The functional equation
$$f(x) = \frac{x}{\sqrt{2}} f(2x)+1$$
doesn't uniquely define a function. I could choose to define $f(1)$ however I want, and then propagate the value to all $f(2^m), m \in \mathbb{Z}$ in a way that satisfies the functional. However, I could choose to rewrite the functional equation as
$$ f(x)-1 = \frac{x}{\sqrt{2}} e^{\ln(2)x D} f(x) = \frac{x}{\sqrt{2}}\left( 1 + \ln(2)xD +\ln(2)^2 \frac{ (xD)^2}{2!} + \ln(2)^3\frac{ (xD)^3}{3!}+\dots\right)f(x)$$
I am curious, with this formulation, is $f$ uniquely defined (perhaps, as the limit of solutions to differential equations up to order n, or maybe something else)? What initial conditions are needed to define the function fully (do we need all the derivatives at 0?)?
Thoughts
One consideration is that $f(x)$ is formally equal to the (divergent) sum $\sum 2^{n^2} x^n$, so perhaps there is a unique way to regularize that series which provides a unique answer.
Mathematica can solve the first two differential equations. $f(x)-1 = \frac{x}{\sqrt{2}}f(x)$ has $$f(x) = \frac{-2}{-2 + \sqrt{2}x}$$ For $f(x)-1 = \frac{x}{\sqrt{2}}(f(x)+ \ln(2) x f'(x))$ with the initial condition $f(0)=1$ we obtain $$\log ^{-\frac{1}{\log (2)}}(2) e^{-\frac{\sqrt{2}}{x \log (2)}} x^{-\frac{1}{\log (2)}} \left(\sqrt{e} \left(-\frac{1}{x}\right)^{\frac{1}{\log (2)}} x^{\frac{1}{\log (2)}} \Gamma \left(1-\frac{1}{\log (2)},-\frac{\sqrt{2}}{x \log (2)}\right)+c_1 \sqrt[\log (2)]{\log (2)}\right) $$ The $e^{\frac{-1}{x}}$ term suggests the Taylor series expansion is probably not unique around $0$, since, at 0, $e^{\frac{-1}{x}}$ is invisible to a Taylor series.