Consider the function

$$F(x):=\int_{\mathbb R} f(t+x)f(t-x) \ dt .$$

Clearly, we have by Cauchy-Schwarz

$$\vert F(x) \vert\le \Vert f \Vert^2_{L^2} $$ $$\vert F'(x)\vert\le 2\Vert f' \Vert_{L^2} \Vert f \Vert_{L^2} \le 2\Vert f \Vert_{H^1}^2$$

where $H^1$ is the $L^2$-Sobolev space of order 1.

This shows that to bound the sup norm of $F$ we require $f \in L^2$ and to bound the $C^1$ norm of $F$ it suffices to have $f \in H^1.$

I wonder now if it is true that the Hölder norm $C^{\gamma}$ with $\gamma \in (0,1)$ can be bounded by the $H^{\gamma}$ norm of $f$ and if that is not the case, I would be curious to learn what the correct interpolation space for $f$ is to bound the Hölder norm of $F$.

  • 1
    $\begingroup$ Nitpick: the estimates you describe don't show that you require $f\in L^2$ or $f\in H^1$, they merely show that such conditions are sufficient. $\endgroup$
    – Yemon Choi
    Aug 22, 2019 at 23:25
  • $\begingroup$ Also, in your first integral, surely you mean $dt$ not $dx$ $\endgroup$
    – Yemon Choi
    Aug 22, 2019 at 23:26
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    $\begingroup$ Finally: if you just want sufficient conditions on $f$ to ensure $F$ is in a certain H\"older class, then a natural idea would be to look at conditions on the decay rate of the Fourier transform of $F$ which ensure $F$ itself has the right H\"older continuity, then use the fact that $F(x) = (f*g)(2x)$ where $g(t)=f(-t)$ $\endgroup$
    – Yemon Choi
    Aug 22, 2019 at 23:30
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    $\begingroup$ @YemonChoi thank you for the first two corrections. Regarding your third comment, I am not quite sure I understand how to apply it. So roughly the Fourier transform of $F$ is the product of $\widehat{f}$ with itself up to some factors. What is then the condition for the Hölder space? $\endgroup$
    – user144765
    Aug 22, 2019 at 23:34
  • 2
    $\begingroup$ @YemonChoi Sorry, the point I tried to make was that the OP asks specifically whether it is possible to bound the Hölder norm in terms of the fractional Sobolev norm. Although you are correct that you can get a bound using the methods you describe, I am not sure it is that one. $\endgroup$
    – user69109
    Aug 23, 2019 at 7:16

1 Answer 1


Your conjecture is correct. Define the bilinear operator $T$ by $$ T(f,g)(x)=\int_{\mathbb{R}}f(t+x)g(t-x)\,dt. $$ As you have shown, this operator is bounded from $L^2\times L^2\to C^0$ and from $H^1\times H^1\to C^1$. It follows from bilinear real interpolation theory (see e.g. Zafran, A multilinear interpolation theorem) that $T$ is also bounded from $(L^2,H^1)_{\theta,p}\times (L^2,H^1)_{\theta,q}\to (C^0,C^1)_{\theta,r}$ whenever $\theta\in (0,1)$ and $p,q,r\in [1,\infty]$ satisfiy $1/r+1=1/p+1/q$.

If we take $r=\infty$ and $p=q=2$ and use $(L^2,H^1)_{\gamma,2}=H^\gamma$, $(C^0,C^1)_{\gamma,\infty}=C^\gamma$, we obtain the desired boundedness from $H^\gamma\times H^\gamma\to C^\gamma$.


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