Half-Quadratic Image Restoration with a Non-parallelism Constraint

Abstract

The problem of image restoration from blur and noise is studied. By regularization techniques, a solution of the problem is found as the minimum of a primal energy function, which is formed by two terms. The former deals with faithfulness to the data, and the latter is associated with the smoothness constraints. We impose that the obtained results are images piecewise continuous and with thin edges. In correspondence with the primal energy function, there is a dual energy function, which deals with discontinuities implicitly. We present a unified approach of the duality theory, also to consider the non-parallelism constraint. We construct a dual energy function, which is convex and imposes such a constraint. To reconstruct images with Boolean discontinuities, the proposed energy function can be used as an initial approximation in a graduated non-convexity algorithm. The experimental results confirm that such a technique inhibits the formation of parallel lines.

Appendices

Appendix 1

In this appendix, we see how the assumptions in Theorems 4, 5 and 6 generalize the conditions of the other dual theorems presented in the literature.

In the following examples, we take $\lambda ^2=1$.

(a) Observe that, with the same hypotheses and notations as in Theorem 6, if f is concave and non-decreasing on ${\mathbb {R}}^+_0$, $f(0)=0$ and f is differentiable on $(0,+\infty )$, then, by de L’Hôpital’s rule, the limit $\displaystyle {\ell _0=\lim _{t\rightarrow + \infty }\frac{f(t)}{t}}$ is equal to the limit $\displaystyle {\ell =\lim _{t\rightarrow + \infty }f^{\prime }(t)}$. Moreover note that, since $g(t)=f(t^2)$ for each $t\in {\mathbb {R}}^+_0$, we get $\displaystyle {f^{\prime }(t^2)=\frac{g^{\prime }(t)}{2 t}}$ for each $t\in (0, +\infty )$, and thus $\displaystyle {\lim _{t\rightarrow +\infty }\frac{g^{\prime }(t)}{2 t}}= \ell =\ell _0$. In the duality theorems, there are many conditions involving the limit $\displaystyle {\lim _{t\rightarrow +\infty }\frac{g^{\prime }(t)}{2 t}}$ (see also [23, Theorem 1], [43, 58, 59]). Therefore, these conditions, when f is non-decreasing, concave on ${\mathbb {R}}^+_0$, $f(0)=0$ and f is differentiable on $(0,+\infty )$, are equivalent to the corresponding ones involving the limit $\ell _0$.

(b) Theorem 4 is a strict generalization of Theorem 3. Indeed, observe that every function $g \in {\mathrm{Lip}}_{\mathrm{loc}}({\mathbb {R}})$ is also continuous on ${\mathbb {R}}$, and a fortiori u.s.c. Moreover, given a fixed real number $\displaystyle { a\in \Bigl ( 0,\frac{1}{2}\Bigr )}$, the functions $g:{\mathbb {R}} \rightarrow {\mathbb {R}}$, $f:{\mathbb {R}} \rightarrow {\mathbb {R}} \cup \{ - \infty \}$, defined by $g(t)=t^{2a}$, $t\in {\mathbb {R}}$,

$$\begin{aligned} f(t)=\left\{ \begin{array}{l@{\quad }l} t^a, &{} \mathrm{if }\; t \ge 0,\\ -\infty , &{} \mathrm{if }\; t < 0, \end{array}\right. \end{aligned}$$

satisfy the hypotheses (4.1), (4.2) of Theorem 4, (5.1) of Theorem 5 and (6.1) of Theorem 6, but since $2a<1$, g is not Lipschitz on [0, 1], and hence $g \not \in {\mathrm{Lip}}_{\mathrm{loc}}({\mathbb {R}})$. Furthermore, with the same notations as in Theorem 7, we get $B= (0,+\infty )$,

$$\begin{aligned} \eta (b)= & {} \Bigl (\frac{b}{a}\Bigr )^{1/(a-1)},\\ \beta (b)&{=}&\left\{ \begin{array}{l@{\quad }l} \displaystyle {b^{a/(a-1)} (a^{a/(1-a)}{-}a^{1/(1-a)}){=}(1-a) \Bigl (\frac{b}{a}\Bigr )^{a/(a-1)}}, &{} \mathrm{if }\; b > 0,\\ {+}\infty , &{} \mathrm{if }\; b \le 0 \end{array}\right. \end{aligned}$$

(see also [43, Table II (d)]).

Furthermore, another reason for which Theorem 4 strictly extends Theorem 3 is that a concave function f does not need to be differentiable in the complement of a finite set, similarly as seen before.

