[proofplan] We prove the contrapositive. If some approximant $q \in X$ had uniform error strictly smaller than $m$, then the difference $h := q-p$ would inherit a strict alternating sign pattern at the points $x_0,\dots,x_n$. Since $h$ is continuous, the intermediate value property forces one zero of $h$ in each interval $(x_i,x_{i+1})$, giving at least $n$ distinct zeros. This contradicts the Haar property, because $h$ is a nonzero element of the $n$-dimensional Haar space $X$. [/proofplan] [step:Assume a strict improvement and compare it with the alternating error] Suppose, toward a contradiction, that $E^* < m$. By the definition of the infimum, there exists $q \in X$ such that $\|f-q\|_\infty < m$. Define the function $h: [a,b] \to \mathbb{R}$ by $h(x) = q(x)-p(x)$ for $x \in [a,b]$. Since $p,q \in X$ and $X$ is a linear subspace of $C([a,b];\mathbb{R})$, we have $h \in X \subset C([a,b];\mathbb{R})$. For each $i \in \{0,\dots,n\}$, write $h(x_i) = q(x_i)-p(x_i) = \bigl(f(x_i)-p(x_i)\bigr)-\bigl(f(x_i)-q(x_i)\bigr)$. Multiplying by $\sigma(-1)^i$ gives $\sigma(-1)^i h(x_i) = \sigma(-1)^i\bigl(f(x_i)-p(x_i)\bigr) - \sigma(-1)^i\bigl(f(x_i)-q(x_i)\bigr)$. Using the assumed alternation inequality and the bound $|f(x_i)-q(x_i)| \leq \|f-q\|_\infty < m$, we obtain $\sigma(-1)^i h(x_i) \geq m - |f(x_i)-q(x_i)| > 0$. Thus $h(x_i) \neq 0$ for every $i$, and the sign of $h(x_i)$ is the sign of $\sigma(-1)^i$. [guided] We want to compare the alleged better approximant $q$ with the candidate $p$. The natural object is their difference, so define the function $h: [a,b] \to \mathbb{R}$ by $h(x) = q(x)-p(x)$ for $x \in [a,b]$. Because $X$ is a linear subspace and $p,q \in X$, the difference $h=q-p$ also belongs to $X$. Since $X \subset C([a,b];\mathbb{R})$, the function $h$ is continuous on $[a,b]$. The point of assuming $\|f-q\|_\infty < m$ is that $q$ is uniformly closer to $f$ than the guaranteed alternating error size $m$. At a point $x_i$, we rewrite $h(x_i)$ as $h(x_i) = q(x_i)-p(x_i) = \bigl(f(x_i)-p(x_i)\bigr)-\bigl(f(x_i)-q(x_i)\bigr)$. Multiplying by the prescribed sign $\sigma(-1)^i$, we get $\sigma(-1)^i h(x_i) = \sigma(-1)^i\bigl(f(x_i)-p(x_i)\bigr) - \sigma(-1)^i\bigl(f(x_i)-q(x_i)\bigr)$. The first term on the right is at least $m$ by hypothesis. The second term has absolute value at most $|f(x_i)-q(x_i)|$, and this is bounded by $\|f-q\|_\infty < m$. Therefore $\sigma(-1)^i h(x_i) \geq m - |f(x_i)-q(x_i)| > 0$. So $h(x_i)$ is never zero, and its sign alternates as $i$ increases because the factor $(-1)^i$ alternates. [/guided] [/step] [step:Use the alternating signs to force one zero in each interval] Fix $i \in \{0,\dots,n-1\}$. From the previous step, $\sigma(-1)^i h(x_i) > 0$ and $\sigma(-1)^{i+1} h(x_{i+1}) > 0$. Multiplying the [second inequality](/theorems/2136) by $-1$ gives $\sigma(-1)^i h(x_{i+1}) < 0$. Hence $h(x_i)$ and $h(x_{i+1})$ have opposite signs. Since $h$ is continuous