Setting: Let $(M,g)$ be a compact Riemannian manifold without boundary. Let $\Delta_g$ be the Laplacian and $L=\Delta_g-\partial_t$ the heat operator. Let $0<\alpha<1$, $0<t_0<T$.

Let $$u\in C^{2,1}(M\times[0,T))\cap C^\infty(M\times(0,T))$$ solve $$Lu=f$$ where $$f\in C^0(M\times[0,T))\cap C^\infty(M\times(0,T))$$ is such that $f(\cdot,t)\in C^\alpha(M)$ for any $t\in[0,T)$ and $$\sup_{t\in[0,T)}\|f(\cdot,t)\|_{C^\alpha(M)}<\infty.$$

QUESTION: Do we have the follow a priori estimate? $$\sup_{t\in[t_0,T)}\|u(\cdot,t)\|_{C^{2+\alpha}(M)}\leq C\bigg(\sup_{s\in[t_0,T]}\|f(\cdot,s)\|_{C^\alpha(M)}+\|u\|_{L^\infty(M\times[0,T])}\bigg)$$

This estimate was needed for proving the global existence of a solution for the harmonic map heat flow, which is a quasilinear parabolic equation: $$(\Delta_g-\partial_t)u=\Pi(u)(du,du).$$ Basically it says that $u(\cdot,t)$ does not blow up as $t\nearrow T$.

This is stated on page 182 of the book Variational Problems in Geometry by Seiki Nishikawa, where the author calls it Schauder estimate. However, the standard Schauder estimate, which involves the Hölder seminorm of $f$ in $t$ as well. The author lists Gilbarg–Trudinger (which obviously do not talk about parabolic equations) and a Japanese book (I don't speak Japanese) for such Schauder estimates.

So is the above estimate correct? If so, where can I find a proof?

Thanks in advance!

Edit: A possible approach might be to use the heat kernel, so that $u$ has an inegral expression which we can analyze. However I'm totally unfamiliar with estimates on the heat kernel. For example, is the estimate true for $M=\mathbb{R}^n$?

Edit: On page 316 of Partial Differential Equations III: Nonlinear Equations (Second Edition) by Michael E. Taylor, $(1.13)$ is the following $$\|e^{t\Delta}\|_{\mathcal{L}(C^r,C^{r+s})}\leq C_st^{-s/2},\quad0<t\leq1,$$ where $\Delta$ is the Laplacian on a complete Riemannian manifold. This implies my wanted estimate, but unfortunately no proof was given. I skimmed the earlier parts of the book and think maybe this follows from the estimates of pseudo-differential operators acting on Hölder–Zygmund spaces that is established earlier. Am I right?

  • $\begingroup$ Maybe this can help? mathoverflow.net/questions/328081/… $\endgroup$
    – Onil90
    Dec 11, 2019 at 12:50
  • $\begingroup$ @Onil90 The estimate here is different (and could not be deduced) from the standard Schauder estimate. $\endgroup$
    – Colescu
    Dec 11, 2019 at 14:04

1 Answer 1


For the heat equation in $\mathbb R^n$ the estimate holds. The main thing is the smoothness of the volume potential $g=\Gamma*f$, where $\Gamma$ is the fundamental solution. It is obtained in O. A. Ladyzenskaja, V. A. Solonnikov, and N. N. Ural'ceva, Linear and Quasi-linear Equations of Parabolic Type, ch.4, $\S2$. Though in the beginning of the paragraph the assumption on $f$ is Holder continuity wrt $x$ and $t$ both, in the process it is proven that $$ |D^2_xg(x,t)-D^2_xg(x',t)|\le C|x-x'|\sup_{s\in[0,T]}\|f(\cdot,s)\|_{C^\alpha(\mathbb R^n)},\quad x,x'\in \mathbb R^n,\ t>0. $$

  • $\begingroup$ Thanks! If this is true, could it be possible that the manifold case can be obtained by "freezing the coefficients"? $\endgroup$
    – Colescu
    Dec 11, 2019 at 16:26
  • $\begingroup$ I think that probably yes, but don't know how exactly. May be it would bу easier to obtain estimates for volume potential. Since the issue of smoothness is local it boils down to estimates of the volume potential in $\mathbb R^n$ for a uniformly parabolic operator with smooth coefficients. In $\S11$ of the same chapter such estimates for Holder $f$ are obtained. One has to check that no smoothness wrt $t$ is used when the required estimates $x$ are derived. $\endgroup$
    – Andrew
    Dec 11, 2019 at 19:31
  • $\begingroup$ I tried and I think it works: Take a normal coordinate neighborhood $U$ and a cutoff function $\zeta$ supported in $U$. We only need to estimate the $C^\alpha$ norm of $(\Delta-\partial_t)(\zeta u)$ in the space variable, where $\Delta$ is the standard Laplacian on $\mathbb{R}^n$. Then the proof is routine, using some basic interpolation inequalities for Hölder spaces. This is pretty much the same proof as the standard Schauder estimate, obtained by freezing the coefficients. However I'm not an expert at PDE so I'm not sure whether I've made any serious mistakes here. $\endgroup$
    – Colescu
    Dec 12, 2019 at 5:23

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