[proofplan] Fix $f \in C([0,1])$. [Uniform continuity](/page/Uniform%20Continuity) gives a quadratic majorant for the oscillation $|f(t)-f(x)|$, uniformly in both variables $x,t \in [0,1]$. For each fixed evaluation point $x$, positivity of $L_n$ converts the two-sided pointwise inequality for the function of $t$ into a two-sided inequality for $L_n f(x)-f(x)L_n e_0(x)$. The quadratic term expands into the three Korovkin test functions $e_0,e_1,e_2$, so the assumed convergence on those test functions controls the full error uniformly in $x$. [/proofplan]

[step:Build a quadratic modulus bound for $f$]Let $f \in C([0,1])$ be fixed. Since $[0,1]$ is compact and $f$ is continuous, $f$ is uniformly continuous on $[0,1]$ (citing a result not yet in the wiki: [Heine-Cantor theorem](/theorems/280)). Fix $\varepsilon>0$. Choose $\delta \in (0,1]$ such that, for all $s,t \in [0,1]$, \begin{align*} |s-t|<\delta \implies |f(s)-f(t)|<\varepsilon. \end{align*} Define the constant \begin{align*} M := \|f\|_{C([0,1])} \end{align*} and define \begin{align*} A := \frac{2M}{\delta^2}. \end{align*} Then, for all $x,t \in [0,1]$, \begin{align*} |f(t)-f(x)| \leq \varepsilon + A(t-x)^2. \end{align*} Indeed, if $|t-x|<\delta$, the bound follows from the choice of $\delta$. If $|t-x|\geq \delta$, then \begin{align*} |f(t)-f(x)| \leq |f(t)|+|f(x)| \leq 2M \leq A(t-x)^2. \end{align*}[/step]

[guided]We need an estimate that positivity can act on. Positivity compares functions pointwise, so we want to dominate the function $t \mapsto |f(t)-f(x)|$ by a simple nonnegative polynomial in $t$ whose image under $L_n$ is controlled by the hypotheses. Because $f$ is continuous on the compact interval $[0,1]$, it is uniformly continuous on $[0,1]$ (citing a result not yet in the wiki: Heine-[Cantor theorem](/theorems/759)). Thus, for the fixed number $\varepsilon>0$, there exists $\delta \in (0,1]$ such that, whenever $s,t \in [0,1]$ satisfy $|s-t|<\delta$, we have \begin{align*} |f(s)-f(t)|<\varepsilon. \end{align*} Let \begin{align*} M := \|f\|_{C([0,1])} \end{align*} and set \begin{align*} A := \frac{2M}{\delta^2}. \end{align*} We claim that, for every pair $x,t \in [0,1]$, \begin{align*} |f(t)-f(x)| \leq \varepsilon + A(t-x)^2. \end{align*} There are two cases. If $|t-x|<\delta$, uniform continuity gives $|f(t)-f(x)|<\varepsilon$, so the desired inequality follows because $A(t-x)^2 \geq 0$. If $|t-x|\geq \delta$, boundedness gives \begin{align*} |f(t)-f(x)| \leq |f(t)|+|f(x)| \leq 2M. \end{align*} Since $|t-x|\geq \delta$, we also have $(t-x)^2 \geq \delta^2$, hence \begin{align*} A(t-x)^2 = \frac{2M}{\delta^2}(t-x)^2 \geq 2M. \end{align*} Therefore $|f(t)-f(x)| \leq A(t-x)^2 \leq \varepsilon + A(t-x)^2$. This proves the uniform quadratic oscillation bound.[/guided]

[step:Apply positivity at a fixed evaluation point] Fix $x \in [0,1]$. Define the function $q_x \in C([0,1])$ by $q_x(t)=(t-x)^2$. The previous step gives the pointwise inequalities \begin{align*} -\varepsilon e_0(t)-Aq_x(t) \leq f(t)-f(x)e_0(t) \leq \varepsilon e_0(t)+Aq_x(t) \end{align*} for every $t \in [0,1]$. Since $L_n$ is linear and positive, applying $L_n$ to these inequalities gives, pointwise at $x$, \begin{align*} -\varepsilon L_n e_0(x)-A L_n q_x(x) \leq L_n f(x)-f(x)L_n e_0(x) \leq \varepsilon L_n e_0(x)+A L_n q_x(x). \end{align*} Hence \begin{align*} |L_n f(x)-f(x)L_n e_0(x)| \leq \varepsilon L_n e_0(x)+A L_n q_x(x). \end{align*} [/step]

[step:Rewrite the quadratic term using the three test functions] For the fixed $x \in [0,1]$, the function $q_x$ satisfies \begin{align*} q_x = e_2 - 2x e_1 + x^2 e_0. \end{align*} By linearity of $L_n$, \begin{align*} L_n q_x(x)=L_n e_2(x)-2xL_n e_1(x)+x^2L_n e_0(x). \end{align*} Since $x^2-2x^2+x^2=0$, we may rewrite this as \begin{align*} L_n q_x(x)=\bigl(L_n e_2(x)-e_2(x)\bigr)-2x\bigl(L_n e_1(x)-e_1(x)\bigr)+x^2\bigl(L_n e_0(x)-e_0(x)\bigr). \end{align*} Because $0 \leq x \leq 1$, taking absolute values gives \begin{align*} |L_n q_x(x)| \leq \|L_n e_2-e_2\|_{C([0,1])}+2\|L_n e_1-e_1\|_{C([0,1])}+\|L_n e_0-e_0\|_{C([0,1])}. \end{align*} Moreover, positivity gives $L_n q_x(x)\geq 0$, because $q_x \geq 0$ pointwise. [/step]

[step:Take the supremum and pass to the limit] For each $x \in [0,1]$, \begin{align*} |L_n f(x)-f(x)| \leq |L_n f(x)-f(x)L_n e_0(x)|+|f(x)|\,|L_n e_0(x)-e_0(x)|. \end{align*} Using the estimates above, $L_n e_0(x)\leq 1+\|L_n e_0-e_0\|_{C([0,1])}$, and $|f(x)|\leq M$, we obtain \begin{align*} |L_n f(x)-f(x)| \leq \varepsilon\bigl(1+\|L_n e_0-e_0\|_{C([0,1])}\bigr)+A\bigl(\|L_n e_2-e_2\|_{C([0,1])}+2\|L_n e_1-e_1\|_{C([0,1])}+\|L_n e_0-e_0\|_{C([0,1])}\bigr)+M\|L_n e_0-e_0\|_{C([0,1])}. \end{align*} Taking the supremum over $x \in [0,1]$ yields the same bound for $\|L_n f-f\|_{C([0,1])}$. By the three hypotheses, the three error terms \begin{align*} \|L_n e_0-e_0\|_{C([0,1])},\qquad \|L_n e_1-e_1\|_{C([0,1])},\qquad \|L_n e_2-e_2\|_{C([0,1])} \end{align*} converge to $0$. Therefore \begin{align*} \limsup_{n\to\infty}\|L_n f-f\|_{C([0,1])}\leq \varepsilon. \end{align*} Since $\varepsilon>0$ was arbitrary, it follows that \begin{align*} \|L_n f-f\|_{C([0,1])}\to 0. \end{align*} This proves the theorem. [/step]