$\langle Tf, f \rangle =0 \implies \langle Tg, f \rangle =0$

Question

The usual inner product on $L^2$ is $\langle f, h \rangle = \int f(x)\overline{h}(x) dx.$ Assume that $T:L^2(\mathbb R)\to L^2(\mathbb R)$ be an (linear) operator.

Assume that $\langle Tf, f \rangle =0$ for all $f\in L^2(\mathbb R).$

Can we say that $\langle Tg, f \rangle =0$ for all compactly supported function $g$ and $f$?

Side thought: I might need to use polarization identity

@Adayah: The set of compactly supported functions are subset of $L^2$ — Math Learner, Jun 29 '18 at 16:02
@Adayah: Yes. Thanks. You are right. I've edited the question. — Math Learner, Jun 29 '18 at 16:06
This was already discussed here: https://math.stackexchange.com/questions/1307747/prove-that-if-langle-tx-x-rangle-0-for-all-x-in-x-then-t-0?rq=1 — supinf, Jun 29 '18 at 16:39

score 2 · Answer 1 · answered Jun 29 '18 at 16:12

2

Define $T: L^2 \to L^2 ,$ $f=\sum_i a_i e_i \in L^2,$ $ Tf =-a_2 e_1 +a_1 e_2 $. Then $<Tf , f > =0 $ for all $f .$ But it is not true that $<Tf , g>=0$ for all $f, g.$

answered Jun 29 '18 at 16:12

1

OP seems to be using complex-valued functions on $\mathbb{R}$, and in that case, $\langle Tf, f \rangle = \pm 2 \operatorname{im}(a_1 \bar a_2)$ (with $\pm$ depending on the convention). – Daniel Schepler Jun 29 '18 at 16:19

score 2 · Accepted Answer · answered Jun 29 '18 at 16:36

Based on the statement of the question, I'll assume that $L^2(\mathbb{R})$ is the space of square-integrable complex-valued functions on $\mathbb{R}$, as a complex Hilbert space. In that case, you can indeed use an argument similar to the polarization identity as follows:

$$\langle T(f+g), f+g \rangle = \langle Tf, f \rangle + \langle Tf, g \rangle + \langle Tg, f \rangle + \langle Tg, g \rangle = \langle Tf, g \rangle + \langle Tg, f \rangle = 0.$$

Similarly,

$$\langle T(f+ig), f+ig \rangle = \langle Tf, f \rangle - i \langle Tf, g \rangle + i \langle Tg, f \rangle + \langle Tg, g \rangle = -i \langle Tf, g \rangle + i \langle Tg, f \rangle = 0.$$

It follows that $\langle Tf, g \rangle = \langle Tg, f \rangle = 0$.

score 1 · Answer 3 · answered Jun 29 '18 at 17:33

If you are looking at $L^2(\mathbb{R})$ as a complex Hilbert space then the answer is yes:

$$\langle (T - T^*)f,f\rangle = \langle Tf,f\rangle - \langle f,Tf\rangle = 0, \forall f \in L^2(\mathbb{R})$$

Now, $T - T^*$ is a normal operator so

$$\|T - T^*\| = \sup_{\|f\| = 1} \left|\langle (T - T^*)f,f\rangle\right| = 0 \implies T - T^* = 0$$

Hence $T = T^*$ so the same argument gives

$$\|T\| = \sup_{\|f\| = 1} \left|\langle Tf,f\rangle\right| = 0 \implies T = 0$$

$\langle Tf, f \rangle =0 \implies \langle Tg, f \rangle =0$

3 Answers3

Linked