Ulam’s Type Stability and Generalized Norms

Abstract

A symmetric functional equation is one whose form is the same regardless of the order of the arguments. A remarkable example is the Cauchy functional equation: $f(x+y)=f(x)+f(y).$ Interesting results in the study of the rigidity of quasi-isometries for symmetric spaces were obtained by B. Kleiner and B. Leeb, using the Hyers-Ulam stability of a Cauchy equation. In this paper, some results on the Ulam’s type stability of the Cauchy functional equation are provided by extending the traditional norm estimations to ther measurements called generalized norm of convex type (v-norm) and generalized norm of subadditive type (s-norm).

1. Introduction

Ulam [1] posed the following problem concerning group homomorphism when presenting a talk at the University of Wisconsin:

Let $(G,\circ )$ be a group, $(G_2,\ast )$ be a metric group with the metric $d(·,·)$ and $\epsilon >0.$ If there exists a $\delta >0$ such that $f:G_1\to G_2$ satisfies

$$d(f(x\circ y),f(x)\ast f(y))\le \delta ,\mathrm{for}\mathrm{all}x,y\in G_1$$

then there exists a homomorphism $h:G_1\to G_2$ with

$$d(f(x),h(x))\le \epsilon ,\mathrm{for}\mathrm{all}x\in G_1?$$

The first affirmative answer to this problem was the one provided by Hyers [2], who solved the problem for additive mappings in Banach spaces.

A remarkable generalization of this result was given by Rassias [3]. A further generalization was obtained by Găvruţa [4], where he also introduced the concept of generalized Hyers–Ulam–Rassias stability in the spirit of Rassias’s approach. Using the idea of [4], Kim [5] proved the modified Hyers–Ulam–Rassias stability for the family of functional equations

$$f(x\circ y)=H\left(f\right(x),f(y))(x,y\in S)$$

where H is a homogeneous function and ∘ is a square-symmetric operation on the set S. For an instructive treatment of Găvruţa’ s stability theorem, see the paper [6].

For other results and generalizations, see the books [7,8,9,10,11] and their references.

In 1997, Kleiner and Leeb used the Hyers–Ulam stability of a Cauchy equation in the study of the rigidity of quasi-isometries for symmetric spaces (see [12]).

In [13], we proved a fixed-point theorem for a class of operators with suitable properties in very general conditions and some corollaries, which showed that our main result is a useful tool for proving properties of generalized Hyers–Ulam stability for some functional equations in a single variable.

More recently, in the survey [14], the authors presented applications of different fixed point theorems to the theory of stability of functional equations.

In [15], we investigated the approximation of functions by additive and by quadratic mappings, and we discussed the approximation of functions by cubic mappings [16].

In this paper, we extend the main result of [4] in the context of generalized norms.

2. Results

We consider the notion of generalized norm of convex type (v-norm), inspired by [17].

Let $(X,\parallel ·\parallel )$ be a normed space.

Definition 1.

A mapping ${\parallel ·\parallel }_v:X\to {\mathbb{R}}_{+}=[0,\infty )$ is called a generalized norm of convex type or a v-norm if

(V1) 

${\parallel 0\parallel }_v=0$

(V2) 

$\parallel {\displaystyle \frac{x+y}{2}}{\parallel }_v\le {\displaystyle \frac{{\parallel x\parallel }_v+{\parallel y\parallel }_v}{2}},(\forall )x,y\in X$; (property of convexity)

(V3) 

For any $\epsilon >0,$ there exists $\delta =\delta (\epsilon )>0$, so that if ${\parallel x\parallel }_v<\delta$, then $\parallel x\parallel <\epsilon$;

(V4) 

The mapping $x\mapsto {\parallel x\parallel }_v$ from X to ${\mathbb{R}}_{+}$ is lower semicontinuous.

Example 1.

We provide some examples of $v\text{-}$norms on X.

(a1) 

The norm of X is a $v\text{-}$norm.

