Let $L_{e}(\mathbb{R})$ denote the Hilbert space with inner product $\langle f, g \rangle = \int_{-\infty}^{\infty}f(x)\overline{g(x)}e^{-x^2}dx$.
Let $T: L_{e}(\mathbb{R}) \to L_{e}(\mathbb{R})$ be an operator defined by $Tf(x) = f(x+1)$. I need to find the adjoint operator of $T$, $T^{*}$, and find $||T||$ in the operator norm.
I was able to find $T^{*}$, but unable to find the norm. I'll share my proof for $T^{*}$ and the progress I made on finding the norm (which might be useless towards finding an actual solution, but it's all I have).
Finding $T^{*}$:
$$ \langle Tf, g \rangle = \int_{-\infty}^{\infty} f(x+1)\overline{g(x)} e^{-x^2}dx =^{1} \int_{-\infty}^{\infty} f(t)\overline{g(t-1)}e^{-(t-1)^2}dt = \int_{-\infty}^{\infty} f(t)\overline{g(t-1)e^{t^2-(t-1)^2}} e^{-t^2}dt = \int_{-\infty}^{\infty} f(t) \overline{g(t-1)e^{2t-1}} e^{-t^2}dt = \langle f, T^{*}g \rangle.$$
In the equality marked by footnote $1$, I made a substitution $t = x+1$. Therefore: $$T^{*}g(t) = g(t-1)e^{2t-1}.$$
Now, for my progress on finding the norm (and some random ideas):
$$||Tf||^2 = \int_{-\infty}^{\infty}|f(x+1)|^{2}e^{-x^2}dx = \int_{-\infty}^{\infty}|f(t)|^{2} e^{-(t-1)^2}dt = \int_{-\infty}^{\infty} |f(t)|^{2}e^{2t-1}e^{-t^2}dt = \frac{1}{e} \int_{-\infty}^{\infty}|f(t)e^{t}|^{2}e^{-t^2}dt = \frac{1}{e}||fe^{t}||^2.$$ Is there an some kind of submultiplicativity inequality that I could use here?
Also, some other ideas: finding $||T^{*}||$ instead of $||T||$; also, if I could prove $T$ is compact (which I also am not able to; I don't even know if it is), I could then use the fact that $TT^{*}$ is compact and self-adjoint, and therefore $||TT^{*}|| = ||T||^2 = \max_{\lambda} |\lambda|$, where the $\lambda$'s are the eigenvalues of $TT^{*}$.
However, I haven't made any progress on any of these.
We will prove that the operator $T$ is not bounded.
Consider $f(x)=e^{a x^2/2}$ with $0<a<1$ then $$\Vert f\Vert^2=\int_{-\infty}^\infty e^{(a-1)x^2}dx=\sqrt{\frac{\pi}{1-a}}<+\infty$$ And $$\eqalign{\Vert Tf\Vert^2&=\int_{-\infty}^\infty e^{a(x+1)^2}e^{-x^2}dx\\ &=e^{a}\int_{-\infty}^\infty e^{-(1-a)x^2+2ax}dx \\ &=e^{a+a^2/(1-a)}\int_{-\infty}^\infty \exp\left({-(1-a)\left(x-\frac{a}{1-a}\right)^2}\right)dx \\ &=e^{a/(1-a)}\int_{-\infty}^\infty e^{-(1-a)t^2}dt =e^{a/(1-a)}\Vert f\Vert^2 }$$ So, if $T$ is bounded then we must have $$\Vert T\Vert \ge \exp\left(\frac{a}{2(1-a)}\right)\quad\text{for $a\in(0,1)$}$$ This leads to a contradiction when $a$ approaches $1$. So, $T$ is not bounded.