(c) We now give an example in which the conditions (6.1) and (6.2) of Theorem 6 do not hold (see also [6, Table 3.2, Example 2]). Put

$$\begin{aligned} g(t)=\left\{ \begin{array}{l@{\quad }l} \log (t^2+1), &{} \mathrm{if }\; t \in [-1,1],\\ \displaystyle { \frac{t^2}{2}+\log 2 -\frac{1}{2}}, &{} \mathrm{if }\; t \in (-\infty , 1) \cup (1,+\infty ). \end{array}\right. \end{aligned}$$

It is not difficult to check that g is even, $g(0)=0$,

$$\begin{aligned} g^{\prime }(t)= \left\{ \begin{array}{l@{\quad }l} \displaystyle {\frac{2t}{t^2+1}}, &{} \mathrm{if }\; t \in [-1,1],\\ t &{} \mathrm{if }\; t \in (-\infty ,-1) \cup (1,+\infty ), \end{array}\right. \end{aligned}$$

and hence $g\in C^1({\mathbb {R}})$, and a fortiori $g\in C^1({\mathbb {R}}^+_0)$, g is strictly increasing in $[0, +\infty )$, $g \in {\mathrm{Lip}}_{\mathrm{loc}}({\mathbb {R}})$. Set

$$\begin{aligned} f(t)=\left\{ \begin{array}{l@{\quad }l} -\infty , &{} \mathrm{if }\; t \in (-\infty ,0), \\ \log (t+1), &{} \mathrm{if }\; t \in [0,1],\\ \displaystyle { \frac{t}{2}+\log 2 -\frac{1}{2}}, &{} \mathrm{if }\; t \in (1,+\infty ). \end{array}\right. \end{aligned}$$

It is not difficult to see that $f(t)=g(\sqrt{t})$ for every $t\ge 0$, $f(0)=0$, f is concave and strictly increasing on ${\mathbb {R}}_0^+$,

$$\begin{aligned} f^{\prime }(t)=\left\{ \begin{array}{l@{\quad }l} \displaystyle {\frac{1}{t+1}}, &{} \mathrm{if }\; t \in [0,1],\\ \\ \displaystyle {\frac{1}{2}}, &{} \mathrm{if }\; t \in (1,+\infty ), \end{array}\right. \end{aligned}$$

$f\in C^1({\mathbb {R}}^+_0)$, $B=[1/2,1]$. Moreover, $\displaystyle {\eta (b)=\frac{1}{b}-1}$ for every $b\in [1/2,1]$,

$$\begin{aligned} \beta (b)=\left\{ \begin{array}{l@{\quad }l} +\infty , &{} \mathrm{if }\; b < 1/2, \\ 0, &{} \mathrm{if }\; b>1,\\ -\log b - 1 + b , &{} \mathrm{if }\; b \in [1/2,1], \end{array}\right. \end{aligned}$$

and so the condition (6.2) of Theorem 6 is not satisfied. Furthermore, it is not difficult to check that $\beta$ is decreasing and convex on B.

(d) We now show that, under the same hypotheses and notations as in Theorem 7, in general the set $(f^{\prime })^{-1}(\{b\})$ can have more than one element for some $b\in B$. For instance, set

$$\begin{aligned} g(t){=}\left\{ \begin{array}{l@{\quad }l} 2 \sqrt{2}(\sqrt{t^2+1}-1), &{} \mathrm{if }\; t \in (-1,1),\\ t^2+3-2\sqrt{2}, &{} \mathrm{if }\; t \in [-\sqrt{2},-1] \cup [1,\sqrt{2}],\\ 2\sqrt{2}\,|t|{+}1{-}2\sqrt{2}, &{} \mathrm{if }\; t \in ({-}\infty ,{-}\sqrt{2}) \cup (\sqrt{2},{+}\infty ). \end{array}\right. \end{aligned}$$