(a2) 

Let $\lambda :X\to [0,\infty )$ be a convex and continuous function with $\lambda (0)=0.$ Then

$${\parallel x\parallel }_v:=max\{\parallel x\parallel ,\lambda (x)\}$$

is a $v\text{-}$norm.

(a3) 

Let $h:[0,\infty )\to [0,\infty )$ be a continuous, nondecreasing mapping so that

$$(\forall )a>0,{\int }_0^{a}h(t)dt>0.$$

Then

$${\parallel x\parallel }_v:={\int }_0^{\parallel x\parallel }h(t)dt\mathrm{is}\mathrm{a}\mathrm{v}\text{-}\mathrm{norm}.$$

We prove that the function defined in $(a3)$ is a $v\text{-}$norm.

$(V1)$${\parallel 0\parallel }_v={\int }_0^{0}h(t)dt=0$.

$(V2)$ We prove that:

$${\int }_0^{\parallel \frac{x+y}{2}\parallel }h(t)dt\le \frac{1}{2}\left({\int }_0^{\parallel x\parallel }h(t)dt+{\int }_0^{\parallel y\parallel }h(t)dt\right).$$

We have

$$\parallel {\displaystyle \frac{x+y}{2}}\parallel \le \frac{1}{2}(\parallel x\parallel +\parallel y\parallel )$$

Let ${\displaystyle F(u)={\int }_0^{u}h(t)dt.}$ We prove that:

$$F\left(\frac{u+v}{2}\right)\le \frac{1}{2}(F(u)+F(v)).$$

For $u\ge v$, we have:

$$\begin{array}{cc}\hfill F(u)+F(v)-2F\left(\frac{u+v}{2}\right)& =F(u)-F\left(\frac{u+v}{2}\right)+F(v)-F\left(\frac{u+v}{2}\right)\hfill \\ \hfill & =F^{\prime }(c_1)\left(u-\frac{u+v}{2}\right)+F^{\prime }(c_2)\left(v-\frac{u+v}{2}\right)\hfill \\ \hfill & =\frac{u-v}{2}(h(c_1)-h(c_2))\ge 0,\hfill \end{array}$$

since $u\ge c_1\ge {\displaystyle \frac{u+v}{2}}\ge c_2\ge v$ and h is nondecreasing.

$(V3)$ We prove that $(\forall )\epsilon >0,$$(\exists )\delta (\epsilon )>0$, so that if x is

$${\displaystyle {\int }_0^{\parallel x\parallel }h(t)dt<\delta (\epsilon )\Rightarrow \parallel x\parallel <\epsilon .}$$

Conversely, there exists ${\epsilon }_0>0$ such that for all $\delta >0,$ there exists $x_{\delta }$ such that:

$${\int }_0^{\parallel x_{\delta }\parallel }h(t)dt<\delta \mathrm{and}\parallel x_{\delta }\parallel \ge {\epsilon }_0.$$

It follows that:

$${\int }_0^{{\epsilon }_0}h(t)dt\le {\int }_0^{\parallel x_{\delta }\parallel }h(t)dt<\delta ,(\forall )\delta >0.$$

It follows that ${\int }_0^{{\epsilon }_0}h(t)dt=0$, in contradiction with the hypothesis on h.

$(V4)$ The mapping $x\mapsto {\parallel x\parallel }_v$ is continuous. If $x_n\to x$$(n\to \infty ),$ implies that $\parallel x_n\parallel \to \parallel x\parallel$$(n\to \infty ).$ We have:

$$|\parallel x_n{\parallel }_v-{\parallel x\parallel }_v|=|{\int }_{\parallel x_n\parallel }^{\parallel x\parallel }h(t)dt|=|\parallel x_n\parallel -\parallel x\parallel |·|h(c_n)|,$$

where $c_n$ is between $\parallel x_n\parallel$ and $\parallel x\parallel ,$ by the integral mean value theorem. It follows:

$$\underset{n\to \infty }{lim}|\parallel x_n{\parallel }_v-{\parallel x\parallel }_v|=0·h(\parallel x\parallel )=0.$$

So, $\parallel x_n{\parallel }_v\to {\parallel x\parallel }_v.$

Next, we provide some elementary results on convex norms.

