I was wondering whether the following interpolation between $L^1$ and $L^2$ spaces is true:

Let $f \in \mathbb{R}^n$ be such that

$$ \alpha_1:= \int_{\mathbb{R}} \left\lVert f(x_1,\cdot,....\cdot) \right\rVert_{L^2(\mathbb{R}^{n-1})} dx_1$$

up to

$$ \alpha_n:= \int_{\mathbb{R}} \left\lVert f(\cdot,\cdot,....,x_n) \right\rVert_{L^2(\mathbb{R}^{n-1})} dx_n$$ are finite.

Does this give us an upper bound for the $L^1$ norm of $f$ in the sense that

$$ \int_{\mathbb{R}^n} \left\lvert f(x) \right\rvert dx \le \alpha_1 + ...+ \alpha_n?$$

  • $\begingroup$ The two sides of your inequality have different scaling, it can not be true $\endgroup$ Commented Mar 3, 2017 at 9:21
  • $\begingroup$ @PieroD'Ancona I do not see that....to me it looks as if both scale like first order terms $\endgroup$ Commented Mar 3, 2017 at 9:26
  • $\begingroup$ I guess now you see that $\endgroup$ Commented Mar 4, 2017 at 1:53

1 Answer 1


Yes for $n = 1$, and no for $n \geq 2$. Indeed, if we had an inequality $||f||_1 \leq C(\alpha_1 + \dots + \alpha_n)$ then we could set $f = \mathbb{1}_{[0,K]^n}$ and get

$$ K^n \leq C n K^{\frac{n+1}{2}}, $$ and we get $n \leq 1$ by letting $K$ tend to infinity.


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