Let $f \in L^p(\mathbb R^d)$ for every $p \in (1, \infty]$. Does this imply $f \in L^1(\mathbb R^d)$? My intuition tells me yes, because
$$ \lVert f \rVert _{1+\frac{1}{n}} = \left( \int _{\mathbb{R}^d}\lvert f \rvert ^{1+ \frac{1}{n}} \, d\lambda_{d} \right)^{\frac{1}{1+1/n}} < \infty $$
for every $n \in \mathbb{N}$. Therefore, I can approximate $\lVert f \rVert _{1}$ with finite values, but I am struggling to conclude that $\lVert f \rVert _{1} < \infty$.
I tried using the dominated convergence theorem with $\lvert f \rvert^{2} \in L^1(\mathbb{R}^d)$. I want to show
$$ \lvert f \rvert ^{1+\frac{1}{n}} \stackrel{?}{\leq} \lvert f \rvert^{2} \quad \forall n \in N $$
but the function $p \mapsto \lvert x \rvert^p$ is monotically decreasing for values $x \in (0, 1)$. Even then, if
$$ \lVert f \rVert _{1} \stackrel{?}{=} \lim_{ n \to \infty } \underbrace{ \lVert f \rVert _{1+\frac{1}{n}} }_{ \text{finite} } $$
would be true, I cannot say that the limit converges to a finite value.
Is the assumption $f \in L^1(\mathbb{R}^d)$ even correct? If yes, is the dominated convergence theorem the way to go to prove it? If no, can you give an example of a function $f \in L^p(\mathbb{R}^d) \,\forall p \in (1,\infty]$, but $f \notin L^1(\mathbb{R}^d)$?
This is false. Take $d=1$, and $f:\mathbb{R}\to\mathbb{R}$ defined as follows: if $x\leq 0$ then $f(x):=0$. If $x>0$, then $f(x):=1/\lceil x\rceil$. Then for every $p>1$ we have $$ ||f||_p=\sum_{n\geq 1}\frac{1}{n^p}<+\infty, $$ but $$ ||f||_1=\sum_{n\geq 1}\frac{1}{n}=+\infty. $$