Reading about the Nash-Kuiper theorem, I found the following statement in Y. Eliashberg and N. Mishachev's book Introduction to the $h$-Principle:
Is there a regular homotopy $f_t:S^2\rightarrow\mathbb R^3$ which begins with the inclusion $f_0$ of the unit sphere and ends with an isometric immersion $f_1$ into the ball of radius $\frac{1}{2}$? Here the word 'isometric' means preserving length of all curves. The answer is, of course, negative if $f_1$ is required to be $\mathcal C^2$-smooth. Indeed, in this case the Gaussian curvature of the metric on $S^2$ should be at least $\geq 4$ somewhere.
I am puzzled about the last affirmation. More precisely, suppose we have a regular surface immersed inside a ball of radius $1/2$. Why should its Gaussian curvature be at least 4 at some point? This is intuitively clear, but I am not familiar with results about isometric immersions, so I don't know how one should stablish such a conclusion.
Let $f(p) \in f(S^2)$ be a point whose norm $R = |f(p)|$ is maximal, so $R \le 1/2$ and $f(S^2) \subset B(R) =$ the ball of radius $R$ centered at the origin.
Consider the sphere $\partial B(R)$. It follows that $f(S^2)$ is tangent to $\partial B(0,R)$. Let $f(p)$ be a point of tangency, $p \in S^2$.
Since $f(S^2)$ is entirely contained in $B(0,R)$, it follows that the osculating sphere to $f(S^2)$ at $p$ has radius $\le R \le 1/2$. The Gaussian curvature at $p$ is therefore $\ge (1/2)^{-2} = 4$.
By the way, this has nothing in particular to do with isometric immersions, except insofar as every isometric immersion is a local embedding at each point of its domain. Once one knows that $f(S^2)$ is tangent to $\partial B(R)$ at $f(p)$, the argument becomes entirely local, taking place on (the immersed image of) some neighborhood of $p \in S^2$ on which $f$ restricts to an embedding.