Computing Hermitian Conjugate for an Operator on a Function The operator $\hat D$ is defined by $(\hat D f)(x) = \sqrt 2 f(2x)$. Show that $\hat D$ is a linear transformation, compute its hermitian conjugate and show it is unitary. Determine all eigenfunctions of $\hat D$.
It is not stated in the given problem explicitly, but I assume it operates on infinite dimensions, as this is actually a problem from a quantum mechanics course.
I know what conditions must be satisfied for a linear transformation: $$(\hat D(f + g))(x) = (\hat D f)(x) + (\hat D g)(x)$$ $$\hat D(cf)(x) = c(\hat D f)(x)$$ but I'm not sure how to prove it holds true for all functions. And then I am totally lost on computing the hermitian conjugate, as I have only done that with matrices, not an operator on a function. Any help pointing me in the right direction would be appreciated.

We assume the linear operator $\hat{D}$ is defined on the Hilbert space $H = \mathcal{L}^2(\mathbb{R})$ equipped with the usual inner product $(f,g)=\int_{\mathbb{R}} f(x)g(x) \, \textrm{d}x$ since it is not specified. 
The linearity just follows from definition. Indeed, for $f, g \in H$, $\alpha, \beta \in \mathbb{C}$,
$$(\hat{D}(\alpha f + \beta g))(x) = \sqrt{2}(\alpha f + \beta g)(2x) = \sqrt{2} \alpha f(2x) + \sqrt{2} \beta g(2x) = \alpha (\hat{D}f)(x) + \beta (\hat{D}g)(x)$$
Define $\hat{D}^{*}: H \rightarrow H$ by $(\hat{D}^{*}g)(x) = \frac{1}{\sqrt{2}}g(\frac{1}{2}x)$. Then for $f,g \in H$, we have
$$ (\hat{D}f, g) = \int_{\mathbb{R}} \sqrt{2}f(2x)g(x) \, \textrm{d}x $$
$$ (f, \hat{D}^{*}g) = \int_{\mathbb{R}} f(x)\frac{1}{\sqrt{2}}g(\frac{1}{2}x) \, \textrm{d}x $$
Letting $y = \frac{1}{2}x$ in the second formula, we have
$$ (f, \hat{D}^{*}g) = \int_{\mathbb{R}} f(2y) \sqrt{2} g(y) \, \textrm{d}y = (\hat{D}f, g) $$
Therefore $\hat{D}^{*}$ is the Hermitian conjugate of $\hat{D}$.
It is now easy to check the operator is unitary.
$$(\hat{D}^{*}\hat{D}f)(x) = \frac{1}{\sqrt{2}}(\hat{D}f)(\frac{1}{2}x) = \frac{1}{\sqrt{2}} \times \sqrt{2} f(2 \times \frac{1}{2}x) = f(x)$$
Hence $\hat{D}^{*}\hat{D} = I$. The other side $\hat{D}\hat{D}^{*} = I$ is symmetric.