It is not difficult to see that g is even, $g\not \equiv 0$, $g(0)=0$, $f(t)=g(\sqrt{t})$ for each $t\ge 0$,

$$\begin{aligned} g^{\prime }(t)=\left\{ \begin{array}{l@{\quad }l} \displaystyle { -2 \sqrt{2}}, &{} \mathrm{if }\; t \in (-\infty , -\sqrt{2}), \\ \displaystyle {\frac{2\sqrt{2}\,t}{\sqrt{t^2+1}}}, &{} \mathrm{if }\; t \in (-1,1),\\ 2t &{} \mathrm{if }\; t \in [-\sqrt{2},-1] \cup [1,\sqrt{2}],\\ \displaystyle { 2 \sqrt{2}}, &{} \mathrm{if }\; t \in (\sqrt{2},+\infty ), \end{array}\right. \end{aligned}$$

and so $g\in C^1({\mathbb {R}})$, and a fortiori $g\in C^1({\mathbb {R}}^+_0)$. Put

$$\begin{aligned} f(t)=\left\{ \begin{array}{l@{\quad }l} -\infty , &{} \mathrm{if }\; t \in (-\infty ,0), \\ 2 \sqrt{2}(\sqrt{t+1}-1), &{} \mathrm{if }\; t \in [0,1),\\ t+3-2\sqrt{2}, &{} \mathrm{if }\; t \in [1,2],\\ 2\sqrt{2t}+1-2\sqrt{2}, &{} \mathrm{if }\; t \in (2,+\infty ). \end{array}\right. \end{aligned}$$

It is not difficult to check that $f(t)=g(\sqrt{t})$ for every $t \ge 0$, $f(0)=0$, f is concave and strictly increasing on ${\mathbb {R}}_0^+$,

$$\begin{aligned} f^{\prime }(t)=\left\{ \begin{array}{l@{\quad }l} \displaystyle {\frac{\sqrt{2}}{\sqrt{t+1}}}, &{} \mathrm{if }\; t \in [0,1),\\ 1, &{} \mathrm{if }\; t \in [1,2],\\ \displaystyle { \frac{\sqrt{2}}{\sqrt{t}}}, &{} \mathrm{if }\; t \in (2,+\infty ), \end{array}\right. \end{aligned}$$

$f\in C^1({\mathbb {R}}^+_0)$, $\overline{a}=\sqrt{2}$, $B=(0,\sqrt{2}]$. Moreover, note that $f^{\prime }(t)>1$ if $t\in [0,1)$ and $f^{\prime }(t)<1$ if $t\in (2,+\infty )$. Hence, the function $h_1(t)=f(t)-t$ is increasing on [0, 1], decreasing on $[2, +\infty [$, $h_1$ assumes the maximum value on [1, 2], and $h_1([1,2])=3-2\sqrt{2}$. Furthermore, we get

$$\begin{aligned} \eta (b)=\left\{ \begin{array}{l@{\quad }l} \displaystyle {\frac{2}{b^2} }, &{} \mathrm{if }\; b\in (0,1), \\ \\ \displaystyle {\frac{2}{b^2} - 1 }, &{} \mathrm{if }\; b \in (1, \sqrt{2}], \end{array}\right. \end{aligned}$$

$(f^{\prime })^{-1}(\{1\})=[1,2]$,

$$\begin{aligned} \beta (b)=\left\{ \begin{array}{l@{\quad }l} + \infty , &{} \mathrm{if }\; b \le 0, \\ \\ \displaystyle {\frac{2}{b}+1-2\sqrt{2} }, &{} \mathrm{if }\; b\in (0,1), \\ \\ \displaystyle {\sup _{t\in {\mathbb {R}}} (f(t)-t)=3 - 2 \sqrt{2}}, &{} \mathrm{if }\; b=1,\\ \\ \displaystyle {\frac{2}{b} + b - 2 \sqrt{2} }, &{} \mathrm{if }\; b \in (1, \sqrt{2}), \\ \\ 0, &{} \mathrm{if }\; b \ge \sqrt{2}. \end{array}\right. \end{aligned}$$

It is not difficult to check that $\beta$ is decreasing and convex on B. Finally, we get that $\beta (1)$ is well-defined, since $f(t)-t=3 - 2\sqrt{2}$ for every $t\in (f^{\prime })^{-1}(\{1\})$.

(e) In this example, the hypotheses of Theorem 4 are satisfied, but g does not fulfil the conditions (5.1) and (6.1), and g is not continuous at 0. Here, the corresponding function $\beta$ does not fulfil (5.2), but satisfies (6.2), since the implication (6.2) $\Longrightarrow$ (6.1) does not hold when, for every $\delta >0$, the function f assumes some negative real values on $[\delta , + \infty )$. Put

$$\begin{aligned} g(t)= & {} \left\{ \begin{array}{l@{\quad }l} 0, &{} \mathrm{if }\; t =0,\\ -t^4 -1, &{} \mathrm{if }\; t \in {\mathbb {R}} \setminus \{ 0 \}, \end{array}\right. \\ \\ f(t)= & {} \left\{ \begin{array}{l@{\quad }l} -\infty , &{} \mathrm{if }\; t \in (-\infty ,0), \\ 0, &{} \mathrm{if }\; t =0,\\ -t^2 -1, &{} \mathrm{if }\; t \in (0,+\infty ). \end{array}\right. \end{aligned}$$

