Let *C* be a nonempty, closed and convex subset of *E*. Let {A}_{i}:C\to {E}^{\ast}, for i=1,2,\dots ,N, be continuous monotone mappings. For r>0, define {T}_{i,r}x:=\{z\in C:\u3008{A}_{i}z,y-z\u3009+\frac{1}{r}\u3008\mathrm{\nabla}f(z)-\mathrm{\nabla}f(x),y-z\u3009\ge 0,\mathrm{\forall}y\in C\} for all x\in E and i\in \{1,2,\dots ,N\}, where *f* is a Legendre and convex function from *E* into (-\mathrm{\infty},+\mathrm{\infty}). Then, in what follows, we shall study the following iteration process:

\{\begin{array}{l}{x}_{0}=u\in C\phantom{\rule{1em}{0ex}}\text{chosen arbitrarily},\\ {w}_{n}={T}_{N,{r}_{n}}\circ {T}_{N-1,{r}_{n}}\circ \cdots \circ {T}_{1,{r}_{n}}{x}_{n},\\ {x}_{n+1}={P}_{C}^{f}\mathrm{\nabla}{f}^{\ast}({\alpha}_{n}\mathrm{\nabla}f(u)+(1-{\alpha}_{n})\mathrm{\nabla}f({w}_{n})),\phantom{\rule{1em}{0ex}}\mathrm{\forall}n\ge 0,\end{array}

(3.1)

where \{{\alpha}_{n}\}\subset (0,1) satisfies {lim}_{n\to \mathrm{\infty}}{\alpha}_{n}=0 and {\sum}_{n=1}^{\mathrm{\infty}}=\mathrm{\infty}, and \{{r}_{n}\}\subset [{c}_{1},\mathrm{\infty}) for some {c}_{1}>0.

**Theorem 3.1** *Let* f:E\to \mathbb{R} *be a strongly coercive Legendre function which is bounded*, *uniformly Fréchet differentiable and totally convex on bounded subsets of* *E*. *Let* *C* *be a nonempty*, *closed and convex subset of* int(domf) *and* {A}_{i}:C\to {E}^{\ast}, *for* i=1,2,\dots ,N, *be a finite family of continuous monotone mappings with* \mathcal{F}:={\bigcap}_{i=1}^{N}\mathit{VI}(C,{A}_{i})\ne \mathrm{\varnothing}. *Let* {\{{x}_{n}\}}_{n\ge 0} *be a sequence defined by* (3.1). *Then* \{{x}_{n}\} *converges strongly to* {x}^{\ast}={P}_{\mathcal{F}}^{f}(u).

*Proof* By Lemma 2.7 we have that each \mathit{VI}(C,{A}_{i}) for each i\in \{1,2,\dots ,N\} and hence ℱ are closed and convex. Thus, we can take {x}^{\ast}:={P}_{\mathcal{F}}^{f}u. Let {u}_{n,1}={T}_{1,{r}_{n}}{x}_{n},{u}_{n,2}={T}_{2,{r}_{n}}{u}_{n,1},\dots ,{u}_{n,N-1}={T}_{N-1,{r}_{n}}{u}_{n,N-2} and {u}_{n,N}={T}_{N,{r}_{n}}{u}_{n,N-1}={w}_{n}. Then, from (3.1), Lemmas 2.2, 2.3 and the property of *ϕ*, we get that

