Question of an inequality from curve-shortening flow I am reading a paper about the curve shortening flow which make use of one inequality but I don't know where does it come from where f(x,t) is a smooth function and C is a constant depending on time t. Since the curve $\gamma$ is closed and bounded, it also compact,$$\underset{\gamma}{\max}|f|^2\le C\int_\gamma|f'|^2+f^2$$ My friend told me it maybe from Morrey's inequality but I cant see it.Can anyone tell me what it is? Paper:https://projecteuclid.org/journals/journal-of-differential-geometry/volume-23/issue-1/The-heat-equation-shrinking-convex-plane-curves/10.4310/jdg/1214439902.full [Corollary 4.4.4]
 A: $\newcommand{\thh}{\theta}$The inequality in the proof of Corollary 4.4.4 in the linked paper is stated there (without proof) literally as follows:
\begin{equation}
    \max|f|^2\le C\int|f'|^2+f^2. \tag{1}\label{1}
\end{equation}
The paper does not appear to specify the meaning of $C$ or $\int$ or $f$. The proof only says that \eqref{1} is then to apply to $k''$, where $k$ is the curvature of a convex planar curve; $k$ is a function of "the angle $\thh$ of the tangent line" (with values in $[0,2\pi)$) and a parameter $t$. The prime $'$ denotes the partial derivative with respect to $\thh$.
So, apparently $\int|f'|^2+f^2$ means $\int_0^{2\pi}(|f'|^2+f^2)$, and it is enough to show that
\begin{equation}
    \max|f|^2\le C\int_0^{2\pi}(|f'|^2+f^2), \tag{2}\label{2}
\end{equation}
where $C$ is a universal positive real constant and $f\in C^1[0,2\pi]$.
Without loss of generality, $\max|f|=\max f$ (otherwise, replace $f$ by $-f$). So,
\begin{equation}
    \max|f|=\max f\le\min f+\int_0^{2\pi}|f'|
    =\int_0^{2\pi}\Big(|f'|+\frac{\min f}{2\pi}\Big)
\le \int_0^{2\pi}\Big(|f'|+\frac{|f|}{2\pi}\Big)
    \le\int_0^{2\pi}(|f'|+|f|)
\end{equation}
and hence, by the Cauchy--Schwarz inequality,
\begin{equation}
    \max|f|^2
    \le2\pi\int_0^{2\pi}(|f'|+|f|)^2
    \le4\pi\int_0^{2\pi}(|f'|^2+f^2).
\end{equation}
So, \eqref{2} follows.
