$\langle Tf, f \rangle =0 \implies \langle Tg, f \rangle =0$ 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 
 A: 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 $f, g.$
A: 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 $\langle Tf, g \rangle = \langle Tg, f \rangle = 0$.
A: 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
$$\|T\| = \sup_{\|f\| = 1} \left|\langle Tf,f\rangle\right| = 0 \implies T = 0$$ 