\begin{array}{rcl}{D}_{f}({x}^{\ast},{x}_{n+1})& =& {D}_{f}({x}^{\ast},{P}_{C}^{f}\mathrm{\nabla}{f}^{\ast}({\alpha}_{n}\mathrm{\nabla}f(u)+(1-{\alpha}_{n})\mathrm{\nabla}f({w}_{n})))\\ \le & {D}_{f}({x}^{\ast},\mathrm{\nabla}{f}^{\ast}({\alpha}_{n}\mathrm{\nabla}f(u)+(1-{\alpha}_{n})\mathrm{\nabla}f({w}_{n})))\\ \le & {\alpha}_{n}{D}_{f}({x}^{\ast},u)+(1-{\alpha}_{n}){D}_{f}({x}^{\ast},{w}_{n})\\ =& {\alpha}_{n}{D}_{f}({x}^{\ast},u)+(1-{\alpha}_{n}){D}_{f}({x}^{\ast},{T}_{N,{r}_{n}}\circ {T}_{N-1,{r}_{n}}\circ \cdots \circ {T}_{1,{r}_{n}}{x}_{n})\\ \le & {\alpha}_{n}{D}_{f}({x}^{\ast},u)+(1-{\alpha}_{n}){D}_{f}({x}^{\ast},{x}_{n}).\end{array}

(3.2)

Thus, by induction,

{D}_{f}({x}^{\ast},{x}_{n+1})\le max\{{D}_{f}({x}^{\ast},{x}_{n}),{D}_{f}({x}^{\ast},u)\},\phantom{\rule{1em}{0ex}}\mathrm{\forall}n\ge 0,

which implies by Lemma 2.4 that \{{x}_{n}\} and hence \{{w}_{n}\} are bounded. Now let {z}_{n}=\mathrm{\nabla}{f}^{\ast}({\alpha}_{n}\mathrm{\nabla}f(u)+(1-{\alpha}_{n})\mathrm{\nabla}f({w}_{n})). Then we have from (3.1) that {x}_{n+1}={P}_{C}^{f}{z}_{n}. Using Lemmas 2.2, 2.3, 2.7(3), (2.3) and (2.4), we obtain that

\begin{array}{rcl}{D}_{f}({x}^{\ast},{x}_{n+1})& =& {D}_{f}({x}^{\ast},{P}_{C}^{f}{z}_{n})\le {D}_{f}({x}^{\ast},{z}_{n})=V({x}^{\ast},\mathrm{\nabla}f({z}_{n}))\\ \le & V({x}^{\ast},\mathrm{\nabla}f({z}_{n})-{\alpha}_{n}(\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right)))+\u3008{\alpha}_{n}(\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right)),{z}_{n}-{x}^{\ast}\u3009\\ =& {D}_{f}({x}^{\ast},\mathrm{\nabla}{f}^{\ast}({\alpha}_{n}\mathrm{\nabla}f\left({x}^{\ast}\right)+(1-{\alpha}_{n})\mathrm{\nabla}f({w}_{n})))\\ +{\alpha}_{n}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right),{z}_{n}-{x}^{\ast}\u3009\\ \le & {\alpha}_{n}\varphi ({x}^{\ast},{x}^{\ast})+(1-{\alpha}_{n}){D}_{f}({x}^{\ast},{w}_{n})+{\alpha}_{n}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right),{z}_{n}-{x}^{\ast}\u3009\\ \le & (1-{\alpha}_{n}){D}_{f}({x}^{\ast},{T}_{N,{r}_{n}}{u}_{n,N-1})+{\alpha}_{n}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right),{z}_{n}-{x}^{\ast}\u3009,\end{array}

which implies that

\begin{array}{rcl}{D}_{f}({x}^{\ast},{x}_{n+1})& \le & (1-{\alpha}_{n})[{D}_{f}({x}^{\ast},{u}_{n,N-1})-{D}_{f}({w}_{n},{u}_{n,N-1})]\\ +{\alpha}_{n}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right),{z}_{n}-{x}^{\ast}\u3009\\ \le & (1-{\alpha}_{n})[{D}_{f}({x}^{\ast},{u}_{{n}_{N}-2})-{D}_{f}({u}_{{n}_{N}-1},{u}_{{n}_{N}-2})]\\ -(1-{\alpha}_{n}){D}_{f}({w}_{n},{u}_{n,N-1})+{\alpha}_{n}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right),{z}_{n}-{x}^{\ast}\u3009\\ \cdots \\ \le & (1-{\alpha}_{n}){D}_{f}({x}^{\ast},{x}_{n})-(1-{\alpha}_{n})[{D}_{f}({u}_{n,1},{x}_{n})+{D}_{f}({u}_{n,2},{u}_{n,1})\\ +\cdots +{D}_{f}({u}_{n,N-1},{u}_{n,N-2})+{D}_{f}({w}_{n},{u}_{n,N-1})]\\ +{\alpha}_{n}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right),{z}_{n}-{x}^{\ast}\u3009\\ \le & (1-{\alpha}_{n}){D}_{f}({x}^{\ast},{x}_{n})+{\alpha}_{n}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right),{z}_{n}-{x}^{\ast}\u3009.\end{array}

