Consider the following operator $T:L^2([0,1]) \rightarrow L^2([0,1])$ given by $$Tu(x) = \int_{x^2}^{x} u(t) dt$$ They ask me to dermine whether or not the set $T(B_1)$ is precompact (or, equivalently, relatively compact) in $L^2([0,1])$, where $B_1$ denotes the unit ball in $L^2([0,1])$.
Any hint is greatly appreciated, thank you.
Here is a short direct proof of that the operator $Tu(x)=\int_0^1 1_{[x^2,x]}(t)u(t)\,\mathrm{d}t,$ is compact. We need the following statement: if $e_n(x)$ forms orthonormal basis in $L_2[0,1]$, then $e_n(x)e_m(t)$ forms orthonormal basis in $L_2([0,1]\times[0,1])$. This is a well-known simple statement, but if you wish, you can prove it yourself as an exercise.
Now, consider the generating function $K(x,t)=1_{[x^2,x]}(t)$ of our operator $T$. For all $n\in\mathbb{N}$ consider the integral operator $T_n$ with generating function $K_n(x,t)=\sum\limits_{k,l=1}^n c_{k,l}e_k(x)e_l(t)$, where $c_{k,l}=(K,e_k(x)e_l(t))$ are Fourier coefficients of $K$. It's obvious that $T_n$ has a finite rank (it's range are linear combination of functions $e_k(x)$, $k=1,2,...,n$). Hence $T_n$ -- is compact. Further, $T-T_n$ -- is a integral operator with generating function $K-K_n$. By Holder's inequality it's easy to proof that $\|T-T_n\|\leq \|K-K_n\|\to0$ according to the previous statement. Thus $T$ also compact.
Note that it is also possible to proof according to the criterion of precompactness, but it is much longer.
The operator can be written in the form $$ Tu(x)=\int_0^1 1_{[x^2,x]}(t)u(t)\,\mathrm{d}t, $$ where $1_{[x^2,x]}(t)$ is the characteristic function. This is a Hilbert-Schmidt integral operator with $K(x,t)=1_{[x^2,x]}(t)\in L^2([0,1]\times[0,1]).$ This operator is compact, meaning the image of bounded sets is precompact. Hence, $T(B_1)$ is precompact.
For more information, perhaps consider Proof of compactness. This proof assumes $K$ is continuous, but you may be able to tweak the proof.
