[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*}
[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]
[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]