It is not difficult to check that $f(t)=g(\sqrt{t})$ for any $t \ge 0$. Note that g is strictly decreasing on ${\mathbb {R}}^+_0$ and $\displaystyle {\lim _{t\rightarrow +\infty } \frac{f(t)}{t}=-\infty }$. For $b\in {\mathbb {R}}$, we get

$$\begin{aligned}&\beta (b)= \Bigl (\sup _{t>0}(-bt-t^2-1)\Bigr ) \vee 0\\&\quad = \left\{ \begin{array}{l@{\quad }l} \displaystyle {\frac{b^2}{4}-1}, &{} \mathrm{if }\; b \in (-\infty , -2), \\ \\ 0, &{} \mathrm{if }\; b \in [-2, +\infty ), \end{array}\right. \end{aligned}$$

and thus $\beta (b)<+\infty$ for every $b\in {\mathbb {R}}$.

(f) Note that it may happen that g and f take the value $-\infty$ on some positive real numbers and assume real values on other points of the positive half line, and $\beta$ is not constant on ${\mathbb {R}}$. For example, set

$$\begin{aligned} g(t)= & {} \left\{ \begin{array}{l@{\quad }l} t^2, &{} \mathrm{if }\; t \in [0,1],\\ -\infty , &{} \mathrm{otherwise}, \end{array}\right. \\ \\ f(t)= & {} \left\{ \begin{array}{l@{\quad }l} t, &{} \mathrm{if }\; t \in [0,1],\\ -\infty , &{} \mathrm{otherwise}. \end{array}\right. \end{aligned}$$

It is not difficult to see that $f(t)=g(\sqrt{t})$ for each $t \ge 0$. For $b\in {\mathbb {R}}$, we get

$$\begin{aligned} \beta (b)= & {} \sup _{t\in [0,1]}(-bt+t) = \left\{ \begin{array}{l@{\quad }l} 1-b, &{} \mathrm{if }\; b \in (-\infty , 1), \\ \\ 0, &{} \mathrm{if }\; b \in [1, +\infty ). \end{array}\right. \end{aligned}$$

(g) Observe that, if f is non-negative in a suitable interval $(0,t_0]$ and $\displaystyle { \limsup _{t\rightarrow + \infty }\frac{f(t)}{t}>0}$, then, thanks to the last part of Theorem 6, there are some positive real numbers b with $\beta (b)=+\infty$. This means to impose a positive weight to the values of the finite difference which appear in the expression of the primal energy. This is too restrictive, since in the cliques having strong discontinuities it is not advisable to impose regularity constraints. So, the condition (6.1) is not restrictive for our purposes.

(h) Observe that it may happen that f, g and $\beta$ satisfy the properties in (a) and (b) of Theorem 4, and that $g(0) < 0$ and $\beta (\overline{b})<0$ for some $\overline{b} \in {\mathbb {R}}$. Indeed, it is enough to take $g(t)=-1$ for every $t \in {\mathbb {R}}$, and

$$\begin{aligned} \beta (b)= \left\{ \begin{array}{l@{\quad }l} + \infty , &{} \mathrm{if}\; b < 0, \\ -1, &{} \mathrm{if}\; b \ge 0. \end{array} \right. \end{aligned}$$

Note that, in this case, the conditions of Theorems 5 and 6 are satisfied.

(i) If $g:{\mathbb {R}} \rightarrow \widetilde{{\mathbb {R}}}$ is convex and even on ${\mathbb {R}}$, $g(0)\in {\mathbb {R}}$ and g satisfies the condition (4.2), then g is real-valued on ${\mathbb {R}}$ and $g\in C^1({\mathbb {R}} \setminus \{0\})$ (see also [43, Lemma 3]).

(j) If $g\in C^1({\mathbb {R}}^+_0)$, then the function $\beta$ in Theorem 7 coincides with the corresponding one investigated by D. Geman and G. Reynolds in [34]. This function is defined in a closed interval $B^*$, while, in our setting, $\beta$ is defined on the whole real line.