Proposition 1.

We have:

(i) 

${\parallel x+y\parallel }_v\le {\displaystyle \frac{1}{2}}{(\parallel 2x\parallel }_v+{\parallel 2y\parallel }_v)$;

(ii) 

${\parallel x\parallel }_v\le {\displaystyle \frac{1}{2^n}}{\parallel 2^{n}x\parallel }_v$;

(iii) 

${\parallel x+y\parallel }_v\le {\displaystyle \frac{1}{2^n}}(\parallel 2^n{x\parallel }_v+\parallel 2^{n}y{\parallel }_v)$;

(iv) 

If $\parallel x_n{\parallel }_v\le a_n,$$n\in \mathbb{N},$ for ${(x_n)}_{n=1}^{\infty }\subset X$ and ${(a_n)}_{n=1}^{\infty }\subset {\mathbb{R}}_{+}$ with ${\displaystyle \underset{n\to \infty }{lim}{a}_n=0,}$ then ${\displaystyle \underset{n\to \infty }{lim}{x}_n=0;}$

(v) 

If $\parallel x_n-x_m{\parallel }_v\le b_m,$$n>m$${\displaystyle \underset{m\to \infty }{lim}{b}_m=0,}$ then $(x_n)$ is a Cauchy sequence;

(vi) 

If $\parallel x-x_n{\parallel }_v\le a_n$ and $\parallel y-x_n{\parallel }_v\le b_n,$ for $a_n\to 0$ and $b_n\to 0$, then $x=y.$

Proof. 

$(i)$ In the condition $(V2)$ of Definition 1, we take $2x$ instead of x, and instead of y, we take $2y$. $(ii)$ In the condition $(V2)$ of Definition 1, we take $y=0$ and using the condition $(V1)$, we obtain

$$\parallel \frac{x}{2}{\parallel }_v\le \frac{{\parallel x\parallel }_v+{\parallel 0\parallel }_v}{2}=\frac{{\parallel x\parallel }_v}{2}$$

If we take $2x$ instead of x in the above inequality, we obtain:

$${\parallel x\parallel }_v\le \frac{1}{2}{\parallel 2x\parallel }_v.$$

Now, by induction: if ${\parallel x\parallel }_v\le \frac{1}{2^n}{\parallel 2^{n}x\parallel }_v,$ then:

$${\parallel x\parallel }_v\le \frac{1}{2^n}\parallel 2^n{x\parallel }_v\le \frac{1}{2^n}\frac{1}{2}{\parallel 2^{n+1}x\parallel }_v.$$

$(iii)$ We use induction. For $n=1$, we have $(i).$ If

$${\parallel x+y\parallel }_v\le \frac{1}{2^n}(\parallel 2^n{x\parallel }_v+\parallel 2^{n}y{\parallel }_v)$$

then, with $(ii),$ we obtain:

$$\begin{array}{cc}\hfill {\parallel x+y\parallel }_v& \le \frac{1}{2^n}\left(\frac{1}{2}\parallel 2^{n+1}{x\parallel }_v+\frac{1}{2}{\parallel 2^{n+1}\parallel }_v\right)\hfill \\ & =\frac{1}{2^{n+1}}\left(\parallel 2^{n+1}{x\parallel }_v+{\parallel 2^{n+1}\parallel }_v\right).\hfill \end{array}$$

$(iv)$ We use condition $(V3)$ of Definition 1. Let $\epsilon >0.$ From $(V3)$, it follows that there exists $\delta =\delta (\epsilon )>0$, so that if ${\parallel x\parallel }_v<\delta$, then $\parallel x\parallel <\epsilon$. Since ${\displaystyle \underset{n\to \infty }{lim}{a}_n=0}$, it follows that there exists $n_0(\epsilon )$ such that for all $n\ge n_0(\epsilon )$, we have $a_n<\delta .$ It follows