(3.3)

Now, we consider two possible cases.

Case 1. Suppose that there exists {n}_{0}\in N such that \{{D}_{f}({x}^{\ast},{x}_{n})\} is decreasing. Then we obtain that \{{D}_{f}({x}^{\ast},{x}_{n})\} is convergent. Thus, from (3.3) we have that {D}_{f}({u}_{n,1},{x}_{n}),{D}_{f}({u}_{n,2},{u}_{n,1}),\dots ,{D}_{f}({w}_{n},{u}_{n,N-1})\to 0 as n\to \mathrm{\infty}, and hence by Lemma 2.1 we get that

{u}_{n,1}-{x}_{n}\to 0,\phantom{\rule{2em}{0ex}}{u}_{n,2}-{u}_{n,1}\to 0,\phantom{\rule{2em}{0ex}}\dots ,\phantom{\rule{2em}{0ex}}{w}_{n}-{u}_{n,N-1}\to 0\phantom{\rule{1em}{0ex}}\text{as}n\to \mathrm{\infty}.

(3.4)

Furthermore, from the property of {D}_{f}(\cdot ,\cdot ) and the fact that {\alpha}_{n}\to 0 as n\to \mathrm{\infty}, we have that

\begin{array}{rcl}{D}_{f}({w}_{n},{z}_{n})& =& {D}_{f}({w}_{n},\mathrm{\nabla}{f}^{\ast}({\alpha}_{n}\mathrm{\nabla}f(u)+(1-{\alpha}_{n})\mathrm{\nabla}f({w}_{n})))\\ \le & {\alpha}_{n}{D}_{f}({w}_{n},u)+(1-{\alpha}_{n}){D}_{f}({w}_{n},{w}_{n})\\ \le & {\alpha}_{n}{D}_{f}({w}_{n},u)+(1-{\alpha}_{n}){D}_{f}({w}_{n},{w}_{n})\to 0\phantom{\rule{1em}{0ex}}\text{as}n\to \mathrm{\infty},\end{array}

and hence from Lemma 2.1 we have that {w}_{n}-{z}_{n}\to 0 and this with (3.4) implies that

{z}_{n}-{u}_{n,N-1}\to 0,\phantom{\rule{2em}{0ex}}{z}_{n}-{u}_{n,N-2}\to 0,\phantom{\rule{2em}{0ex}}\dots ,\phantom{\rule{2em}{0ex}}{z}_{n}-{u}_{n,1}\to 0\phantom{\rule{1em}{0ex}}\text{as}n\to \mathrm{\infty}.

(3.5)

Since \{{z}_{n}\} is bounded and *E* is reflexive, we choose a subsequence \{{z}_{{n}_{k}}\} of \{{z}_{n}\} such that {z}_{{n}_{k}}\rightharpoonup z and {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty}}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f({x}^{\ast}),{z}_{n}-{x}^{\ast}\u3009={lim}_{k\to \mathrm{\infty}}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f({x}^{\ast}),{z}_{{n}_{k}}-{x}^{\ast}\u3009. Then, from (3.5) and (3.4), we get that {u}_{{n}_{k},i}\rightharpoonup z for each i\in \{1,2,\dots ,N\}.

