Young's inequality for integral operators

Remarks

Proof idea for $1 < p < \infty$: let $p'$ be the conjugate exponent and write $\abs{K} = \abs{K}^{1/p'} \abs{K}^{1/p}$. Hölder's inequality in $y$ and the first hypothesis give $\abs{T f(x)}^p \le C^{p/p'} \int_Y \abs{K(x, y)} \abs{f(y)}^p \, d\nu(y)$. Integrating in $x$, Tonelli's theorem and the second hypothesis give $\norm{T f}_p^p \le C^{p/p'} \cdot C \norm{f}_p^p = C^p \norm{f}_p^p$. The case $p = \infty$ uses only the first hypothesis, and the case $p = 1$ uses only the second.

Example: for $X = Y = \R^n$ with Lebesgue measure and $K(x, y) = g(x - y)$ with $g \in L^1(\R^n)$, both hypotheses hold with $C = \norm{g}_1$. The result is $\norm{g * f}_p \le \norm{g}_1 \norm{f}_p$, the case $q = 1$ of Young's convolution inequality.

Both hypotheses are needed for the full range of $p$. On $X = Y = (0, \infty)$, the Hardy kernel $K(x, y) = 1/x$ for $y < x$ and $K(x, y) = 0$ otherwise has $\int_0^\infty K(x, y) \, dy = 1$ for every $x$, but $\int_0^\infty K(x, y) \, dx = \infty$ for every $y$. The operator $T f(x) = \frac{1}{x} \int_0^x f(y) \, dy$ is bounded on $L^p$ for $1 < p \le \infty$ by Hardy's inequality, but not on $L^1$: for $f$ the indicator function of $(0, 1]$, $T f(x) = 1/x$ for $x > 1$, which is not integrable.

Going deeper

For a student

This result is sometimes called the generalized Young inequality, since it contains the case $\norm{g * f}_p \le \norm{g}_1 \norm{f}_p$ of Young's convolution inequality. It is the quickest way to show that an integral operator with an integrable kernel acts boundedly on all $L^p$ spaces at once. For $p = 2$ it is the Schur test with constant weights; the Schur test allows other weights and so applies to more kernels. In the theory of integral equations it gives the bounds that the contraction principle needs for second-kind equations in $L^p$.