Let $f$ be a $C^2$ function defined on $[0,1] \times [0,1]$. Let $0 \leq f(x) < 1$ on $[0,1] \times [0,1] \setminus \left(\frac{1}{2},\frac{1}{2}\right)$ and $f(\frac{1}{2},\frac{1}{2})=1$. It has no critical point except $(\frac{1}{2},\frac{1}{2})$. I wonder there exists a number $0<a<1$ such that $f^{-1}(a)$ to be a circle.
By the smooth manifold theory, $f^{-1}(a)$ is always a 1-dimensional submanifold. Since $f$ is continuous, it is closed. Thus, it is a compact. But it can still be a line segment.
Any reference are welcome. Thanks in advance.
There's one way to do it:
Assume the contrary that $f^{-1}(a)$ is a line segment for all $ a\in [0,1)$. Then there is $c_a \in \partial ([0,1]\times [0,1])$ so that $f(c_a) = a$. As the boundary is compact, there is $a_n \to 1$ so that $c_{a_n} \to b$, where $b$ lies in the boundary. Then $$f(b) = \lim_{n\to \infty} f(c_{a_n}) = \lim_{n\to \infty} a_n = 1$$ and this cannot be true. As the level set can only be line segments or circles, you are done.
There's another way:
As $f$ has the maximum only at $(\frac 12, \frac 12)$, there is $\epsilon>0$ so that $f(x) <1-\epsilon$ for all $x$ closed to the boundary. That means $f^{-1}(a)$ for $1>a>1-\epsilon$ will be a closed one dimensional manifold without boundary: that is, a circle.