Let \(\Delta \) be the unit disc, \(\mathbf{T} \) its boundary and consider the disk algebra \(\mathcal{A}\) of those continuous functions on \(\overline{\Delta }\) that are holomorphic in \(\Delta \). In \(\mathcal{A}\) the norm is the supremum norm

$$\begin{aligned} \Vert f\Vert _\infty =\sup _t |f(e^{it})|, \end{aligned}$$

but we shall also consider the \(L^1\) norm given by

$$\begin{aligned} \Vert f\Vert _1=\frac{1}{2\pi }\int _0^{2\pi } |f(e^{it})|dt. \end{aligned}$$

In connection with approximation in \(L^1\)-norm by elements of a uniform algebra D. Khavinson, F. Pérez-González and H. Shapiro proved the following theorem (see [2, Theorem 3.3]).

FormalPara Theorem

Let f be a continuous function on \(\mathbf{T} \) with \(\Vert f\Vert _\infty =1\). Assume there exists an \(H^1\)-function G such that

$$\begin{aligned} \Vert f-G\Vert _1\le \varepsilon . \end{aligned}$$

Then there exists a function \(G^*\) in the disk algebra \(\mathcal{A}\) such that \(\Vert G^*\Vert _\infty \le 1\) and

$$\begin{aligned} \Vert f-G^*\Vert _1\le C\varepsilon \log \frac{1}{\varepsilon }, \end{aligned}$$
(1)

where C is a constant independent of f.

See [2] for the motivation of this result and its connection to a theorem of Hoffman and Wermer on homomorphisms of uniform algebras.

The authors of [2] also verified that in (1) the bound \(C\varepsilon \log \frac{1}{\varepsilon }\) cannot be replaced by \(C\varepsilon \) ( [2, Theorem 3.4]), but the problem if the order \(O(\varepsilon \log \frac{1}{\varepsilon })\) in (1) can be improved at all, i.e., if it is precise or not, remained open and stated explicitly in Remark (i) in [2]. That problem was communicated to us by D. Khavinson [1]. In this note we prove that the stated order is, indeed, precise.

FormalPara Theorem 1

There is a constant \(c>0\) with the property that for every sufficiently small \(\varepsilon >0\) there is a continuous function \(f=f_\varepsilon \), \(\Vert f\Vert _\infty =1\), such that

$$\begin{aligned} \Vert f-G\Vert _1\le \varepsilon \end{aligned}$$

for some \(G\in \mathcal{A}\), but for any \(G^*\) in \(\mathcal{A}\) with \(\Vert G^*\Vert _\infty \le 1\) we have

$$\begin{aligned}\Vert f-G^*\Vert _1\ge c\varepsilon \log \frac{1}{\varepsilon }.\end{aligned}$$
FormalPara Proof

It will be convenient to verify the claim with \(\varepsilon \) replaced by \(\varepsilon ^2\).

Let

$$\begin{aligned} u(z)+iv(z)=\frac{\varepsilon ^2}{1+\varepsilon -z}, \end{aligned}$$

where u(z) and v(z) are real. Using that

$$\begin{aligned} (1+\varepsilon -\cos t)^2+\sin ^2 t =\varepsilon ^2+4(1+\varepsilon )\sin ^2(t/2), \end{aligned}$$

for \(z=e^{it}\) we haveFootnote 1

$$\begin{aligned} \mathfrak {R}\frac{1}{1+\varepsilon -z}=\frac{1+\varepsilon -\cos t}{(1+\varepsilon -\cos t)^2+\sin ^2 t} \sim \left\{ \begin{array}{ll} 1/\varepsilon &{} \text{ if } |t|\le \varepsilon , \\ 1+\varepsilon /t^2 &{} \text{ if } \varepsilon \le |t|\le \pi , \end{array}\right. \end{aligned}$$

while

$$\begin{aligned} \left| \mathfrak {I}\frac{1}{1+\varepsilon -z}\right| =\frac{|\sin t|}{(1+\varepsilon -\cos t)^2+\sin ^2 t} \sim \left\{ \begin{array}{ll} |t|/\varepsilon ^2 &{} \text{ if } |t|\le \varepsilon , \\ 1/|t| &{} \text{ if } \varepsilon \le |t|\le \pi /2,\\ \pi -|t| &{}\text{ if } \pi /2\le |t|\le \pi . \end{array}\right. \end{aligned}$$

Indeed, these are easy consequences of the inequality

$$\begin{aligned} \frac{2}{\pi } u\le \sin u\le u,\qquad 0\le u\le \pi /2, \end{aligned}$$

i.e. of

$$\begin{aligned} \sin u\sim u, \qquad 0\le u\le \pi /2. \end{aligned}$$

For example, for \(z=e^{it}\), \(\varepsilon \le |t|\le \pi \) (\(0<\varepsilon \le 1\)), we obtain

$$\begin{aligned} \mathfrak {R}\frac{1}{1+\varepsilon -z}= & {} \frac{1+\varepsilon -\cos t}{(1+\varepsilon -\cos t)^2+\sin ^2 t} =\frac{2\sin ^2(t/2)+\varepsilon }{\varepsilon ^2+4(1+\varepsilon )\sin ^2(t/2)} \\\sim & {} \frac{t^2+\varepsilon }{\varepsilon ^2+t^2}\sim \frac{t^2+\varepsilon }{t^2}= 1+\varepsilon /t^2 \end{aligned}$$

as was claimed above.

