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5. Comments on Homework 2, Problem 3

≪ 4. Integration Review | Table of Contents | 6. Differentiating Power Series, Taylor Series ≫

After grading the second homework a few days ago, I noticed that a lot of people had missed the third problem of the homework. I’ll provide a (more or less) full solution here, and I’ll point out the common mistakes made along the way.

For ease of notation, I’ll denote by $\left\lVert f \right\rVert_\infty := \sup _{x\in \left[ 0,1 \right]} \left\lvert f(x) \right\rvert$ (the “sup-norm” of $f$), and I’ll abbreviate $C ^ \alpha$ for $C ^ \alpha \left( \left[ 0,1 \right] \right)$.

For the first part of the problem, it’s good to verify that this is indeed a metric; however, I didn’t take points off if you didn’t verify this fact. Ultimately it’s not a critical part of the problem, and the completeness of the metric space is more important.

Let $\left\lbrace f_n \right\rbrace$ be a Cauchy sequence in $C ^{\alpha }$ with respect to $d_ \alpha$. We need to find some $f\in C ^{\alpha }$ such that $\left\lVert f_n - f \right\rVert_ \alpha \to 0$ as $n\to \infty$. Thus, we not only need to verify that $\left\lVert f \right\rVert _ \alpha$ is finite, but also that the norms converge to zero. This was the most common mistake: half of the submissions omitted this last step.

First, we observe for any $g\in C^ \alpha$, we have $\left\lVert g \right\rVert_\infty \leq \left\lVert g \right\rVert_ \alpha$. Thus, by Cauchiness in $C^ \alpha$, for every $\epsilon >0$ there exists some $N_ \epsilon$ such that $n, m > N_ \epsilon$ implies $$\left\lVert f_n-f_m \right\rVert _ \infty \leq \left\lVert f_n-f_m \right\rVert _ \alpha < \epsilon .$$ In particular, $\left\lbrace f_n \right\rbrace$ is a uniformly Cauchy sequence of functions on $\left[ 0,1 \right]$, and thus they have a uniform limit $f$.

Now, we’ll verify that $\left\lVert f \right\rVert _ \alpha < \infty$. Certainly $f$ is continuous, so $\left\lVert f \right\rVert_\infty < \infty$; we need only handle the second term of the $C^ \alpha$ norm.

Since $\left\lbrace f_n \right\rbrace$ are a Cauchy sequence in $C^ \alpha$, they are bounded; in particular, there exists some $M\geq 0$ such that $$\sup _{\substack{x,y\in \left[ 0,1 \right] \\ x\neq y}} \frac{\left\lvert f_n(x) - f_n(y) \right\rvert}{\left\lvert x-y \right\rvert^ \alpha }\leq M$$ for all $M$. Then, for any $x\neq y$ in $\left[ 0,1 \right]$, we have that $$\frac{\left\lvert f(x)-f(y) \right\rvert}{\left\lvert x-y \right\rvert^ \alpha } = \lim _{n\to\infty} \frac{\left\lvert f_n(x)-f_n(y) \right\rvert}{\left\lvert x-y \right\rvert^ \alpha }\leq M.$$ Thus, $\left\lVert f \right\rVert_ \alpha \leq \left\lVert f \right\rVert_ \infty + M$, hence $f\in C^ \alpha$. Note we can freely move the limit inside and outside the absolute values here because absolute values are continuous.

Next, we need to verify that $\left\lVert f-f_n \right\rVert_ \alpha \to 0$ as $n\to\infty$. Certainly $\sup _{x\in \left[ 0,1 \right]} \left\lvert f(x)-f_n(x) \right\rvert\to 0$ as $n\to\infty$, so we need only show $$\sup _{\substack{x,y\in \left[ 0,1 \right] \\ x \neq y}} \frac{f(x)-f_n(x) - f(y) + f_n(y)}{ \left\lvert x-y \right\rvert^ \alpha }\to 0$$ as $n\to\infty$.

By Cauchiness, for any $\epsilon > 0$, there is some $N_ \epsilon$ such that for all $m>n>N_ \epsilon$, $\left\lVert f_m-f_n \right\rVert_ \alpha < \epsilon$. Then, we have for any $x\neq y$ in $\left[ 0,1 \right]$ that

$$\begin{align*} \frac{\left\lvert f(x)-f_n(x) - f(y) + f_n(y) \right\rvert}{\left\lvert x-y \right\rvert^ \alpha } & \\ \leq \frac{\left\lvert f(x)-f_m(x)-f(y) + f_m(y) \right\rvert}{\left\lvert x-y \right\rvert^ \alpha } &+ \frac{\left\lvert f_m(x)-f_n(x)-f_m(y)+f_n(y) \right\rvert}{\left\lvert x-y \right\rvert^ \alpha }. \end{align*}$$

The second term is bounded by $\left\lVert f_m-f_n \right\rVert_ \alpha < \epsilon$ when $m$ is sufficiently large. Moreover, since $x$ and $y$ are fixed, we can make the first term as small as we want to, i.e. by finding $m$ so large that $\left\lvert f(x)-f_m(x)-f(y)+f_m(y) \right\rvert < \epsilon \left\lvert x-y \right\rvert^ \alpha$ by uniform convergence. Note here we have the freedom to choose $m$ as large as we want.

