Question: **Problem:** Let $g\_t(x)=\frac{1}{\sqrt{2\pi}} e^{ \frac{- (x-t)^2}{2}}$ be the Gaussian function centered at $t\in \mathbb R$. Let $I$ be a subset of $\mathbb R$. Consider the "convolution operator"
\begin{align}
\mathcal A: L^1(I) & \longrightarrow L^2(I)\\
f & \longmapsto \left(x\longmapsto \int\_{I} g\_t(x)f(t) dt \right).
\end{align}
How can I prove (or disprove) that the operator is well-defined and injective?

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**My attempts:** Here are some of my attempts to prove the well-defined property.

* I have tried to use Cauchy inequality but with no success:
\begin{align}
\int\_{I} \left( \int\_{I} g\_t(x)f(t)dt\right)^2 dx \leq \int\_{I} \int\_{I} g\_t^2(x)dx dt \int\_{I}f^2(t)dt.
\end{align}
as the right-hand-side may be infinite as $f$ does not belong to $L^2(I)$.
* I have tried to use the boundedness of Gaussian function but nothing better. We have
\begin{align}
\int\_{I} \left( \int\_{I} g\_t(x)f(t)dt\right)^2 dx \leq
M \left( \int\_{I} f(t)dt \right)^2 \int\_I dx.
\end{align}
and the right-hand-side may be infinite as $I$ is unbounded.

Answer: 
I'm going to unclutter the situation and leave out some constants. If it works in the simplified case, you can throw constants at it later.

Suppose $f\in L^1(\mathbb R)$ and $g(t) = e^{-t^2}.$ For $x\in \mathbb R,$ define

$$(g\*f)(x) = \int\_{\mathbb R}g(x-t)f(t)\,dt.$$

That is certainly well defined for any $x,$ since $g\in L^\infty.$ To show $g\*f\in L^2,$ note

$$|g\*f(x)|^2 \le (\int\_{\mathbb R}g(x-t)|f(t)|\,dt)^2$$ $$ = \|f\|\_1^2(\int\_{\mathbb R}(g(x-t)(|f(t)|/\|f\|\_1)\,dt)^2.$$

Use Jensen to see the last expression is bounded above by

$$\|f\|\_1^2\int\_{\mathbb R}(g(x-t))^2(|f(t)|/\|f\|\_1)\,dt.$$

Thus, using Fubini, we have

$$\int |g\*f(x)|^2\,dx \le \|f\|\_1^2 \int g(x)^2\,dx.$$

This proves $\|g\*f\|\_2 \le \|f\|\_1\|g\|\_2.$ Thus $f\to g\*f$ is a bounded linear operator from $L^1$ to $L^2.$

For injectivity, I'll use some Fourier analysis. Let $F$ denote the Fourier transform. Then $F$ is an isometry on $L^2,$ and is injective on $L^1.$ Thus $g\*f$ is the zero function iff $F(g\*f)$ is the zero function. Now

$$F(g\*f)(x) =F(g)(x)\cdot F(f)(x).$$

And as is well known, $F(g)>0$ everywhere. (In fact for the Gaussian $g,$ $F(g)$ is essentially $g,$ meaning $F(g)(x)= ag(bx)$ for some nonzero constants $a,b.$) Thus $F(g\*f)=0$ iff $F(f)=0.$ And the later happens iff $f=0.$ It follow that $f\to g\*f$ is injective.
\begin{align\*}
\left|\int\_I \left(\int\_I e^{-(x-t)^2}f(t)dt\right)^2dx\right| &= \left|\int\_I \int\_I \int\_I e^{-(x-t)^2}f(t)e^{-(x-s)^2}f(s)dsdtdx\right| \\ &= \left|\int\_I \int\_I f(t)f(s)\left(\int\_I e^{-(x-t)^2}e^{-(x-s)^2}dx\right)dsdt\right| \\ &\le \int\_I\int\_I |f(t)|\hspace{1mm}|f(s)| \left(\int\_\mathbb{R} e^{-(x-t)^2}e^{-(x-s)^2}dx\right)dsdt \\ &\le C\int\_I \int\_I |f(t)|\hspace{1mm} |f(s)|dsdt \\ &= C\left(\int\_I |f(t)|dt\right)^2
\end{align\*}