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Suppose for contradiction that some $q\in {\mathcal{P}}_n$ satisfiesAt $t_i$,Because $|f(t_i)-q(t_i)|<a_i$, the values $p(t_i)-q(t_i)$ have strictly alternating signs. The intermediate value theorem therefore gives at least one root of $p-q$ in each of the $n+1$ intervals $(t_i,t_{i+1})$. A nonzero polynomial of degree at most $n$ cannot have $n+1$ distinct roots. If $p-q$ were identically zero, its error at $t_i$ would be $a_i\ge a_{\ast }$, also a contradiction. Hence

For any nonzero positive drag, the cubic dust gravitational dispersion relation with gas drag is negative at $s=0$ if ${\omega }^2<{\Omega }^2$ and positive at sufficiently large positive $s$. The intermediate value theorem therefore guarantees a growing real root. This existence argument is independent of the weak-drag expansion and includes the dynamically unstable ${\omega }^2<0$ regime.

Every nonconstant harmonic function $u:\mathbb{C}\to \mathbb{R}$ is unbounded above and below by the harmonic Liouville theorem. The intermediate value theorem then gives $u(\mathbb{C})=\mathbb{R}$.

If a twice differentiable real function on $\mathbb{R}$ satisfies $f^{\prime \prime }(x)\le -c$ for some fixed $c>0$, then it has a unique global maximum. The mean value theorem shows that $f^{\prime }$ is strictly decreasing, and that $f^{\prime }(x)+cx$ is nonincreasing. Hence $f^{\prime }(x)\le f^{\prime }(0)-cx$ for $x>0$ and $f^{\prime }(x)\ge f^{\prime }(0)-cx$ for $x<0$. These inequalities force opposite derivative signs at the two ends. The intermediate value theorem supplies a unique zero of $f^{\prime }$, at which $f$ changes from increasing to decreasing. Merely having $f^{\prime \prime }<0$ does not guarantee existence: $f(x)=-e^{-x}$ is increasing and approaches its unattained supremum zero.

For real continuous functions $f,g$ on a compact interval with $g\ge 0$, some $\alpha$ in the interval satisfies $\int fg=f(\alpha )\int g$. If $\int g>0$, the weighted average lies between the minimum and maximum of $f$, and the intermediate value theorem supplies $\alpha$. If $\int g=0$, the bound $|\int fg|\le \sup |f|\int g$ makes both sides zero. Nonnegativity of the weight is essential to this argument.