on $[x_i,x_{i+1}]$, the [Intermediate Value Theorem](/theorems/632) applies to the restricted map $h|_{[x_i,x_{i+1}]}: [x_i,x_{i+1}] \to \mathbb{R}$. Because $h(x_i)$ and $h(x_{i+1})$ are nonzero and have opposite signs, there exists $y_i \in (x_i,x_{i+1})$ such that $h(y_i)=0$. The intervals $(x_i,x_{i+1})$ are pairwise disjoint, so the points $y_0,\dots,y_{n-1}$ are distinct. Therefore $h$ has at least $n$ distinct zeros in $[a,b]$. [guided] Fix $i \in \{0,\dots,n-1\}$. From the previous step, $\sigma(-1)^i h(x_i)>0$ and $\sigma(-1)^{i+1}h(x_{i+1})>0$. Multiplying the [second inequality](/theorems/2899) by $-1$ gives $\sigma(-1)^i h(x_{i+1})<0$. Thus $h(x_i)$ and $h(x_{i+1})$ are nonzero numbers with opposite signs. Now we use the continuity of $h$ on the closed interval $[x_i,x_{i+1}]$. The restricted map $h|_{[x_i,x_{i+1}]}: [x_i,x_{i+1}] \to \mathbb{R}$ is continuous because $h \in C([a,b];\mathbb{R})$. Since its endpoint values have opposite signs, the [Intermediate Value Theorem](/theorems/632) gives a point $y_i \in (x_i,x_{i+1})$ with $h(y_i)=0$. The point lies in the open interval because both endpoint values are nonzero. This construction gives one zero $y_i$ for each interval $(x_i,x_{i+1})$. These intervals are pairwise disjoint, so no two constructed zeros can be equal. Hence $y_0,\dots,y_{n-1}$ are $n$ distinct zeros of $h$ in $[a,b]$. [/guided] [/step] [step:Contradict the Haar property and conclude the lower bound] For each $i \in \{0,\dots,n\}$, the inequality $\sigma(-1)^i h(x_i) > 0$ shows in particular that $h(x_i) \neq 0$. Hence $h$ is not the zero function. Thus $h$ is a nonzero element of $X$ with at least $n$ distinct zeros in $[a,b]$. This contradicts the Haar-space property of $X$, which says that every nonzero element of an $n$-dimensional Haar space has at most $n-1$ zeros in $[a,b]$. Therefore no $q \in X$ can satisfy $\|f-q\|_\infty < m$. Hence $\|f-q\|_\infty \geq m$ for every $q \in X$. Taking the infimum over $q \in X$ gives $E^* = \inf_{q \in X}\|f-q\|_\infty \geq m$. This proves the theorem. [guided] The previous step produced $n$ distinct points $y_0,\dots,y_{n-1} \in [a,b]$ such that $h(y_i)=0$ for every $i \in \{0,\dots,n-1\}$. We also know that $h$ is not the zero function: for each $i \in \{0,\dots,n\}$, the inequality $\sigma(-1)^i h(x_i)>0$ implies $h(x_i) \neq 0$. Now we use the defining property of a Haar space. Since $X$ is an $n$-dimensional Haar space on $[a,b]$, every nonzero element of $X$ has at most $n-1$ zeros in $[a,b]$. But $h=q-p$ belongs to $X$, is nonzero, and has at least the $n$ distinct zeros $y_0,\dots,y_{n-1}$. This is impossible. The contradiction came from assuming $E^*<m$. Therefore no $q \in X$ can satisfy $\|f-q\|_\infty<m$. Equivalently, $\|f-q\|_\infty \geq m$ for every $q \in X$. Taking the infimum over all $q \in X$ gives $E^* = \inf_{q \in X}\|f-q\|_\infty \geq m$, which is the desired conclusion. [/guided] [/step]