In a question which was recently asked, the goal was to (dis)prove that the set of functions which satisfy a certain equation is trivial. There is already a very neat solution there, but I came up with an alternative idea to solve it, purely based on functional analysis arguments, but there remains a gap I can't seem to fix.
Here is the setup : let $0<\gamma\le 1 $ and consider the sets of real-valued functions $$X_1:=\{f\in C^{0,\gamma}(\mathbb R),\text{ s.t. } f(x)=0\ \forall x\not\in[0,1]\},$$ where $C^{0,\gamma}(\mathbb R)$ denotes the space of $\gamma$-Hölder continuous on $\mathbb R$, and $$X_2:=\left\{g\in C(\mathbb R_{>0}),\text{ s.t. } \sup_{a>0}g(a)<\infty\right\}.$$ That is, $X_1$ is the space of $\gamma$-Hölder functions on $\mathbb R$ which vanish outside of $[0,1]$, and $X_2$ is the space of bounded continuous functions on $\mathbb R_{>0}$. We respectively equip $X_1$ and $X_2$ with the norms $\|\cdot\|_{X_1}$ and $\|\cdot\|_{X_2}$ defined for all $(f,g)\in X_1\times X_2 $ by $$\|f\|_{X_1}:= \sup_{x\in\mathbb R}|f(x)| + \sup_{x,y\in\mathbb R,x\ne y}\frac{|f(x)-f(y)|}{|x-y|^\gamma}, $$ $$\|g\|_{X_2}:= \sup_{x>0}|g(x)|. $$ Equipped with those norms, $X_1$ and $X_2$ are both Banach spaces, and we can define the linear operator $T:X_1\to X_2$ by $$Tf : a\mapsto \int_{\mathbb{R}}f(ax)\exp \left(-\frac{1}{2}(x-1)^2\right)dx. $$ It is is not hard to show that $T$ is a well-defined bounded linear operator from $X_1$ to $X_2$ (details below if interested), and my goal is to show that $\operatorname{ker}(T)\ne\{0\}$, which would give a negative answer to the original question I mentioned.
To show that $T$ has non-trivial spectrum, I proved that $T$ is compact (details below, again), and I thought that it followed from "standard theory" that the kernel of $T$ could not be trivial in that case. However I found out that this is wrong, as there are compact operators on infinite-dimensional spaces with trivial kernels (I could only find Hilbert space examples though...).
I know that $T$ being compact implies that its range is separable and thus has cardinality at most $2^{\aleph_0}$, but unfortunately $X_1$ also has cardinality at most $2^{\aleph_0}$ so I can't conclude anything...
Question : How to show that the kernel of $T$ is (not ?) trivial ? Ideally I would like to see solutions based on my approach so far, but alternative ideas are also welcome.
Many thanks for reading thus far !
Proof that $T$ is well-defined : $T$ is clearly linear, and for all $f\in X_1,$ we have that for any $a_1,a_2>0$ $$\begin{align} |Tf(a_1)-Tf(a_2)|&\le\int_{\mathbb{R}}|f(a_1x)-f(a_2x)|\exp \left(-\frac{1}{2}(x-1)^2\right)dx\\ &\le \int_{\mathbb{R}}\|f\|_{X_1}|x|^\gamma|a_1-a_2|^\gamma\exp \left(-\frac{1}{2}(x-1)^2\right)dx\\ &\le|a_1-a_2|^\gamma\|f\|_{X_1}\int_{\mathbb{R}}|x|^\gamma\exp \left(-\frac{1}{2}(x-1)^2\right)dx\\ &\le |a_1-a_2|^\gamma C(f),\tag1\end{align} $$ where $C(f)<\infty$ is a finite number which depends only on $f$, hence $Tf$ is (Hölder) continuous. Likewise, it's easy to see that for all $a>0$, $|Tf(a)|\le \|f\|_{X_1}\int_{\mathbb{R}}\exp \left(-\frac{1}{2}(x-1)^2\right)dx$, hence $T$ is indeed a well defined (and bounded) linear map from $X_1$ to $X_2$.
Proof that $T$ is compact : to see this let $M>0$ be a fixed constant and consider a family $\{f_n,n\in\mathbb N\}\subseteq X_1$ such that for all $n$, $\|f_n\|_{X_1}\le M $. We claim that the sequence $(Tf_n)_{n\ge 1}$ is Cauchy in $X_2$, which will immediately imply that $T$ is compact. First observe that for any $a>0$ and any $n\ge 1$,
$$|Tf_n(a)| \le \int_0^{1/a} |f_n(ax)|\exp \left(-\frac{1}{2}(x-1)^2\right)dx\le M/a. $$ Hence for any $\varepsilon>0$ and $a>4M/\varepsilon $, we have $|Tf_n(a)|\le \varepsilon/4 $ uniformly over $n\in\mathbb N$. But now note that for any $n,m\in\mathbb N$, triangle inequality yields \begin{align}\|Tf_n - Tf_m\|_{X_2}&\le \sup_{0<a\le 4M/\varepsilon}|Tf_n(a) - Tf_m(a)| + \sup_{a> 4M/\varepsilon}|Tf_n(a) - Tf_m(a)| \\ &\le \sup_{0<a\le 4M/\varepsilon}|Tf_n(a) - Tf_m(a)| + \varepsilon/2.\end{align} So denote $I_\varepsilon:=[0,4M/\varepsilon]$ : we need to find a $N(\varepsilon)$ such that $\sup_{a\in I_\varepsilon}|Tf_n(a) - Tf_m(a)| \le \varepsilon/2$ whenever $n,m\ge N(\varepsilon)$. But this follows directly from Arzelà-Ascoli theorem applied to the restriction of each $Tf_n$ to $I_\varepsilon$ : uniform boundedness is clear and uniform equicontinuity follows from the Hölder property (the same way we derived inequality $(1)$).
We conclude that for any $\varepsilon>0$, we can find $N(\varepsilon)$ such that $\|Tf_n-Tf_m\|_{X_2}\le \varepsilon $ whenever $n,m>N(\varepsilon)$, that is, $ (Tf_n)_{n\ge1}$ is Cauchy. Hence $T$ maps bounded sets in $X_1$ to relatively compact sets in $X_2$ and is therefore compact.
