Is convolution a coercive bilinear form in $L^2$ -space?

Question

This is one of the problems in functional analysis course I'm having. Suppose $f,g \in L^2(0,10)$. Then define a bilinear form $$ B:(f,g)\mapsto \int_0^{10} f(x)g(10-x) dx. $$ Now I have to find out whether this is coercive, eg. it fulfills the condition $$ B(f,f) \geq C \| f \| ^2 $$ when $C>0$ and $f \in L^2(0,10)$, or not.

My attempt: I guess it is coercive. Since $L^2(0,10)$ is a Hilbert space we can write bilinear forms as $$ B(f,g) = (f,Tg) $$ where $T$ is a continuous linear operator. Here it is $Tg = g(10-x)$. I have a feeling that it is possible to find a constant $c>$ such that $Tg\geq c g$. Thus the coerciveness follows. I have not found such constant. My best guess is around $g(10-x) \geq \frac{1}{10} g(x)$ but I'm not sure about it.

Edit: I assume the scalar field is $\mathbb{R}$.

Answer 1

Take $f(x) = \chi_{[0,5)}$. Then $f(x)f(10-x) \equiv 0$, so $B(f,f)=0 < 5=\lVert f \rVert_2^2$