What Is the Adjoint of an Integral Operator on $L^p$?

An integral operator sends a function on one measure space to a function on another. Its adjoint acts in the reverse direction, and its kernel is transposed: the output variable and integration variable exchange roles. The precise statement depends on which dual spaces are being used, especially at $p=1$ and $p=\infty$.

The general formula

Let $(Y,\nu)$ and $(X,\mu)$ be measure spaces, and suppose

$$(Tf)(x)=\int_Y K(x,y)f(y)\,d\nu(y)$$

maps $L^p(Y,\nu)$ boundedly into $L^p(X,\mu)$. The Banach-space adjoint, also called the transpose, is the map

$$T':(L^p(X,\mu))^*\longrightarrow (L^p(Y,\nu))^*.$$

It reverses the direction of the operator. When $1<p<\infty$ and the usual $L^p$ duality applies, set $q=p/(p-1)$, so that $1/p+1/q=1$. Identifying the duals with $L^q$, the adjoint is represented by

$$(T'g)(y)=\int_X K(x,y)g(x)\,d\mu(x).$$

This is the formula for the bilinear dual pairing $\int f g$. For real-valued functions, it is the familiar formula without any convention issue.

For complex-valued functions, many texts use the sesquilinear pairing $\langle f,g\rangle=\int f\overline{g}$. With this convention, the adjoint formula is

$$(T^*g)(y)=\int_X \overline{K(x,y)}g(x)\,d\mu(x).$$

The bar is complex conjugation. On $L^2$, this is the usual Hilbert-space adjoint. The two formulas express the same reversal of variables; the apparent difference comes from how the duality pairing is defined.

These formulas require hypotheses ensuring that the operator is bounded and the integrals and change of order are legitimate. For example, a bounded measurable kernel on finite measure spaces gives a bounded operator on each $L^p$, $1\le p\le\infty$. More general kernels can be handled under suitable integrability or boundedness conditions, such as Schur-test hypotheses.

Why the kernel is transposed

Using the bilinear pairing, take $f\in L^p(Y)$ and $g\in L^q(X)$. If Fubini’s theorem applies, then

$$\begin{aligned} \int_X (Tf)(x)g(x)\,d\mu(x) &=\int_X\left(\int_Y K(x,y)f(y)\,d\nu(y)\right)g(x)\,d\mu(x)\\ &=\int_Y f(y)\left(\int_X K(x,y)g(x)\,d\mu(x)\right)\,d\nu(y). \end{aligned}$$

The final expression is the dual pairing of $f$ with $T'g$. Thus the adjoint integrates over the old output variable $x$, leaving a function of the old input variable $y$.

Worked example: the Volterra operator

On $[0,1]$ with Lebesgue measure, define

$$(Tf)(x)=\int_0^x f(y)\,dy.$$

Its kernel is $K(x,y)=1$ when $0\le y\le x\le1$, and $0$ otherwise. For $1<p<\infty$, its adjoint on $L^q([0,1])$ is

$$(T'g)(y)=\int_y^1 g(x)\,dx.$$

Indeed, the condition $y\le x$ in the original integral becomes the integration range $x\in[y,1]$ after exchanging the order of integration. With the complex sesquilinear convention, the kernel is real, so the formula is unchanged. The same integral formula defines the corresponding dual operator at the endpoint $p=1$ as a map on $L^\infty$.

What happens at $p=1$ and $p=\infty$?

For standard measure spaces, the dual of $L^1$ is represented by $L^\infty$. Thus, if $T:L^1(Y)\to L^1(X)$ is bounded, its Banach adjoint maps $L^\infty(X)$ to $L^\infty(Y)$, and the transposed-kernel formula often applies when the integrals are justified.

The dual of $L^\infty$ is generally not just $L^1$. It is a larger space of bounded linear functionals (often described using finitely additive measures). Consequently, for an operator on $L^\infty$, the Banach adjoint always exists as a map between the full dual spaces, but it need not be representable by an ordinary $L^1$ function or by the displayed kernel formula alone. Additional structure may yield a useful representation, but it should not be assumed automatically.

Common mistakes

• Keeping the same direction: If $T$ maps functions on $Y$ to functions on $X$, its adjoint maps functionals on $X$ back to functionals on $Y$.
• Using $p$ instead of its dual exponent: For $1<p<\infty$, the adjoint acts on $L^q$, where $1/p+1/q=1$.
• Forgetting the convention for complex functions: The Banach transpose under a bilinear pairing has no conjugate; the Hilbert-space adjoint under a sesquilinear pairing conjugates the kernel.
• Treating $L^\infty$ like a reflexive $L^p$ space: In general, $(L^\infty)^*$ cannot be identified with $L^1$.
• Ignoring integrability conditions: Exchanging integrals requires suitable hypotheses, such as absolute integrability or a theorem that otherwise justifies the exchange.

Does compactness change the formula?

No. Compactness is not needed to derive the transposed-kernel formula; boundedness and valid duality and integration arguments are the relevant issues. Separately, the Schauder theorem says that the adjoint of a compact bounded operator between Banach spaces is compact. This is a property of the operator, not a change to how its kernel is transposed.