The preceding relations show that

$$\begin{aligned} u(e^{it})\sim \left\{ \begin{array}{ll} \varepsilon &{} \text{ if } |t|\le \varepsilon , \\ \varepsilon ^2+\varepsilon ^3/t^2 &{} \text{ if } \varepsilon \le |t|\le \pi , \end{array}\right. \end{aligned}$$
(2)

while

$$\begin{aligned} |v(e^{it})|\sim \left\{ \begin{array}{ll} |t| &{} \text{ if } |t|\le \varepsilon , \\ \varepsilon ^2/|t| &{} \text{ if } \varepsilon \le |t|\le \pi /2\\ \varepsilon ^2(\pi -|t|) &{} \text{ if } \pi /2\le |t|\le \pi . \end{array}\right. \end{aligned}$$
(3)

Therefore, if we set \(F_\varepsilon =F(z)=\exp (\varepsilon ^2/(1+\varepsilon -z))\), \(|z|\le 1\), then for small \(\varepsilon \)

$$\begin{aligned} \mathfrak {R}F(e^{it})-1= & {} e^{u(e^{it})}\cos v(e^{it})-1\ge (1+u(e^{it}))\cos v(e^{it})-1\\= & {} u(e^{it})\cos v(e^{it})-2\sin ^2(v(e^{it})/2)>0. \end{aligned}$$

This is so, because we subtract from a term \(\sim \varepsilon \) (\(|t|\le \varepsilon \)) resp \(\succeq \varepsilon ^3/t^2\) (\(|t|\ge \varepsilon \)) a term that is at most \(\preceq \varepsilon ^2\) resp. \(\preceq \varepsilon ^4/t^2\). Now the preceding inequality implies in view of the maximum principle that \(\mathfrak {R}F(z)-1\) is positive in the unit disk.

Let \(f_\varepsilon =f=F/|F|\), for which we have for small \(\varepsilon \)

$$\begin{aligned} \Vert f-F\Vert _1= & {} \left\| \frac{F}{|F|}(|F|-1)\right\| _1= \Vert |F|-1\Vert _1=\frac{1}{2\pi }\int _{-\pi }^\pi (e^{u(e^{it})}-1)dt\nonumber \\\sim & {} \int _{-\pi }^\pi u(e^{it})dt\sim \varepsilon ^2, \end{aligned}$$
(4)

where we used that \(u\le e^u-1\le 2u\) provided \(0\le u\le 1/2\) (cf. (2)).

Note that F is in the disk algebra and f is a continuous function with \(\Vert f\Vert _\infty = 1\). Now let \(G^*\in \mathcal{A}\), \(\Vert G^*\Vert _\infty \le 1\), be any function. We are going to show that

$$\begin{aligned} \Vert f-G^*\Vert _1\ge c\varepsilon ^2\log \frac{1}{\varepsilon } \end{aligned}$$
(5)

with some \(c>0\) independent of \(\varepsilon \), and that will prove the theorem (with \(\varepsilon \) replaced by \(\varepsilon ^2\) and \(f_{\alpha \varepsilon }\) resp. \(F_{\alpha \varepsilon }\) replacing f resp. G in it, where \(\alpha \) is a constant for which \(\Vert f_{\alpha \varepsilon }-F_{\alpha \varepsilon }\Vert _1\le \varepsilon ^2\); see (4)).

For the \(L^1\) distance of F and \(G^*\) we have

$$\begin{aligned} \Vert F-G^*\Vert _1\le \Vert F-f\Vert _1+\Vert f-G^*\Vert _1\le C_1\varepsilon ^2+\Vert f-G^*\Vert _1. \end{aligned}$$
(6)

The real part of

$$\begin{aligned} g_1(z):=1-G^*(z)+i\mathfrak {I}G^*(0) \end{aligned}$$

is clearly nonnegative. Now to the pairs \(g_1(z)\) and \(g_2(z):=F(z)-1\) with nonnegative real part in \(\Delta \) and with imaginary part \(=0\) at the origin we can apply the “reverse triangle inequality"

$$\begin{aligned} \Vert g_1\Vert _1+\Vert g_2\Vert _1\le C_0\Vert g_1+g_2\Vert _1, \end{aligned}$$

proved in [2, Lemma 3.5], where \(C_0\) is an absolute constant. This yields

$$\begin{aligned} \Vert F-1\Vert _1\le & {} \Vert g_1\Vert _1+\Vert g_2\Vert _1\le C_0 \Vert (F-1)+(1-G^*+i\mathfrak {I}G^*(0))\Vert _1\\= & {} C_0\Vert F-G^*+i\mathfrak {I}G^*(0)\Vert _1\le C_0(\Vert F-G^*\Vert _1+|i\mathfrak {I}G^*(0)|). \end{aligned}$$

On the right

$$\begin{aligned} |i\mathfrak {I}G^*(0)|=|\mathfrak {I}(F-G^*)(0)|\le \Vert F-G^*\Vert _1, \end{aligned}$$

so, in view of (6),

$$\begin{aligned} \Vert F-1\Vert _1\le 2C_0\Vert F-G^*\Vert _1\le 2C_0C_1\varepsilon ^2+2C_0\Vert f-G^*\Vert _1 \end{aligned}$$
(7)

follows. Since on the left

$$\begin{aligned} \Vert F-1\Vert _1\ge \Vert \mathfrak {I}F\Vert _1= & {} \frac{1}{2\pi }\int _{-\pi }^\pi e^{u(e^{it})}|\sin v(e^{it})|dt \ge \frac{1}{2\pi }\int _{-\pi }^\pi |\sin v(e^{it})|dt\\\succeq & {} \frac{1}{2\pi }\int _{\varepsilon }^{\pi /2}|v(e^{it})|dt \sim \varepsilon ^2\log \frac{1}{\varepsilon }=\frac{1}{2}\varepsilon ^2\log \frac{1}{\varepsilon ^2} \end{aligned}$$

(where, for the \(\sim \) relation we used (3)), the inequality (5) follows from (7) for all sufficiently small \(\varepsilon \). \(\square \)

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