Correction: Singh, Y. Mahendra, et al. F-Convex Contraction via Admissible Mapping and Related Fixed Point Theorems with an Application. Mathematics 2018, 6, 105

We found some errors in Lemma 1 of our paper [1], thus, we would like to make the following corrections:

Instead of the following Lemma 1 [1]:

Lemma 1.

Let $(X,d)$ be a metric space and $T:X\to X$ be an α-F-convex contraction satisfying the following conditions:

(i) 

T is α-admissible;

(ii) 

there exists $x_0\in X$ such that $\alpha (x_0,T{x}_0)\ge 1$.

Define a sequence $\left\{x_n\right\}$ in X by $x_{n+1}=T{x}_n=T^{n+1}{x}_0$ for all $n\ge 0$. Then $\left\{d^p(x_n,x_{n+1})\right\}$ is strictly non-increasing sequence in X.

It should read:

Lemma 2.

Let $(X,d)$ be a metric space and $T:X\to X$ be an $\alpha -F$-convex contraction satisfying the conditions:

(i) 

T is α-admissible;

(ii) 

there exists $x_0\in X$ such that $\alpha (x_0,T{x}_0)\ge 1$.

Define a sequence $\left\{x_n\right\}$ in X by $x_{n+1}=T{x}_n=T^{n+1}{x}_0$ for all $n\ge 0$, then $F\left(d^p(x_n,x_{n+1})\right)\le F\left(v\right)-l\tau$, whenever $n=2l$ or $n=2l+1$ for $l\ge 1$.

Proof. 

Following the same steps as in Lemma 1, the last paragraph was replaced with the following: Therefore, $v>d^p(x_2,x_3)$ and hence $F\left(d^p(x_2,x_3)\right)\le F\left(v\right)-\tau$. By a similar argument, we obtain $F\left(d^p(x_3,x_4)\right)\le F\left(v\right)-\tau ;$ continuing in these way, we arrive at $F\left(d^p(x_n,x_{n+1})\right)\le F\left(v\right)-l\tau ,$ whenever $n=2l$ or $n=2l+1$ for $l\ge 1$. ☐

In the proof of the Theorem 2 [1], instead of the following:

“By Lemma 1, $\left\{d^p(x_n,x_{n+1})\right\}$ is strictly non-increasing sequence. Therefore,

$$\begin{array}{c}\hfill F\left(d^p(x_n,x_{n+1})\right)\le F\left(d^p(x_{n-2},x_{n-1})\right)-\tau \le \dots \le F\left(v\right)-l\tau \end{array}$$

whenever $n=2l$ or $n=2l+1$ for $l\ge 1$”.

It should be: By Lemma 1, we obtain:

$$\begin{array}{c}\hfill F\left(d^p(x_n,x_{n+1})\right)\le F\left(v\right)-l\tau ,\end{array}$$

whenever $n=2l$ or $n=2l+1$ for $l\ge 1$. The rest of the proof is unaltered.

The authors apologize to all the readers for any inconvenience this may have caused.