Infinite intersection of $L^p$ space Consider the space
$$L^{2-}(\mathbb{R}):=\bigcap_{1\leq p<2}L^{p}(\mathbb{R})$$
equipped with the norm
$$\Vert f\Vert_{L^{2-}(\mathbb{R})}:=\sup_{1\leq p<2}\Vert f\Vert_{L^{p}(\mathbb{R})}.$$
Is it true that $(L^{2-}(\mathbb{R}),\Vert\cdot\Vert_{L^{2-}(\mathbb{R})})$ is a Banach space?
 A: No, as your norm is not well-defined on the space. Consider the function $f(x) = x^{-1/2} 1_{(0,1)}(x),$ then for $p<2$ we have,
$$ \lVert f \rVert_{L^p(\mathbb R)} = \left(\int_0^1 x^{-p/2} dx\right)^{1/p} = \frac1{(1-p/2)^{1/p}},$$
so $f \in L^p(\mathbb R)$ for each $1 \leq p <2,$ but $\lVert f \rVert_{L^{2-}(\mathbb R)} = \infty.$
Note if you instead consider the space
$$ X = \left\{ f \in \bigcap_{1\leq p < 2} L^p(\mathbb R) : \lVert f \rVert_{L^{2-}(\mathbb R)} < \infty \right\},$$
then this is a Banach space with the given norm and moreover it coincides with $L^1(\mathbb R) \cap L^2(\mathbb R).$
Edit: To elaborate on my comment on the end, we can show that $X = L^1(\mathbb R) \cap L^2(\mathbb R)$ as follows:
First if $f \in X,$ then noting $f^p \rightarrow f^2$ as $p \rightarrow 2$ pointwise, we get by Fatou's lemma,
$$ \int_{\mathbb R} |f|^2 dx \leq \liminf_{p \rightarrow 2^-} \int |f|^p dx \leq \lVert f \rVert_{L^{2-}(\mathbb R)}^2, $$
so $f \in L^2(\mathbb R).$ Conversely if $1<p<2$ and $f \in L^1(\mathbb R) \cap L^2(\mathbb R)$ choosing $\alpha \in (0,1)$ such that $\frac1p = \frac{\alpha}{1} + \frac{1-\alpha}{2}$ we get by Hölder's inequality,
$$ \lVert f \rVert_{L^p(\mathbb R)} \leq \lVert f \rVert_{L^1(\mathbb R)}^{\alpha} \lVert f \rVert_{L^2(\mathbb R)}^{1-\alpha}. $$
So $f \in X.$ So $X = L^1(\mathbb R) \cap L^2(\mathbb R).$
A: Consider a Cauchy sequence $\{f_n\}$. Then for every $\varepsilon$ there is $n_\varepsilon$ such that $\Vert f_n-f_m\Vert_{L^{2-}}\le \varepsilon$ for all $n,m\ge n_\varepsilon$. This implies that  $\{f_n\}$ is a Cauchy sequence in every $L^p$ and so there exists $g_p\in L^p$ such that $f_n\to g_p$ in $L^p$. But in any compact set $[-L,L]$ you have that $L^p([-L,L])\subset L^1([-1,1])$ and so by uniqueness of the limit $g_p=g_1$ a.e. in $[-L,L]$.  Since this is true for every $L>0$, you can assume that $g_p=g_1$. Hence, $g_1\in L^{2-}$. Take $m\ge n_\varepsilon$ and using the definition of sup find $p_m$ such that 
$$\Vert f_m-g_1\Vert_{L^{2-}}\le  (2-p_m)^{1/p_m}\Vert f_m-g_1\Vert_{L^{p_m}}+\varepsilon.$$
Since $f_n\to g_1$ in $L^{p_m}$ there exists $n_m\ge n_\varepsilon$ (depending on $m$ ) such that
$$ (2-p_m)^{1/p_m}\Vert f_n-g_1\Vert_{L^{p_m}}\le\varepsilon$$ for all $n\ge n_m$.
Hence,
\begin{align}\Vert f_m-g_1\Vert_{L^{2-}}&\le   (2-p_m)^{1/p_m}\Vert f_m-g_1\Vert_{L^{p_m}}+\varepsilon\\&\le  (2-p_m)^{1/p_m}\Vert f_m-f_n\Vert_{L^{p_m}}+ (2-p_m)^{1/p_m}\Vert f_n-g_1\Vert_{L^{p_m}}+\varepsilon\\&\le \Vert f_n-f_m\Vert_{L^{2-}}+2\varepsilon\le 3\varepsilon.\end{align}
Edit These types of spaces are called the grand Lebesgue spaces. You define the space as all $f$ for which 
$$\Vert f\Vert_{L^{2-}}=\sup_{1< p<2}\left((2-p)\int|f|^pdx\right)^{1/p}<\infty.$$
The standard notation is $L^{2)}$.
