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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. \(\square\)