I am new to math-exchange, so please let me know if I state my question in an incomprehensible way. My problem comes from the field of wave optics but can be stated in a quite general setting:
Problem statement: Given a real-valued function $f \in L^2(\mathbb R)$, that vanishes on the negative half-axis, $f|_{\mathbb R_ -} = 0$, consider the convolution $S(f):= s \ast f$ with a real-valued and symmetric kernel $s \in L^1(\mathbb R)$, i.e. $s(x) = s(-x) \in \mathbb R$ for all $x \in \mathbb R$. Is there a general bound of the form
$\|S(f)|_{\mathbb R_-}\|_{L^2} \leq C \|S(f)|_{\mathbb R_+}\|_{L^2}$
for some constant $C \geq 1$? It can be readily seen that this cannot hold for any $C < 1$ but numerical experiments indicate that it might hold for $C = 1$.
The idea behind it: Intuitively, the idea behind my question is whether convolutions with symmetric kernels also transport the mass of $f$ in some sufficiently symmetric manner into the negative- and positive direction.
Preliminary thoughts: I tried to take advantage of the fact that the operator $S$ commutes with the (anti-)symmetrization operations, $g^{\text s} := \frac 1 2 (g + g^-)$ and $g^{\text a} := \frac 1 2 (g - g^-)$ where $g^-(x) := g(-x)$:
$S(g)^{\text s} = S(g^{\text s})$ and $S(g)^{\text a} = S(g^{\text a})$.
Moreover, I think that it may be of use that the decomposition is orthogonal: $\|g\|_{L^2}^2 = \|g^{\text s}\|_{L^2}^2 + \|g^{\text a}\|_{L^2}^2$ for all real-valued $g \in L^2(\mathbb R)$. Finally, the condition $f|_{\mathbb R_-} = 0$ can be equivalently stated as
$f^{\text a}|_{\mathbb R_+} = f^{\text s}|_{\mathbb R_+}$ and $f^{\text a}|_{\mathbb R_-} = - f^{\text s}|_{\mathbb R_-}$.
Summary: I have the feeling that the problem is so general that there could already be an exhaustive and rather well-known answer to it. If so, however, I really do not know where to find it, probably lacking the right buzzwords.
Thanks for help!
The answer is NO. This was pointed out to me be Cong Shi in personal communication.
Counter-example: For simplicity, we take $s(x):= \delta( x + \frac 1 2 ) + \delta ( x - \frac 1 2 )$, where $\delta$ denotes the Dirac-delta-distribution. Then $s$ is real-valued and symmetric but technically $s \notin L^1(\mathbb R)$. However, we can approximate $s$ by a suitable Dirac-sequence $(s_k)_{k \in \mathbb R} \subset L^1(\mathbb R)$ to make the constructed example apply rigorously to the given setting.
Now, for $0 < a < 1$ and $0 < \varepsilon < \frac 1 2$, we define
$f(x) := \sum_{k=0}^\infty (-a)^k \boldsymbol 1_{[0; \varepsilon]} ( x - k), \qquad \boldsymbol 1_{[0; \varepsilon]}(x):= \begin{cases} 1 &\text{if }x \in [0; \varepsilon] \\ 0 &\text{else} \end{cases}$.
Then the convolution with $s$ can be computed analytically:
$(s \ast f)(x) = \sum_{k=0}^\infty (-a)^k \boldsymbol 1_{[0; \varepsilon]} ( x - k + \frac 1 2) + \sum_{k=0}^\infty (-a)^k \boldsymbol 1_{[0; \varepsilon]} ( x - k - \frac 1 2) \\ = \boldsymbol 1_{[0; \varepsilon]} ( x + \frac 1 2) + \sum_{k=0}^\infty ((-a)^k + (-a)^{k+1}) \boldsymbol 1_{[0; \varepsilon]} ( x - k - \frac 1 2) \\ = \boldsymbol 1_{[0; \varepsilon]} ( x + \frac 1 2) + (1-a) \sum_{k=0}^\infty (-a)^k \boldsymbol 1_{[0; \varepsilon]} ( x - k - \frac 1 2)$.
The first and second summand in the bottom line correspond to the parts of $s \ast f$ that are supported in the negative- and positive half axis, respectively. Accordingly, we obtain
$\|S(f)|_{\mathbb R _-}\|_{L^2}^2 = \|(s\ast f)|_{\mathbb R _-}\|_{L^2}^2 = \int_{\mathbb R_-} \boldsymbol 1_{[0; \varepsilon]} ( x + \frac 1 2) ^2 \,\text d x = \int_{-\frac 1 2}^{\varepsilon-\frac 1 2} 1 \, \text d x = \varepsilon \\ \|S(f)|_{\mathbb R _+}\|_{L^2}^2 = \|(s\ast f)|_{\mathbb R _+}\|_{L^2}^2 = (1-a)^2 \sum_{k=0}^\infty a^{2k} \int_{\mathbb R_+} \boldsymbol 1_{[0; \varepsilon]} ( x - k - \frac 1 2)^2 \, \text d x \\ = (1-a)^2 \sum_{k=0}^\infty a^{2k} \int_{k + \frac 1 2}^{k + \frac 1 2 +\varepsilon} 1 \, \text d x \\ = (1-a)^2 \varepsilon \sum_{k=0}^\infty a^{2k} = \varepsilon \cdot \frac{(1-a)^2}{1-a^2} = \varepsilon \frac{ 1-a }{1+a}$.
The result implies that $\|(s\ast f)|_{\mathbb R _+}\|^2/\|(s\ast f)|_{\mathbb R _-}\|^2 = \frac{ 1-a }{1+a}$ is always less than 1 and even goes to 0 for $a \to 1$. Hence, there cannot exist any constant $C > 0$ such that
$\|S(f)|_{\mathbb R _-}\|_{L^2} \leq C \|S(f)|_{\mathbb R _-}\|_{L^2}$
holds true in the general problem setting.