$$\parallel x_n{\parallel }_v<\delta ,n\ge n_0(\epsilon ).$$

Hence, $\parallel x_n\parallel <\epsilon ,n\ge n_0(\epsilon ).$$(v)$ We use $(V3)$ again. We have $b_n<\delta (\epsilon ),$$n\ge n_0(\epsilon )=n_0.$ It follows

$$\parallel x_n-x_{n_0}{\parallel }_v<\delta (\epsilon ),n\ge n_0.$$

From $(V3),$ we obtain:

$$\parallel x_n-x_{n_0}\parallel <\epsilon .$$

Let be $m,n\ge n_0:$

$$\parallel x_n-x_m\parallel \le \parallel x_n-x_{n_0}\parallel +\parallel x_{n_0}-x_m\parallel <2\epsilon .$$

$(vi)$ From $(iv)$, we have $x_n\to x,$$x_n\to y.$ From the uniqueness of the limit, it follows that $x=y.$ □

We prove the stability of additive mappings.

Theorem 1.

Let S be an abelian semigroup, X a Banach space, and $\phi :S\times S\to [0,\infty )$, such that:

$$\mathsf{\Phi }(x,y):=\sum _{n=0}^{\infty }{\displaystyle \frac{\phi (2^{n}x,2^{n}y)}{2^{n+1}}}<\infty ,x,y\in S.$$

(1)

Let ${\parallel ·\parallel }_v$ be a $v\text{-}$norm on $X.$ If $f:S\to X$ is so that:

$${\parallel f(x+y)-f(x)-f(y)\parallel }_v\le \phi (x,y),(\forall )x,y\in S,$$

(2)

then there exists a unique $T:S\to X$ additive so that:

$${\parallel T(x)-f(x)\parallel }_v\le \mathsf{\Phi }(x,x),x\in S.$$

(3)

Proof. 

In $(2)$, we take $x=y$ to obtain:

$${\parallel f(2x)-2f(x)\parallel }_v\le \phi (x,x),$$

(4)

and from $(ii)$:

$$\parallel {\displaystyle \frac{f(2x)}{2}}-f(x){\parallel }_v\le {\displaystyle \frac{1}{2}}{\parallel f(2x)-2f(x)\parallel }_v;$$

hence,

$$\parallel \frac{f(2x)}{2}-f(x){\parallel }_v\le \frac{1}{2}\phi (x,x).$$

(5)

In the above relation, we replace x with $2x$, and we obtain:

$$\parallel \frac{f(2^{2}x)}{2}-f(2x){\parallel }_v\le \frac{1}{2}\phi (2x,2x)$$

(6)

and from $(i)$, and relations (4) and (6), we have:

$$\begin{array}{cc}\hfill \parallel \frac{f(2^{2}x)}{2^2}-f(x){\parallel }_v& =\parallel \frac{f(2^{2}x)}{2^2}-\frac{f(2x)}{2}+\frac{f(2x)}{2}-f(x){\parallel }_v\hfill \\ \hfill & \le \frac{1}{2}\left[\parallel \frac{f(2^{2}x)}{2}-f(2x){\parallel }_v+{\parallel f(2x)-2f(x)\parallel }_v\right]\hfill \\ \hfill & \le \frac{1}{2}\left[\frac{1}{2}\phi (2x,2x)+\phi (x,x)\right],\hfill \end{array}$$

hence,

$$\parallel \frac{f(2^{2}x)}{2^2}-f(x){\parallel }_v\le \frac{1}{2}\phi (x,x)+\frac{1}{2^2}\phi (2x,2x).$$

(7)

We prove that:

$$\parallel \frac{f(2^{n}x)}{2^n}-f(x){\parallel }_v\le \sum _{k=0}^{n-1}\frac{\phi (2^{k}x,2^{k}x)}{2^{k+1}}.$$

(8)

We use the induction. We suppose that relation $(8)$ holds.