(k) The condition of convexity of $\beta$ is not restrictive. Indeed, since $\beta$ is l.s.c., then, by [70, Corollary 12.1.1], we get $\beta ^*=$ (conv $\beta )^*$. This implies that the functions f, g obtained by starting with $\beta$ coincide with the corresponding ones constructed by starting with the convex hull of $\beta$.

Appendix 2

We now sketch the proof of the convexity of the function $\varphi$ defined in (28) on

$$\begin{aligned} \displaystyle {{\mathbb {R}} \times \left[ -\sqrt{\frac{1-\delta }{\varepsilon (3-\delta )}}, \sqrt{\frac{1-\delta }{\varepsilon (3-\delta )}}\,\right] }, \end{aligned}$$

where $\varepsilon \in (0, + \infty )$ and $\delta \in (0,1)$. Note that $\varphi \in C^2((-\infty ,0) \times {\mathbb {R}})$ and $\varphi \in C^2((0,+\infty ) \times {\mathbb {R}})$, since

$$\begin{aligned}&\dfrac{\partial ^2 \varphi }{\partial t_1^2} (t_1,t_2)=\lambda ^2 (\varepsilon t_2^2 +1)(2-\delta )(1-\delta )|t_1|^{-\delta }, \quad t_1 \ne 0,\\&\dfrac{\partial ^2 \varphi }{\partial t_2^2}(t_1,t_2)=2 \lambda ^2 \varepsilon |t_1|^{2-\delta },\\&\dfrac{\partial ^2 \varphi }{\partial t_1 \partial t_2}(t_1,t_2){=}\dfrac{\partial ^2 \varphi }{\partial t_2 \partial t_1} (t_1,t_2){=}2\lambda ^2 \varepsilon t_2(2-\delta )|t_1|^{1{-}\delta }\text {sgn}(t_1), \end{aligned}$$

where the function sgn is defined as in (29). For $t_1\ne 0$ let

$$\begin{aligned} H(t_1, t_2)= \left( \begin{array}{cc} \lambda ^2 (\varepsilon t_2^2+1)(2-\delta )(1-\delta )|t_1|^{-\delta } &{} 2 \lambda ^2 \text {sgn}(t_1)\varepsilon (2-\delta )t_2|t_1|^{1-\delta } \\ 2\lambda ^2\text {sgn}(t_1)\varepsilon (2-\delta )t_2|t_1|^{1-\delta } &{} 2\lambda ^2\varepsilon |t_1|^{2-\delta } \end{array} \right) \end{aligned}$$

be the Hessian matrix associated with $\varphi$.

Note that, for every $(t_1, t_2) \in ({\mathbb {R}} \setminus \{ 0 \}) \times \left[ -\sqrt{\frac{1-\delta }{\varepsilon (3-\delta )}}, \sqrt{\frac{1-\delta }{\varepsilon (3-\delta )}}\,\right]$, H is positive-semidefinite. Furthermore, for every $\overline{t_2}\in {\mathbb {R}}$, the equation of the tangent hyperplane at the point $(0, \overline{t_2})$ is

$$\begin{aligned} z=\varphi (0, \overline{t_2})+\dfrac{\partial \varphi }{\partial t_1}(0,\overline{t_2}) (t_1)+\dfrac{\partial \varphi }{\partial t_2}(0,\overline{t_2}) (t_2-\overline{t_2})=0, \end{aligned}$$

and $\varphi \ge 0$ on ${\mathbb {R}}^2$. Thus, we deduce that $\varphi$ is convex on

$$\begin{aligned} \displaystyle {{\mathbb {R}} \times \left[ -\sqrt{\frac{1-\delta }{\varepsilon (3-\delta )}}, \sqrt{\frac{1-\delta }{\varepsilon (3-\delta )}}\,\right] } \end{aligned}$$

(see also [70, Theorem 25.1], [71, Theorem 2.14, (b) and (c)]).

Now we sketch the proof of the convexity of the function

$$\begin{aligned} \psi (x_1,x_2,x_3,x_4,x_5)= \varphi (x_1-2x_2+x_3,x_4-2x_5+x_1) \end{aligned}$$

on ${\mathbb {R}}^5$. We calculate the Hessian matrix $H(\psi )$ of $\psi$ when $x_1-2x_2+x_3 \ne 0$. Denoting by

