Prove that certain integral operator on $L^1[0,\infty)$ is compact

Question

I met the following exercise in functional analysis, which made me somewhat confused:

Let $f\in L^1[0,\infty)$ (say complex valued, the usual $L^p$ space definition, usual Lebesgue measure on $[0,\infty)$). Consider the integral operator $$H_{f}:L^1[0,\infty)\rightarrow L^1[0,\infty),~\left(H_{f} g\right)(x)=\int_{0}^{\infty} f(x+y) g(y) d y.$$ Prove that $H_f$ is a compact operator. (Of course by Fubini it's bounded.)

I was confused and had no idea how to prove this, since the compactness of integral operator is usually established for the dual index: c.f. Integral operator on $L^p$ is compact . So in the usual context, if the space is $L^p$, then the kernel function is in $L^q$. The exercise looks somewhat similar to something related to convolution, but the interval is not symmetric.

I guess this requires some "ad-hoc" method to deal with. Or is there any background/famous theorem /concrete properties of $L^p$ spaces related to this example? (I'm a beginner in functional analysis course; if this question is too naive, I sincerely apologize.)

Thanks a lot in advance for any suggestion or hint!

Answer 1

Hints:

1. If $(H_f^{n}g)(x)=\int_0^{n} f(x+y)g(y)dy$ then $H_f^{n} \to H_f$ in operator norm (by a Fubini argument) so we may suppose $f$ has compact support.

2. We can approximate $f$ by a continuous function on $[0,n]$ in $L^{1}$ norm and then by a polynomial. [Uniform convergence implies convergence in $L^{1}$ on $[0,n]$).

3. When $f$ is a polynomial the operator $H_f$ has finite rank.

Answer 2

I do not think I can beat the elegance of the solution of geetha290km. However, I would like to point out that there is a characterization of relative compactness in $L^p$-spaces reminiscant of the Arzelà-Ascoli theorem, namely the Kolmogorov-Riesz theorem:

Let $p \in [1, \infty)$, $A \subseteq L^p(\mathbb R^n)$ and consider the translation operator $$\tau_y \colon L^p(\mathbb R^n) \to L^p(\mathbb R^n), \quad \tau_y f(x) = f(x -y).$$ Then $A$ is relatively compact if and only if the following two assertions hold:

1. $A$ is equicontinuous, i.e., $\lVert \tau_y f - f \rVert_p \to 0$ as $y \to 0$ uniformly for $f \in A$.
2. $A$ is equi-tight, i.e., $\displaystyle \int_{\lvert x \rvert > r} \lvert f(x) \rvert^p \to 0$ as $r \to \infty$ uniformly for $f \in A$.

So by this theorem you just need to show that the set $H_f \mathrm B[0, 1]$ is both equicontinuous and equi-tight since then the theorem yields that $H_f \mathrm B[0, 1]$ is relatively compact, i.e., $H_f$ is a compact operator. Implicitely, I used the identification $L^1[0, \infty) \sim \{f \in L^1(\mathbb R) : f \mathbf 1_{(-\infty, 0)} = 0 \text{ a.e.}\}$.

Moreover, the above theorem generalizes in various different directions and there are even versions for weak compactness. This is mainly interesting due to the fact that weakly compact operators play a prominent role in Dunford-Pettis theory.

Answer 3

The space $e^{-x}C_0[0,\infty)$ is dense in $L^1(0,\infty),$ where $C_0[0,\infty)$ denotes the space of continuous functions vanishing at infinity.

The linear span $V$ of the functions $u_n(x)=e^{-nx},$ $n\ge 1,$ is uniformly dense in $C_0[0,\infty),$ by the Weierstrass theorem. Thus $e^{-x}V$ is dense in $e^{-x}C_0[0,\infty)$ with respect to $L^1$-norm, hence it is dense in $L^1(0,\infty).$

For $f\in e^{-x}V$ the corresponding operator $H_f$ has finite rank as a linear combination of rank one operators $$H_{u_n}g(x)=\int\limits_0^\infty e^{-n(x+y)}g(y)\,dy =\left [\int\limits_0^\infty e^{-ny}g(y)\,dy \right ]\, u_n(x)$$