Prove that $(A_0,A_1)_{\theta,q}$ is Banach. Let $A_0$, $A_1$ be two Banach spaces, both embedded continuously in a Hausdorff topological vector space $\mathcal{A}$. Then we can consider the normed spaces $A_0 \cap A_1$ and $A_0 + A_1$ with the norms
$$
||a||_{A_0 \cap A_1} = \max\{ ||a||_{A_0}, ||a||_{A_1}\},
$$
$$
||a||_{A_0 + A_1} = \inf \{ ||a_0||_{A_0} + ||a_1||_{A_1} : a=a_0+a_1, a_j \in A_j\},
$$
and following this, define the real interpolation space $(A_0,A_1)_{\theta,q}$ consisting of those elements $a \in A_0 + A_1$ with finite norm:
$$
||a||_{\theta,q} = \left( \int_0^{\infty} \left(t^{-\theta}K(t,a) \right)^q \frac{dt}{t} \right)^{\frac{1}{q}}, 1\leq q <\infty,
$$
where $K(t,a) = \inf\{ ||a_0||_{A_0} + t||a_1||_{A_1}: a=a_0+a_1, a_j \in A_j\}$ is Peetre K-functional.
My task now is to prove that $(A_0,A_1)_{\theta,q}$ is a Banach space.
I have already proven that, with this construction, $A_0 + A_1$ is a Banach space, as well as the fact that $(A_0,A_1)_{\theta,q}$ is continuously embedded in $A_0 + A_1$ with the norms defined above. So, if I'm not mistaken, it would be enough to prove that the interpolation space $(A_0,A_1)_{\theta,q}$ is a closed subspace of $A_0+A_1$ and using the fact that the latter is Banach, the former would be Banach as well.
However, my efforts have been unfruitful so far. I'd like to show that if I take a succession $\{x_n\} \subset (A_0,A_1)_{\theta,q}$ that converges to $x\in A_0+A_1$, then:
$$
||x||_{\theta,q} = \left( \int_0^{\infty} \left(t^{-\theta}K(t,x) \right)^q \frac{dt}{t} \right)^{\frac{1}{q}} <\infty,
$$
so $x \in (A_0,A_1)_{\theta,q}$, proving it's a closed subspace. Any hints on how to proceed towards my goal would be appreciated.
 A: As you pointed out, the key is to exploit the Banach space structure of $A_0+A_1$, although I will go by definition to show that the space is complete.
Let $(x_n)_n$ be a Cauchy sequence in $(A_0, A_1)_{\theta, q}$. Since the embedding $(A_0, A_1)_{\theta, q}\hookrightarrow A_0+A_1$ is continuous, $(x_n)_n$ is a Cauchy sequence in $A_0+A_1$ as well, therefore it converges to some $x\in A_0+A_1$. Now we want to estimate the difference $\|x_n-x\|_{\theta, q}$ and also show that $x\in (A_0, A_1)_{\theta, q}$.
By the definition of Cauchy sequence, for a fixed $\varepsilon>0$ we have that $\|x_n-x_m\|_{\theta, q}\leq \varepsilon$ for all $n$, $m\geq N$. It is well known that $a\mapsto K(t, a)$ defines a norm in $A_0+A_1$ for all $t>0$, therefore by the triangle inequality
\begin{align*}
    t^{-\theta} K(t, x_n-x)\leq t^{-\theta} K(t, x_n-x_m)+t^{-\theta}\max\{t, 1\}\|x_m-x\|_{A_0+A_1}.
\end{align*}
Now we distinguish two cases: $q=\infty$ and $q<\infty$. In the first case, the above inequality implies that for all $t>0$ and $n$, $m\geq N$
\begin{align*}
    t^{-\theta} K(t, x_n-x)\leq \varepsilon+t^{-\theta}\max\{t, 1\}\|x_n-x\|_{A_0+A_1}.
\end{align*}
Hence by letting $m\to \infty$ we obtain that for all $t>0$ it holds
\begin{align*}
    t^{-\theta} K(t, x_n-x)\leq \varepsilon.
\end{align*}
A simple application of the triangle inequality allows us to deduce that $x\in (A_0, A_1)_{\theta, \infty}$ and hence $x_n\to x$ in $(A_0, A_1)_{\theta, \infty}$. Now for $q<\infty$, with the aim to treat carefully the constants which will appear, we write (continuity of the norm plus monotone convergence theorem if u want)
\begin{align*}
    \|x_n-x\|_{\theta, q}=\lim_{\eta\to 0} \left(\int_\eta^{1/\eta} t^{-\theta q-1}K(t, x_n-x)^qdt\right)^{1/q}.
\end{align*}
The first inequality we obtained allows us to get the following bound for each $\eta\in (0, 1)$ and $n$, $m\geq N$
\begin{align*}
    \left(\int_\eta^{1/\eta} t^{-\theta q-1}K(t, x_n-x)^qdt\right)^{1/q}&\leq \|x_n-x_m\|_{\theta, q}+\left(\int_\eta^{1/\eta} t^{-\theta q-1}\max\{t, 1\}dt\right)^{1/q}\|x_m-x\|_{A_0+A_1}\\
    &\leq \varepsilon+C\|x_m-x\|_{A_0+A_1}.
\end{align*}
The constant $C=C(\eta, q)$ comes directly from the integral, which does not necessarily converge as $\eta$ goes to $\infty$. By letting $m\to \infty$, the second term on the above bound disappears and then we can safely take $\eta\to 0$ to conclude that $x\in (A_0, A_1)_{\theta, q}$ and that $x_n\to x$ in $(A_0, A_1)_{\theta, q}$.