$\kappa =\dfrac{\partial ^2 \varphi }{\partial t_1^2}$, $\upsilon =\dfrac{\partial ^2 \varphi }{\partial t_1 \partial t_2}$ and $\omega =\dfrac{\partial ^2 \varphi }{\partial t_2^2}$, we get

$$\begin{aligned} H(\psi )=\begin{pmatrix} \phantom {a} \kappa +2\upsilon +\omega \phantom {a} &{} \phantom {a} -2(\kappa +\upsilon ) \phantom {a} &{} \phantom {a} \kappa + \upsilon \phantom {a} &{} \phantom {a} \upsilon +\omega \phantom {a} &{} \phantom {a} -2(\upsilon +\omega ) \phantom {a} \\ -2(\kappa +\upsilon ) &{} 4\kappa &{} -2\kappa &{} -2\upsilon &{} 4\upsilon \\ \kappa +\upsilon &{} -2\kappa &{} \kappa &{} \upsilon &{} -2\upsilon \\ \upsilon +\omega &{} -2\upsilon &{} \upsilon &{} \omega &{} -2\omega \\ -2(\upsilon +\omega ) &{} 4\upsilon &{} -2\upsilon &{} -2\omega &{} 4\omega \end{pmatrix}. \end{aligned}$$

We now show that $H(\psi )$ is positive-semidefinite at the points $(x_1, \ldots , x_5)$ such that $x_1 - 2 x_2 +x_3 \ne 0$. To this aim, we use the following result.

Proposition 2

(see also [41, Corollary 7.1.5]) A matrix A of order $n \times n$ is positive-semidefinite if and only if every principal minor of A is non-negative.

Fist of all, we claim that $H(\psi )$ has rank at most two. Indeed, we get

$$\begin{aligned}&H(\psi )= \begin{pmatrix} 1 \\ -2 \\ 1 \\ 0 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} \kappa +\upsilon&-2\kappa&\kappa&\upsilon&-2\upsilon \end{pmatrix} + \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \\ -2 \end{pmatrix}\\&\quad \cdot \begin{pmatrix} \upsilon +\omega&-2\upsilon&\upsilon&\omega&-2\omega \end{pmatrix}. \end{aligned}$$

Since $H(\psi )$ is the sum of two matrices of rank one, then we get the claim. Thus, it is enough for our purposes to prove that every principal minor of $H(\psi )$ of order two is non-negative. We take into account that $H(\varphi )$ is positive-semidefinite, that is $\kappa \, \omega - \upsilon ^2 \ge 0$.

The determinants of the principal minors of order two are

$$\begin{aligned}&\begin{vmatrix} \kappa +2\upsilon +\omega&-2(\kappa +\upsilon ) \\ -2(\kappa +\upsilon )&4\kappa \end{vmatrix} = \begin{vmatrix} \kappa +2\upsilon +\omega&-2(\upsilon +\omega ) \\ -2(\upsilon +\omega )&4\omega \end{vmatrix}= \\&\quad =\begin{vmatrix} 4\kappa&-2\upsilon \\ -2\upsilon&\omega \end{vmatrix} =\begin{vmatrix} \kappa&-2\upsilon \\ -2\upsilon&4\omega \end{vmatrix} =4(\kappa \, \omega - \upsilon ^2 )\ge 0 ;\\&\begin{vmatrix} \kappa +2\upsilon +\omega&\kappa +\upsilon \\ \kappa +\upsilon&\kappa \end{vmatrix}= \begin{vmatrix} \kappa +2\upsilon +\omega&\beta +\omega \\ \upsilon +\omega&\omega \end{vmatrix}=\begin{vmatrix} \kappa&\upsilon \\ \upsilon&\omega \end{vmatrix} \\&\quad =\kappa \, \omega - \upsilon ^2 \ge 0 ;\\&\quad \begin{vmatrix} 4\kappa&-2\kappa \\ -2\kappa&\kappa \end{vmatrix} =0; \quad \begin{vmatrix} 4\kappa&4\upsilon \\ 4\upsilon&4\omega \end{vmatrix} =16(\kappa \, \omega - \upsilon ^2 )\ge 0 . \end{aligned}$$

Now, set $\varPi =\{(x_1,x_2,x_3,x_4,x_5):x_1=2x_2-x_3\}$ and let $P\in \varPi$. Then, $\psi (P)=\varphi (0, x_4-2x_5+2x_2-x_3)=0$. It is possible to see that the equation of the hyperplane tangent to $\psi$ at P is $x_6=0$, and $\psi \ge 0$ on ${\mathbb {R}}^5$. Thus, proceeding similarly as above, it is possible to show that $\psi$ is convex on ${\mathbb {R}}^5$ (see also [70, Theorem 25.1], [71, Theorem 2.14, (b) and (c)]).