Why these operators are not bounded

Question

In the space $L^2(\mathbb{R})$ I'm studying the wave equation $$\frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2}.$$ By taking the second derivative operator $Au=u''$, we can define the associated cosine function as $$(C(t)u)(x):=\frac{1}{2}(u(x+t)+u(x-t)) \quad \forall x,t \in \mathbb{R}, \\$$ where its derivative is $$(C'(t)u)(x)=\frac{1}{2}(u'(x+t)-u'(x-t)), \quad \forall x,t \in \mathbb{R}.$$ Why $C'(t)$ is not bounded in $L^2(\mathbb{R})$? any counterexample ?

Answer 1

So first, not all of these operators are unbounded. If $t=0$, we obtain $A(0)u(x) = \frac{1}{2}(u'(x) - u'(x)) = 0$, which is obviously bounded.

On the other hand, differential operators are rarely bounded. Let's apply your operator to one of the usual candidates, $\sin(x^2)$. For $t \neq 0$, we obtain $A(t)u(x) = \frac{1}{2}(2x\cos((x+t)^2) - 2x \cos((x-t)^2) = x(\cos((x+t)^2) - \cos((x-t)^2))$, and this is not an $L^2(\mathbb{R})$ function for general $t$, since $\int_{\mathbb{R}} |A(t)u(x)|^2 dx$ diverges. And the choice of $p = 2$ isn't really relevant, this integral diverges with any $1 \leq p \leq \infty$