Thus, since $x$ and $y$ were arbitrary, we conclude that for all $n>N _ \epsilon$, $$\sup _{\substack{x, y\in \left[ 0,1 \right]\\x\neq y}} \frac{\left\lvert f(x)-f_n(x)-f(y)+f_n(y) \right\rvert}{\left\lvert x-y \right\rvert^ \alpha } < 2 \epsilon.$$ It follows that $\left\lVert f-f_n \right\rVert_ \alpha \to 0$ as $n\to\infty$.

We have shown that every Cauchy sequence in $C^ \alpha$ converges to a limit with respect to the $C^ \alpha$ norm, hence $C^ \alpha$ is a complete metric space.

For part 2 of this problem, we observe that for all $x, y\in \left[ 0,1 \right]$, we have $\left\lVert f_n \right\rVert _{\frac{1}{2}} \leq 1$ for all $n$ implies $$\left\lvert f_n(x)-f_n(y) \right\rvert \leq \left\lvert x-y \right\rvert ^{\frac{1}{2}}.$$ Thus, for any $\epsilon > 0$, we have $\left\lvert x-y \right\rvert < \epsilon ^2$ implies $\left\lvert f_n(x)-f_n(y) \right\rvert < \epsilon$ for all $x, y\in \left[ 0,1 \right]$; thus $\left\lbrace f_n \right\rbrace$ is an equicontinuous family of functions. Moreover, since $\left\lVert f_n \right\rVert_\infty \leq \left\lVert f_n \right\rVert_ \alpha \leq 1$ for all $n$, they are also uniformly bounded. Thus, by Ascoli’s theorem, there is a uniformly converging subsequence $f _{n_k}$ with some uniform limit $f$. Relabelling, we may assume without loss of generality that $f_n\to f$ uniformly to begin with.

First, we claim that $\left\lVert f \right\rVert _{\frac{1}{2}} < \infty$. Certainly $\left\lVert f \right\rVert_\infty \leq 1$; we need only handle the second term. But for any $x\neq y$ in $\left[ 0,1 \right]$, we have $$\frac{\left\lvert f(x)-f(y) \right\rvert}{\left\lvert x-y \right\rvert ^{\frac{1}{2}}} = \lim _{n\to\infty} \frac{\left\lvert f_n(x)-f_n(y) \right\rvert}{ \left\lvert x-y \right\rvert ^{\frac{1}{2}}} \leq 1,$$ thus $\left\lVert f \right\rVert _{\frac{1}{2}} \leq 2$.

This does not mean that $f_n\to f$ in $C ^{\frac{1}{2}}$. This was the most common mistake in this section. Ascoli’s theorem only affords you uniform convergence, and you have to work harder to control the second term in the $C ^{\frac{1}{2}}$ or $C ^{\frac{1}{3}}$ norm.

What we get out of this is that by the triangle inequality, $\left\lVert f-f_n \right\rVert _{\frac{1}{2}} \leq 3$ for all $n$; we can exploit this boundedness when analysing the $C ^{\frac{1}{3}}$ norm of $f-f_n$. Specifically, we want to show that for all $\epsilon > 0$, there is some $N_ \epsilon$ such that for all $n > N_ \epsilon$, $$\sup _{\substack{x, y\in \left[ 0,1 \right] \\ x\neq y}} \frac{\left\lvert f(x)-f_n(x)-f(y)+f_n(y) \right\rvert}{ \left\lvert x-y \right\rvert ^{\frac{1}{3}}} < \epsilon .$$

Fix an $x\neq y$ in $\left[ 0,1 \right]$. We have that when $\left\lvert x-y \right\rvert$ is quite large, the uniform convergence of $f_n\to f$ will make the numerator of the above quite small. However, when $\left\lvert x-y \right\rvert$ is quite small, we will have to do better. Specifically, we shall decompose $$\frac{\left\lvert f(x)-f_n(x)-f(y)+f_n(y) \right\rvert}{ \left\lvert x-y \right\rvert ^{\frac{1}{3}}} = \frac{\left\lvert f(x)-f_n(x)-f(y)+f_n(y) \right\rvert}{\left\lvert x-y \right\rvert ^{\frac{1}{2}}} \cdot \left\lvert x-y \right\rvert ^{\frac{1}{6}}.$$

When $\left\lvert x-y \right\rvert\leq \left( \frac{\epsilon }{3} \right)^6$, the right hand side is smaller than $\epsilon$. This handles the first case. On the other hand, if $\left\lvert x-y \right\rvert > \left( \frac{\epsilon }{3} \right) ^ 6$, we need $$\left\lvert f(x)-f_n(x)-f(y)+f_n(y) \right\rvert \cdot \left( \frac{3}{ \epsilon } \right) ^{\frac{1}{2}} < \epsilon.$$ But the left side is bounded by $2 \left\lVert f-f_n \right\rVert_\infty \left( \frac{3}{\epsilon } \right) ^{\frac{1}{2}}$, so picking $N_ \epsilon$ large enough so that $n > N_ \epsilon$ implies $$\left\lVert f-f_n \right\rVert_\infty < \frac{\epsilon ^{\frac{3}{2}} }{ 2 \sqrt 3}$$ will do the trick. But this $N_ \epsilon$ is independent of $x$ and $y$, thus we see that $\left\lVert f-f_n \right\rVert _{\frac 13} < \left\lVert f-f_n \right\rVert_ \infty + \epsilon$ for all $n > N _ \epsilon$. By increasing $N_ \epsilon$ if needed, we can make this smaller than say $2 \epsilon$, hence $\left\lVert f-f_n \right\rVert _{\frac{1}{3}} \to 0$, as claimed. $\square$