Using $(i)$ and relations (4) and (8), we have:

$$\begin{array}{cc}\hfill \parallel \frac{f(2^{n+1}x)}{2^{n+1}}-f(x){\parallel }_v& =\parallel \frac{f(2^{n+1}x)}{2^{n+1}}-\frac{f(2x)}{2}+\frac{f(2x)}{2}-f(x){\parallel }_v\hfill \\ \hfill & \le \frac{1}{2}\left[\parallel \frac{f(2^{n+1}x)}{2^n}-f(2x){\parallel }_v+{\parallel f(2x)-2f(x)\parallel }_v\right]\hfill \\ \hfill & \le \sum _{k=0}^{n-1}\frac{\phi (2^{k+1}x,2^{k+1}x)}{2^{k+2}}+\frac{\phi (x,x)}{2}\hfill \end{array}$$

hence,

$$\parallel \frac{f(2^{n+1}x)}{2^{n+1}}-f(x){\parallel }_v\le \sum _{i=0}^n\frac{\phi (2^{i}x,2^{i}x)}{2^{i+1}}.$$

Now, we prove that the sequence ${\displaystyle \left\{\frac{f(2^{n}x)}{2^n}\right\}}$ is a Cauchy sequence. In (8), we take $2^{p}x$ instead of x and we obtain

$$\parallel \frac{f(2^{n+p}x)}{2^n}-f(2^{p}x){\parallel }_v\le \sum _{k=0}^{n-1}\frac{\phi (2^{k+p}x,2^{k+p}x)}{2^{k+1}}$$

and using $(ii)$:

$$\begin{array}{cc}\hfill \parallel \frac{f(2^{n+p}x)}{2^{n+p}}-\frac{f(2^{p}x)}{2^p}{\parallel }_v& \le \frac{1}{2^p}[\parallel \frac{f(2^{n+p}x)}{2^n}-f(2^{p}x){\parallel }_v\hfill \\ \hfill & \le \sum _{k=0}^{n-1}\frac{\phi (2^{k+p}x,2^{k+p}x)}{2^{k+p+1}}\hfill \\ \hfill & =\sum _{i=p}^{n+p-1}\frac{\phi (2^{i}x,2^{i}x)}{2^{i+1}}\hfill \\ \hfill & \le \sum _{i=p}^{\infty }\frac{\phi (2^{i}x,2^{i}x)}{2^{i+1}}.\hfill \end{array}$$

We use $(v)$. Since ${\displaystyle \underset{p\to \infty }{lim}\sum _{i=p}^{\infty }\frac{\phi (2^{i}x,2^{i}x)}{2^{i+1}}=0,}$ it follows that ${\displaystyle \left\{\frac{f(2^{n}x)}{2^n}\right\}}$ is a Cauchy sequence in Banach space $(X,\parallel ·\parallel ).$ Since X is a Banach space, there exists

$$T(x):=\underset{n\to \infty }{lim}\frac{f(2^{n}x)}{2^n},\mathrm{for}\mathrm{all}x\in X.$$

We prove that T is an additive mapping. From (2), we have:

$$\parallel f(2^{n}x+2^{n}y)-f(2^{n}x)-f(2^{n}y){\parallel }_v\le \phi (2^{n}x,2^{n}y),$$

and from $(ii)$ we have:

$$\begin{array}{cc}\hfill \parallel \frac{f(2^{n}x+2^{n}y)}{2^n}-\frac{f(2^{n}x)}{2^n}-\frac{f(2^{n}y)}{2^n}{\parallel }_v& \le \frac{1}{2^n}{\parallel f(2^{n}x+2^{n}y)-f(2^{n}x)-f(2^{n}y)\parallel }_v\hfill \\ \hfill & \le \frac{1}{2^n}\phi (2^{n}x,2^{n}y).\hfill \end{array}$$

From (1), it follows that

$$\underset{n\to \infty }{lim}\frac{\phi (2^{n}x,2^{n}x)}{2^n}=0,$$

and from $(iv)$, we have

$$\underset{n\to \infty }{lim}\left\{\frac{f(2^{n}x+2^{n}y)}{2^n}-\frac{f(2^{n}x)}{2^n}-\frac{f(2^{n}y)}{2^n}\right\}=0;$$

hence, $T(x+y)-T(x)-T(y)=0.$

We prove now the relation (3). We have:

$$\frac{f(2^{n}x)}{2^n}-f(x)⟶T(x)-f(x),n\to \infty .$$

Since $x\mapsto {\parallel x\parallel }_v$ is lower semicontinuous, it follows that

$$liminf\parallel \frac{f(2^{n}x)}{2^n}-f(x){\parallel }_v\ge {\parallel T(x)-f(x)\parallel }_v.$$

