<Home>Qualifying Exams><Analysis>Spring 2021 - Problem 6

Spring 2021 - Problem 6

operator theory

We say that a linear operator T ⁣:C([0,1])→C([0,1])\func{T}{C\p{\br{0,1}}}{C\p{\br{0,1}}} is positive if T(f)(x)≥0T\p{f}\p{x} \geq 0 for all x∈[0,1]x \in \br{0, 1} whenever f∈C([0,1])f \in C\p{\br{0,1}} satisfies f(x)≥0f\p{x} \geq 0 for all x∈[0,1]x \in \br{0, 1}. Let

Tn ⁣:C([0,1])→C([0,1])\func{T_n}{C\p{\br{0,1}}}{C\p{\br{0,1}}}

be a sequence of positive linear operators such that Tn(f)→fT_n\p{f} \to f uniformly on [0,1]\br{0, 1} if ff is a polynomial of degree less than or equal to 22. Show that

Tn(f)→funiformly on [0,1],T_n\p{f} \to f \quad\text{uniformly on } \br{0, 1},

for every f∈C([0,1])f \in C\p{\br{0,1}}.

Hint: Let f∈C([0,1])f \in C\p{\br{0,1}}. Show first that for every ε>0\epsilon > 0 there exists Cε>0C_\epsilon > 0 such that

∣f(x)−f(y)∣≤ε+Cε∣x−y∣2for all x,y∈[0,1].\abs{f\p{x} - f\p{y}} \leq \epsilon + C_\epsilon \abs{x - y}^2 \quad\text{for all } x, y \in \br{0, 1}.
Solution.

We first prove the hint. Let f∈C([0,1])f \in C\p{\br{0, 1}} and ε>0\epsilon > 0. Since [0,1]\br{0, 1} is compact, we see that ff is uniformly continuous, so there exists δ>0\delta > 0 such that if ∣x−y∣<δ\abs{x - y} < \delta, then ∣f(x)−f(y)∣<ε\abs{f\p{x} - f\p{y}} < \epsilon. Set

Cε=1δ2sup⁡x,y∈[0,1] ∣f(x)−f(y)∣.C_\epsilon = \frac{1}{\delta^2} \sup_{x, y \in \br{0,1}} \,\abs{f\p{x} - f\p{y}}.

Thus, if ∣x−y∣<δ\abs{x - y} < \delta, then the inequality in the hint is clear, and otherwise, if ∣x−y∣≥δ\abs{x - y} \geq \delta, then

Cε∣x−y∣2≥sup⁡x,y∈[0,1] ∣f(x)−f(y)∣≥∣f(x)−f(y)∣,C_\epsilon \abs{x - y}^2 \geq \sup_{x, y \in \br{0,1}} \,\abs{f\p{x} - f\p{y}} \geq \abs{f\p{x} - f\p{y}},

so the inequality holds. Next, because TnT_n is positive, we see that if f≤gf \leq g, then g−f≥0g - f \geq 0, so Tn(g−f)≥0T_n\p{g - f} \geq 0 also. By linearity, we get Tn(f)≤Tn(g)T_n\p{f} \leq T_n\p{g}, i.e., TnT_n preserves inequalities. Thus, if x,y∈[0,1]x, y \in \br{0, 1}, we have

−ε−Cε(x−y)2≤f(x)−f(y)≤ε+Cε(x−y)2  ⟹  −Tn(ε+Cε( ⋅ −y)2)(x)≤Tn(f−f(y))(x)≤Tn(ε+Cε( ⋅ −y)2)(x).\begin{gathered} -\epsilon - C_\epsilon\p{x - y}^2 \leq f\p{x} - f\p{y} \leq \epsilon + C_\epsilon\p{x - y}^2 \\ \implies -T_n\p{\epsilon + C_\epsilon\p{\:\cdot\: - y}^2}\p{x} \leq T_n\p{f - f\p{y}}\p{x} \leq T_n\p{\epsilon + C_\epsilon\p{\:\cdot\: - y}^2}\p{x}. \end{gathered}

On the left- and right-hand sides, we have (at most) second degree polynomials, so if nn is large enough, uniform convergence gives

−2ε−Cε(x−y)2≤Tn(f−f(y))(x)≤2ε+Cε(x−y)2-2\epsilon - C_\epsilon\p{x - y}^2 \leq T_n\p{f - f\p{y}}\p{x} \leq 2\epsilon + C_\epsilon\p{x - y}^2

for every x,y∈[0,1]x, y \in \br{0, 1}. In particular, setting x=yx = y, we get

−2ε≤Tn(f−f(y))(y)≤2ε,-2\epsilon \leq T_n\p{f - f\p{y}}\p{y} \leq 2\epsilon,

so Tn(f−f(y))(y)→0T_n\p{f - f\p{y}}\p{y} \to 0 uniformly. Finally, by the triangle inequality, we have for any x∈[0,1]x \in \br{0, 1}

∣Tn(f)(x)−f(x)∣≤∣Tn(f−f(x))(x)∣+∣f(x)⋅Tn(1)(x)−f(x)∣≤∣Tn(f−f(x))(x)∣+∥f∥L∞∣Tn(1)(x)−1∣\begin{aligned} \abs{T_n\p{f}\p{x} - f\p{x}} &\leq \abs{T_n\p{f - f\p{x}}\p{x}} + \abs{f\p{x} \cdot T_n\p{1}\p{x} - f\p{x}} \\ &\leq \abs{T_n\p{f - f\p{x}}\p{x}} + \norm{f}_{L^\infty} \abs{T_n\p{1}\p{x} - 1} \end{aligned}

which tends to 00 uniformly, since Tn(1)→1T_n\p{1} \to 1 uniformly.