Check whether operator $T: L^1[0,1] \to L[0,1]$ is compact.

Question

I need to check whether the integral operator $T:L^1[0,1] \to L^1[0,1]$ that sends $f \to \int_0^x f(y)dy$ is compact or not. I am guessing it is not but I am struggling on proving it. My best shot I think it would be to construct some bounded sequence whose image has no convergent subsequence.

I guess it is not because I have already showed that the same operator but from $T:L^1[0,1] \to C[0,1]$ is not compact. I did it by means of the Arzela Ascoli theorem, where I proved that the image of the unit ball is not equicontinuous. This theorem however can not be applied to the space $L^1[0,1]$ to check compactnes.

What could be a convenient way to prove/disprove compactness of this operator between these spaces? Thanks in advance!

Edit: I have seen the post you guys suggested, however there seems to be a flaw in the argument when $p=1$. This is since, if we use Holder's inequality with such $p$ then of coarse $q$ is infinite. But in such a setting the second line below the (star) seems wrong, since the $q$ norm of the simple function will be $1$ and not $1/n$.

Answer 1

Here is a rather humdrum way of showing $T$ is compact:

Write $(Tf)(x) = \int_0^1 K(x,y) f(y)dy$, where $K(x,y) = 1_{[0,\infty)}(x-y)$.

Let $P_n = \{ ({1 \over 2^n})^2 [j,j+1) \times [k,k+1) \mid j,k=0,...,2^n-1\}$ be a partition of $[0,1)^2$ and define $\Delta = \{ (x,y) \in [0,1]^2 \mid y \le x \}$.

Pick some $R \in P_n$ and note that we can write the characteristic of $R$ as $1_R(x,y) = 1_{R_x}(x) 1_{R_y}(y)$, were $R= R_x \times R_y$.

Define the simple function $K_n = \sum_{R \in P_n, R \subset \Delta} 1_R$. It is straightforward to see that $K_n(x,y) \to K(x,y)$ for all $x,y \in [0,1)$.

Define $(T_n f)(x) = \int_0^1 K_n(x,y) f(y)dy$ and note that we can write $(T_n f)(x) = \sum_{R \in P_n, R \subset \Delta} 1_{R_x}(x) \int_0^1 1_{R_y}(y) f(y)dy$, in particular each $T_n$ has finite range and hence compact.

Since $T = \lim T_n$, we see that $T$ is compact.

Answer 2

The operator is indeed compact. My proof uses the fact that the limit of finite rank operators is compact.

Consider the operator $$T_n(f)(x) = \int_{0}^{\frac{\lfloor nx \rfloor}{n}} f(y) dy.$$ Then we have $$\lvert{(T_n - T)f(x)\rvert} = \int_{\frac{\lfloor nx \rfloor}{n}}^{x} \lvert{f\rvert}(y) dy \leq \int_{\frac{\lfloor nx \rfloor}{n}}^{\frac{\lfloor nx \rfloor}{n} + n^{-1}}\lvert{f\rvert}(x) dx.$$ Thus, on the interval $x \in [\frac{i}{n}, \frac{i + 1}{n}]$, we have $$\lvert{(T_n - T)f(x)\rvert} \leq \int_{i / n}^{(i + 1) / n} \lvert{f\rvert}(x) dx.$$ So we conclude that $$\lVert (T_n - T)f \rVert_{L^1} \leq \frac{1}{n} \sum_{i = 0}^{n-1} \int_{i / n}^{(i + 1) / n} \lvert{f\rvert}(x) dx = \frac{1}{n} \lVert{f\rVert}_{L^1}.$$ Thus $T_n \to T$ in operator norm. Limit of finite rank operator is compact, so $T$ is compact.