From relation (8),

$$liminf\parallel \frac{f(2^{n}x)}{2^n}-f(x){\parallel }_v\le \mathsf{\Phi }(x,x),x\in S;$$

hence,

$${\parallel T(x)-f(x)\parallel }_v\le \mathsf{\Phi }(x,x).$$

We prove that T is a unique additive mapping with the property in relation (3). We suppose that there exists two additive mappings $T_1$ and $T_2$, such that

$$\parallel T_1{(x)-f(x)\parallel }_v\le \mathsf{\Phi }(x,x)\mathrm{and}{\parallel T_2(x)-f(x)\parallel }_v\le \mathsf{\Phi }(x,x),(\forall )x\in S.$$

For any $n\in \mathbb{N}$ and $x\in S,$ we have:

$$\parallel T_1(2^{n}x)-f(2^{n}x){\parallel }_v\le \mathsf{\Phi }(2^{n}x,2^{n}x)$$

$$\parallel T_2(2^{n}x)-f(2^{n}x){\parallel }_v\le \mathsf{\Phi }(2^{n}x,2^{n}x)$$

From condition $(ii)$, we get:

$$\parallel T_1(x)-\frac{f(2^{n}x)}{2^n}{\parallel }_v\le \frac{1}{2^n}\mathsf{\Phi }(2^{n}x,2^{n}x)\to 0$$

$$\parallel T_2(x)-\frac{f(2^{n}x)}{2^n}{\parallel }_v\le \frac{1}{2^n}\mathsf{\Phi }(2^{n}x,2^{n}x)\to 0$$

So, from $(vi)$, it follows that $T_1(x)=T_2(x).$ □

Remark 1.

It is clear that the result in the above Theorem holds if

$$\sum _{n=0}^{\infty }\frac{\phi (2^{n}x,2^{n}x)}{2^{n+1}}<\infty and\underset{n\to \infty }{lim}\frac{\phi (2^{n}x,2^{n}y)}{2^n}=0,$$

for any $x,y\in S.$

The following Corollary is a direct extension of a result of Rassias [3].

Corollary 1.

Let $\psi :{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ be a mapping such that for $0\le p<1,$ we have

$$\psi (2t)\le 2^p\psi (t),t\in {\mathbb{R}}_{+}.$$

Let $(E,\parallel ·{\parallel }_1)$ be a normed space, X a Banach space, $\epsilon >0$, and ${\parallel ·\parallel }_v$ be a $v\text{-}$norm on X. If $f:E\to X$ is so that

$${\parallel f(x+y)-f(x)-f(y)\parallel }_v\le {\epsilon \left\{\psi \right(\parallel x\parallel }_1{)+\psi (\parallel y\parallel }_1)\},(\forall )x,y\in E,$$

then there exists a unique additive mapping T so that

$${\parallel f(x)-T(x)\parallel }_v\le \frac{2\epsilon }{2-2^p}{\psi (\parallel x\parallel }_1),x\in E.$$

Proof. 

We apply Theorem 1, with $\phi (x,y)={\epsilon \left\{\psi \right(\parallel x\parallel }_1{)+\psi (\parallel y\parallel }_1)\}$, $x,y\in E.$ We have:

$$\begin{array}{cc}\hfill \mathsf{\Phi }(x,y)& \le \epsilon \sum _{n=0}^{\infty }\frac{2^{np}}{2^{n+1}}{\left\{\psi \right(\parallel x\parallel }_1{)+\psi (\parallel y\parallel }_1)\}\hfill \\ \hfill & =\frac{\epsilon }{2}·\frac{1}{1-2^{p-1}}{\left\{\psi \right(\parallel x\parallel }_1{)+\psi (\parallel y\parallel }_1)\}\hfill \\ \hfill & =\frac{\epsilon }{2-2^p}{\left\{\psi \right(\parallel x\parallel }_1{)+\psi (\parallel y\parallel }_1)\}.\hfill \end{array}$$

 □

An example of a function that verifies the hypothesis of the Corollary 1 is $\psi (t)=t^p,t\in {\mathbb{R}}_{+}.$ We provide an example of application of Theorem 1, which does not follow from the Corollary 1.

