I consider $\Sigma:=S^2$, as a unit radius sphere in $\mathbb R^3$ so that it has an induced metric, measure etc on it. Is the following statement true?


For a large constant $K$, there exist constants $\epsilon(K)$ and $c(K)$ such that the following is satisfied: for any
smooth positive real-valued function $f$ on $\Sigma$ whose integral over $\Sigma$ is less than $K$, there is a circle of radius $\epsilon$ on which the pointwise norm of $f$ is bounded by $c$.


A circle of radius $\epsilon$ on $\Sigma$ has a center $x \in \Sigma$ and consists of points that are at distance $\epsilon$ with respect to the induced metric mentioned above.



Is there a curve on a surface where an integrable function is pointwise bounded?
In short: There is an integral bound but no pointwise bound.


**Integral bounds:**
Fix any $\epsilon>0$.
Let us denote the circle of radius $\epsilon$ centered at $x\in\Sigma$ by $A\_x=A\_x(\epsilon)$, and let $\sigma=\sigma(\epsilon)$ be the (dimension one Hausdorff) measure of this circle (independent of $x$).
For any smooth (in fact any $L^1$) function $f:\Sigma\to\mathbb R$ we have
$$
\int\_\Sigma f(x)dx
=
\sigma^{-1}\int\_\Sigma\left(\int\_{A\_x}f(y)dy\right)dx.
$$
You can prove this using the [coarea formula](https://en.wikipedia.org/wiki/Smooth_coarea_formula) (and perhaps other ways), but it's easy to see intuitively why it must be true: both sides integrate $f$ over $\Sigma$ and everything is rotation invariant, so they must agree up to a constant (which is easy to fix with a constant function).


This integral identity gives the following claim which gives an integral estimate instead of a pointwise one:



> 
> Fix any $\epsilon>0$ so that the circle of that radius is a circle. Let $f:\Sigma\to[0,\infty)$ be continuous. Then
>  $$
> \min\_{x\in\Sigma}\int\_{A\_x(\epsilon)}f(y)dy\leq\frac{\sigma(\epsilon)}{|\Sigma|}\int\_\Sigma f(x)dx.
> $$
>  The constant $\frac{\sigma(\epsilon)}{|\Sigma|}$ is optimal.
> 
> 
> 


Note that $\epsilon$ is arbitrary.
The only thing you can gain by changing its value is changing the measure $\sigma(\epsilon)$.
In your notation, we get $c(K)=\frac{\sigma(\epsilon)}{|\Sigma|}K$ (and the claim would be false for any smaller constant).


**Pointwise bound:**
If we want to estimate $f$ pointwise on circles, we are interested in the quantity
$$
Q\_\epsilon(f)=\min\_{x\in\Sigma}\max\_{y\in A\_x(\epsilon)}f(y).
$$
If $f\geq0$ and $\int\_\Sigma f\leq K$, the question is whether there is a uniform bound on $Q\_\epsilon(f)$ for some $\epsilon>0$.
The answer is negative.
(Clearly $\min\_\Sigma f\leq Q\_\epsilon(f)\leq\max\_\Sigma f$ but we can't say much more.)


To see this, fix any $\epsilon>0$.
Take a curve $\gamma:S^1\to\Sigma$ so that the $\epsilon/2$-neighborhood of $\gamma(S^1)$ is all of $\Sigma$.
Take a small parameter $\delta>0$ (much smaller than $\epsilon$) and let
$$
f\_\delta(x)=\max(0,1-\delta^{-1}d(x,\gamma(S^1))).
$$
This continuous function is supported in a $\delta$-neighborhood of $\gamma(S^1)$ and it is bounded by one, so $\int\_\Sigma f\_\delta\to0$ as $\delta\to0$.
On the other hand, since the curve $\gamma$ meets every circle $A\_x(\epsilon)$ and $f\_\delta=1$ on this curve, we have $Q\_\epsilon(f\_\delta)=1$.
If we normalize the functions $f\_\delta$ to have integral $K$, we observe that $Q\_\epsilon(f\_\delta)\to\infty$ as $\delta\to0$.
(You can mollify the functions to make them smooth; regularity is no problem.)


