$T(h)=\langle h,u\rangle v$ is a compact operator on Hilbert space $H$

Question

For a Hilbert space $H%$, I'm trying to show that for fixed $u,v\in H$ the operator $T(h)=\langle h,u\rangle v$ is an element of $\mathcal{L}(H)$ and is compact. Clearly $T$ is a map from $H$ to itself and is linear. To show T is bounded, first thing I tried doing was expand out $\lVert Th\rVert $ and only got $\lVert Th\rVert^2=|\langle h,u\rangle |^2\langle v,v\rangle$ which doesn't seem to help. Any help is appreciated.

Answer 1

By C-S inequality $\|Th\| \leq \|h\| \|u\|\|v\|$ which implies $\|T\| \leq \|u\|\|v\|$.

Any operator with finite dimensional range is compact because any closed and bounded set in the range of such an operator is compact. In this case the range of $T$ is contained in the one dimensional space spanned by $v$.

Answer 2

The boundedness follows from Cauchy-Schwarz, as others have mentioned.

To show it's a compact operator, it suffices to show that $T(B)$ is contained in a compact set, where $B$ is the unit ball. But by the aforementioned Cauchy-Schwarz argument you can see that for $\|x\| \le 1$, $T(x)$ is always a scalar multiple of $v$ with norm at most $\|u\|$. So $T(B)$ is contained in the set $\{tv : -\|u\| \le t \le \|u\|\}$. This is a continuous image of the closed interval $[-1, 1]$ and hence compact.

Or to think about it another way: if $x_n$ is a sequence with $\|x_n\| \le 1$ for all $n$, then we have $T(x_n) = t_n v$ where $|t_n| \le \|u\|$. A bounded sequence of real numbers has a convergent subsequence (Bolzano-Weierstrass), call it $t_{n_k}$ and call the limit $t_0$. Then $T(x_{n_k}) = t_{n_k} v \to t_0 v$.

A similar argument shows that every operator whose image is finite dimensional (i.e. every finite rank operator) must be compact. Here, the image is has dimension 1.