Example 2.

We consider $(X,\parallel ·\parallel )$ a Banach space, h a function as in Example 1, a3), $\epsilon >0$ and $0\le p<1$. If $f:[0,\infty )\to X$ is so that

$${\int }_0^{\parallel f(x+y)-f(x)-f(y)\parallel }h(t)dt\le \epsilon (x^p+y^p),(\forall )x,y\in [0,\infty ).$$

Then, there exists a unique additive mapping $T:[0,\infty )\to X$ so that

$${\int }_0^{\parallel f(x)-T(x)\parallel }h(t)dt\le \frac{2\epsilon }{2-2^p}{x}^p,x\in [0,\infty ).$$

Now, we consider the case of generalized norm of subadditive type (or s-norm).

Definition 2.

A mapping ${\parallel ·\parallel }_s:X\to {\mathbb{R}}_{+}$ is called a generalized norm of subadditive type or a s-norm if

(S1) 

${\parallel x+y\parallel }_s\le {\parallel x\parallel }_s+{\parallel y\parallel }_s$;

(S2) 

For any $\epsilon >0,$ there exists $\delta =\delta (\epsilon )>0$, so that if ${\parallel x\parallel }_s<\delta$, then $\parallel x\parallel <\epsilon$;

(S3) 

The mapping $x\mapsto {\parallel x\parallel }_s$ from X to ${\mathbb{R}}_{+}$ is lower semicontinuous.

It is clear that any norm is a generalized norm of the subadditive type. An example of a generalized norm of the subadditive type that is not a norm is the following:

Example 3.

Let $(X,\parallel ·\parallel )$ be a normed space. For all $x\in X,$ we define:

$${\parallel x\parallel }_s:=\frac{\parallel x\parallel }{\parallel x\parallel +1}.$$

We verify the axioms $(S1)-(S3)$ for this function.

Proposition 2.

Elementary properties of $s\text{-}$norms are:

(i) 

${\parallel x\parallel }_s\le 2\parallel {\displaystyle \frac{x}{2}}{\parallel }_s$;

(ii) 

If $\parallel x_n{\parallel }_s\le a_n,$$n\in \mathbb{N},$ for ${(x_n)}_{n=1}^{\infty }\subset X$ and ${(a_n)}_{n=1}^{\infty }\subset {\mathbb{R}}_{+}$ with ${\displaystyle \underset{n\to \infty }{lim}{a}_n=0,}$ then ${\displaystyle \underset{n\to \infty }{lim}{x}_n=0;}$

(iii) 

If $\parallel x_n-x_m{\parallel }_s\le b_m,$$n>m$${\displaystyle \underset{m\to \infty }{lim}{b}_m=0,}$ then $(x_n)$ is a Cauchy sequence;

(iv) 

If $\parallel x-x_n{\parallel }_s\le a_n$ and $\parallel y-x_n{\parallel }_c\le b_n,$ for $a_n\to 0$ and $b_n\to 0$, then $x=y.$

Following the ideas from Theorem 1, we obtain an analogous result for generalized subadditive norms in the context of 2-divisible abelian semigroup.

We recall that an abelian semigroup S is called $2\text{-}$divisible if for all $\epsilon >0$ there exists a unique $x^{\prime }$ such that

$$x^{\prime }+x^{\prime }=x.$$

We denote such an $x^{\prime }$ with ${\displaystyle \frac{x}{2}}:$

Theorem 2.

