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No: pointwise convergence need not be uniform convergence. The continuous function $f(x)=x^2$ maps $(0,1)$ into itself and is strictly below the identity. Its iteration of a map givesFor every finite $n$, these values tend to $1$ as $x\uparrow 1$, so ${\sup }_{0<x<1}{f}^n(x)=1$. With $\epsilon =1/2$, choosing $x>2^{-1/2^n}$ gives $f^n(x)>1/2$. Thus no single iterate works for every starting point, although each fixed starting point tends to zero.

A continuous function at $x_0$ satisfies the following quantified condition:The choice of $\delta$ may depend on $x_0$ and $\epsilon$.

For the bounded example, takeThis is a continuous function on its domain and $|f(x)|\le 1-x<1$. Along $x_n=(\pi /2+2\pi n{)}^{-1}$ its values tend to $1$, while along $y_n=(3\pi /2+2\pi n{)}^{-1}$ its values tend to $-1$. Hence the supremum is $1$ and the infimum is $-1$, and neither is attained. “Upper and lower bound” here means the least upper and greatest lower bounds.

For the nonnegative function on the whole real line, if $f\equiv 0$ the conclusion is immediate. Otherwise choose $x_{\ast }$ with $m=f(x_{\ast })>0$. The limits at both infinities give $R>|x_{\ast }|$ such that $f(x)<m/2$ whenever $|x|>R$. By the extreme value theorem, a continuous function on the compact interval $[-R,R]$ has a maximum $M=f(\alpha )$ there. Since $M\ge m$, it also exceeds every value outside this interval. Thus $f$ is bounded above and attains its global maximum. Nonnegativity ensures that either the function is zero or such a positive comparison value exists; without it, a function such as $-e^{-x^2}$ would have an unattained supremum of zero.

The extreme value theorem ensures that $M$ is finite. Since $f(x)\le |f(x)|\le M$ and $g(x)\ge 0$, multiplication preserves the inequality. Monotonicity of the Riemann integral gives the required boundTo prove the weighted mean value theorem for integrals, put $G={\int }_0^{1}gdx$ and $I={\int }_0^{1}fgdx$. If $G=0$, then $|I|\le {\int }_0^1|f|gdx\le MG=0$, so any $\alpha$ works. This includes $g\equiv 0$.

If $G>0$, the extreme value theorem gives $m_0=\min f$ and $M_0=\max f$. Integrating $m_{0}g\le fg\le M_{0}g$ gives $m_0\le I/G\le M_0$. A continuous function on an interval takes every value between its minimum and maximum by the intermediate value theorem. Choose $\alpha$ with $f(\alpha )=I/G$. Consequently

For $P(x)=x^3+4x-7$, the derivative $P^{\prime }(x)=3x^2+4$ is positive everywhere, so the mean value theorem makes $P$ strictly increasing. Since a polynomial is a continuous function, the intermediate value theorem and the values $P(1)=-2$, $P(2)=9$ locate its unique real root in $(1,2)$. By the rational root theorem, a rational root of a monic integer polynomial must be an integer. There is no integer in $(1,2)$, so the unique real root is an irrational number. The denominator part of that theorem also follows directly: if $a/b$ in lowest terms were a root, $a^3+4a{b}^2-7b^3=0$ would force $b\mid a^3$, hence $b=1$.

The intermediate value theorem states that a real-valued continuous function on a closed interval takes every value between its endpoint values. Explicitly, if $v$ lies between $f(s)$ and $f(t)$, there exists $u\in [s,t]$ with $f(u)=v$. No monotonicity is assumed.

Let $f(x)=-{\sum }_i\log ({\alpha }_i+x_i)$. The feasible probability simplex is a nonempty compact convex set, so the extreme value theorem guarantees a minimizer. Its Hessian matrix is diagonal with positive entries $({\alpha }_i+x_i{)}^{-2}$, making $f$ a strictly convex function. Hence the minimizer is unique.

Use the Lagrange multipliers $\lambda$ for the sum constraint and ${\nu }_i\ge 0$ for $-x_i\le 0$. The Lagrangian isThe KKT conditions areA strictly positive feasible allocation exists, so the Slater condition holds. These conditions are necessary and sufficient for this convex optimization. At least one coordinate is positive, giving $\lambda >0$. Put $\tau =1/\lambda$. For a positive coordinate, complementary slackness gives ${\alpha }_i+x_i=\tau$. At a zero coordinate, stationarity gives ${\nu }_i=1/\tau -1/{\alpha }_i\ge 0$, or ${\alpha }_i\ge \tau$. ThusHere $(u{)}_{+}=\max (u,0)$. This logarithmic water filling raises the smaller baseline values to a common level. The scalar left side is a continuous function and strictly increasing above ${\min }_i{\alpha }_i$, starts at zero, and tends to infinity, so exactly one $\tau$ solves it. One can find it by interval bisection, or sort the baseline values increasingly and find an index $k$ for whichusing ${\alpha }_{(n+1)}=+\infty$. Equality at the next baseline simply gives a zero allocation. Sorting and a running sum implement the water-filling algorithm in $O(n\log n)$ time.

Let $P$ be the row payoff matrix and $Q=P^T$. For $x$ in the probability simplex $\Delta$, write $u(x)=x^{T}Px$ and define the symmetric Nash gain mapThe map is a continuous function, has nonnegative coordinates, and sums to one. The Brouwer fixed-point theorem states that every continuous self-map of a nonempty finite-dimensional compact convex set has a fixed point. Apply it to $T:\Delta \to \Delta$, and let $x=T(x)$.

Set $G={\sum }_i{g}_i(x)$. The fixed-point equation gives $g_i(x)=G{x}_i$. If $G>0$, every positive $x_i$ has strictly positive gain, so $(Px{)}_i-u(x)=g_i(x)=G{x}_i$. Buta contradiction. Thus $G=0$ and every pure payoff $(Px{)}_i$ is at most $u(x)$. Since their $x$-weighted average equals $u(x)$, every supported action attains that maximum. Consequently $x$ is a best response to itself.

The column player's payoff vector against $x$ is $Q^{T}x=Px$, so exactly the same inequalities establish its best response. ThereforeThis supplies the whole fixed-point construction and fixed-point-to-equilibrium argument, rather than assuming Nash's theorem as a black box.

An explicit construction gives Bernstein's lethargy theorem in the requested inequality form. SetStrict decrease makes every coefficient positive. Telescoping and the limit assumption giveThusdefines a continuous function by the Weierstrass M-test and the uniform limit theorem. Apply the positive lacunary Chebyshev series calculation. At $n=0$,For $n\ge 1$, choose $K\ge 1$ so that $3^{K-1}\le n<3^K$. Its error isTherefore the function satisfiesThis explicit Chebyshev construction for Bernstein lethargy also has $E_n(f_{ϵ})\to 0$. It shows that continuity imposes no universal speed of convergence of best polynomial approximation, even though convergence itself follows from the Weierstrass approximation theorem.