$, $q_{0} = 3$, $p_{1} = 5$, $q_{1} = \infty$, $p_{0} = 1$. This makes $q_{\theta} = 7/2$, $p_{1}' = 5/4$, $p_{\theta}' = 35/4$, $p_{\theta} = 35/31$, and $q_{\theta}' = 7/5$. Given these numbers, we cannot find $r_{1}, r_{2}, r_{3}$ as above.

  • $\begingroup$ For a a few long minutes, I was perplexed! But it has been some time since I asked the original question and I noticed, reading the OP again, that $p_0$ is given as 1. In your counter-example, you have taken $p_0=5$. So the counter-example is not applicable. However, I have not checked to see whether we can find valid counter-examples (in which case it would mean that the 3-way Young's inequality approach above fails). $\endgroup$ – Gandhi Viswanathan Apr 30 '14 at 12:37
  • $\begingroup$ I have taken $p_{0} = 1$, $p_{1} = 5$. I think the issue is that in the system of equations above (immediately after you apply Holder's inequality), you also need $q_{\theta}'/r_{2} + q_{\theta}'/r_{3} = 1$. Now you have 4 equations for 3 variables which makes it hard to find solutions. $\endgroup$ – sqw Apr 30 '14 at 14:15
  • $\begingroup$ Sorry for the mistake about $p_0$. You are 100% correct about the missing equation. (But I have not checked whether this leads to overdetermination.) So now my original question again has no answer. If we try to use a 4-way Young's inequality, I expect it will again not work, because there will be, in addition to a new exponent, a new constraint. $\endgroup$ – Gandhi Viswanathan Apr 30 '14 at 15:22

Actually two-way Young's inequality is good enough:

Notice that since $ p_0 =1$ and $ q_1 = \infty $ we have $ p_\theta ( 1 -\theta ) + \frac{\theta p_\theta }{ p_1 } =1$ and $q_\theta' \theta +\frac{(1 -\theta) q_\theta' }{q_0 ' } =1 $. So write $$|f(x) h(y) | = |f(x)|^{p_\theta ( 1 -\theta )} |h(y)|^{\frac{(1 -\theta) q_\theta' }{q_0 ' }} |f(x)|^{\frac{\theta p_\theta }{ p_1 }}|h(y)|^{q_\theta' \theta}$$Thus by Young's inequality we have $$ |f(x) h (y) | \le (1 - \theta ) |f(x) |^{p _\theta } |h (y) |^{ \frac{q_\theta ' }{ q_0 ' }} + \theta |f(x) |^{ \frac{p_\theta}{p_1}} |h(y) |^{ q_\theta ' } $$ Then $$\int_Y \int_X |K(x,y ) f(x) h (y)| d \mu (x) d \mu (y) \le ( 1 - \theta ) \int_Y \int_X |K (x,y )|f(x) |^{p _\theta } |h (y) |^{ \frac{q_\theta ' }{ q_0 ' }} d\mu (x) d \mu (y) + \theta \int _ Y \int_X K(x,y ) |f(x) |^{ \frac{p_\theta}{p_1}} |h(y) |^{ q_\theta ' } d\mu (x) d \mu (y) $$

Since $ K(x, \ast )$ is in $L^{q_0 }$ for almost every $x$ and $ |h(y) |^{ \frac{q_\theta ' }{ q_0 ' }} $ is in $L^{ q_0 ' } $, by holder's inequality we have $$\int_Y |K (x,y )| |h (y) |^{ \frac{q_\theta ' }{ q_0 ' }} d \mu (y) \le 1$$ for almost everywhere $x$. And similarly we have $$\int_X K(x,y ) |f(x) |^{ \frac{p_\theta}{p_1}} d\mu (x) \le 1$$ for almost everywhere $y$.

Thus by Fubini's theorem we have $$ (1 - \theta ) \int _Y \int_X |K (x,y )|f(x) |^{p _\theta } |h (y) |^{ \frac{q_\theta ' }{ q_0 ' }} d\mu (x) d \mu (y) \le (1 - \theta ) $$ and $$\theta \int _ Y \int_X K(x,y ) |f(x) |^{ \frac{p_\theta}{p_1}} |h(y) |^{ q_\theta ' } d\mu (x) d \mu (y) \le \theta $$

Sum up the two parts we get $$ \int_Y \int_X |K(x,y ) f(x) h (y)| d \mu (x) d \mu (y) \le 1- \theta + \theta =1$$ as wanted.

  • $\begingroup$ Very nice. Assuming this answer is right, it is just what I was looking for. I willl check this answer very soon and get back (today hopefully). I had not thought of doing "two rounds" of Young/Holder. $\endgroup$ – Gandhi Viswanathan May 3 '14 at 14:17
  • $\begingroup$ It checks out very nicely. I'll accept this answer and give the bounty. Thank you! $\endgroup$ – Gandhi Viswanathan May 3 '14 at 19:14
  • $\begingroup$ You are welcome! $\endgroup$ – user112564 May 3 '14 at 21:56
  • $\begingroup$ May I ask: how did you know exactly how to split and factor just right, for Young's and Hölder's inequalities? Did you get the information in a book or do you have enough prior experience with this type of calculation to just be able to "do it" without much effort? How did you know, before hand, that the 2-way Young/Hölder method would work? $\endgroup$ – Gandhi Viswanathan May 7 '14 at 21:27
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    $\begingroup$ I didn't know 2-way Young's would work. I felt it might work, and I tried a long time to figure out how to split it. $\endgroup$ – user112564 May 8 '14 at 1:16

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