Let S be an abelian semigroup, $2\text{-}$divisible, X a Banach space and ${\phi }_1:S\times S\to [0,\infty )$, such that

$${\mathsf{\Phi }}_1(x,y):=\sum _{n=1}^{\infty }{2}^{n-1}{\phi }_1\left({\displaystyle \frac{x}{2^n}},{\displaystyle \frac{y}{2^n}}\right)<\infty ,x,y\in S.$$

Let ${\parallel ·\parallel }_s$ be a $s\text{-}$norm on $X.$ If $f:S\to X$ is so that

$${\parallel f(x+y)-f(x)-f(y)\parallel }_s\le {\phi }_1(x,y),(\forall )x,y\in S,$$

then there exists a unique $T:S\to X$ additive so that

$${\parallel T(x)-f(x)\parallel }_s\le {\mathsf{\Phi }}_1(x,x),x\in S.$$

The following application of Theorem 2 is a direct extension of a result by Gajda [18].

Corollary 2.

Let $p>1$ and ${\psi }_1:{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ be a mapping such that

$${\psi }_1(2t)\ge 2^p{\psi }_1(t),t\in {\mathbb{R}}_{+}.$$

Let $(E,\parallel ·{\parallel }_1)$ be a normed space, $\epsilon >0,$$(X,\parallel ·\parallel )$ be a Banach space, and ${\parallel ·\parallel }_s$ be a $s\text{-}$ norm on X.

If $f:E\to X$ is such that

$${\parallel f(x+y)-f(x)-f(y)\parallel }_s\le \epsilon \{{\psi }_1{(\parallel x\parallel }_1)+{\psi }_1{(\parallel y\parallel }_1)\},(\forall )x,y\in E,$$

then, there exists a unique additive mapping T such that

$${\parallel f(x)-T(x)\parallel }_s\le \frac{2\epsilon }{2^p-2}{\psi }_1{(\parallel x\parallel }_1),x\in E.$$

Proof. 

We apply Theorem 2, with ${\phi }_1(x,y)=\epsilon \{{\psi }_1{(\parallel x\parallel }_1)+{\psi }_1{(\parallel y\parallel }_1)\}$, $x,y\in E.$ We have;

$$\begin{array}{cc}\hfill {\mathsf{\Phi }}_1(x,y)& \le \epsilon \sum _{n=1}^{\infty }{2}^{n-1}·\frac{1}{2^{np}}\{{\psi }_1{(\parallel x\parallel }_1)+{\psi }_1{(\parallel y\parallel }_1)\}\hfill \\ \hfill & =\frac{\epsilon }{2}\{{\psi }_1{(\parallel x\parallel }_1)+{\psi }_1{(\parallel y\parallel }_1)\}·\sum _{n=1}^{\infty }{\left(\frac{1}{2^{p-1}}\right)}^n\hfill \\ \hfill & =\frac{\epsilon }{2}\{{\psi }_1{(\parallel x\parallel }_1)+{\psi }_1{(\parallel y\parallel }_1)\}·\frac{1}{2^{p-1}}·\frac{1}{1-\frac{1}{2^{p-1}}}\hfill \\ \hfill & =\frac{\epsilon }{2^p-2}\{{\psi }_1{(\parallel x\parallel }_1)+{\psi }_1{(\parallel y\parallel }_1)\}.\hfill \end{array}$$

 □

An example of function that verifies the hypothesis of the Corollary 2 is $\psi (t)=t^p,t\in {\mathbb{R}}_{+}.$

3. Conclusions

In this paper, on a Banach space, two new measurements were introduced called generalized norm of convex type (v-norm) and generalized norm of subadditive type (s-norm). Hence, the area of applications are extended of the general theorem of Găvruta [4] on the Hyers–Ulam stability.

Among the applications of the Ulam’s type stability, the paper mentioned applications of Hyers–Ulam–Rassias to approximate testing with error relative to input size [19] and applications of the theorem of Găvruta [4] to the study of authomorphism on $C^{*}\text{-}$algebras (see pioneering paper of [20]). The present paper has potential applications in these areas.