Now, we show that z\in \mathit{VI}(C,{A}_{i}) for each i\in \{1,2,\dots ,N\}. But from the definition of {u}_{n,i}, we have that

\u3008{A}_{i}{u}_{n,i},y-{u}_{n,i}\u3009+\u3008\frac{\mathrm{\nabla}f({u}_{n,i})-\mathrm{\nabla}f({x}_{n})}{{r}_{n}},y-{u}_{n,i}\u3009\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in C,

and hence

\u3008{A}_{i}{u}_{{n}_{k},i},y-{u}_{{n}_{k},i}\u3009+\u3008\frac{\mathrm{\nabla}f({u}_{{n}_{k},i})-\mathrm{\nabla}f({x}_{{n}_{k}})}{{r}_{{n}_{k}}},y-{u}_{{n}_{k},i}\u3009\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in C

(3.6)

for each i\in \{1,2,\dots ,N\}. Set {v}_{t}=ty+(1-t)z for all t\in (0,1] and y\in C. Consequently, we get that {v}_{t}\in C. Now, from (3.6) it follows that

\begin{array}{rcl}\u3008{A}_{i}{v}_{t},{v}_{t}-{u}_{{n}_{k},i}\u3009& \ge & \u3008{A}_{i}{v}_{t},{v}_{t}-{u}_{{n}_{k},i}\u3009-\u3008{A}_{i}{u}_{{n}_{k},i},{v}_{t}-{u}_{{n}_{k},i}\u3009\\ -\u3008\frac{\mathrm{\nabla}f({u}_{{n}_{k},i})-\mathrm{\nabla}f({x}_{{n}_{k}})}{{r}_{{n}_{k}}},{v}_{t}-{u}_{{n}_{k},i}\u3009\\ =& \u3008{A}_{i}{v}_{t}-{A}_{i}{u}_{{n}_{k},i},{v}_{t}-{u}_{{n}_{k},i}\u3009\\ -\u3008\frac{\mathrm{\nabla}f({u}_{{n}_{k},i})-\mathrm{\nabla}f({x}_{{n}_{k}})}{{r}_{{n}_{k}}},{v}_{t}-{u}_{{n}_{k},i}\u3009.\end{array}

In addition, since *f* is uniformly Fréchet differentiable and bounded, we have that ∇*f* is uniformly continuous (see [36]). Thus, from (3.4) and the uniform continuity of ∇*f*, we obtain that

\frac{\mathrm{\nabla}f({u}_{{n}_{k},i})-\mathrm{\nabla}f({x}_{{n}_{k}})}{{r}_{{n}_{k}}}\to 0\phantom{\rule{1em}{0ex}}\text{as}k\to \mathrm{\infty},

and since *A* is monotone, we also have that \u3008{A}_{i}{v}_{t}-{A}_{i}{u}_{{n}_{k},i},{v}_{t}-{u}_{{n}_{k},i}\u3009\ge 0. Thus, it follows that

0\le \underset{k\to \mathrm{\infty}}{lim}\u3008{A}_{i}{v}_{t},{v}_{t}-{u}_{{n}_{k},i}\u3009=\u3008{A}_{i}{v}_{t},{v}_{t}-z\u3009,

and hence

\u3008{A}_{i}{v}_{t},y-z\u3009\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in C,\text{for all}i\in \{1,2,\dots ,N\}.

If t\to 0, the continuity of {A}_{i} implies that

\u3008{A}_{i}z,y-z\u3009\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}y\in C.

This implies that z\in \mathit{VI}(C,{A}_{i}) for all i\in \{1,2,\dots ,N\}.

Therefore, we obtain that z\in {\bigcap}_{i=1}^{N}\mathit{VI}(C,{A}_{i}). Thus, by Lemma 2.2, we immediately obtain that {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty}}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f({x}^{\ast}),{z}_{n}-{x}^{\ast}\u3009=\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f({x}^{\ast}),z-{x}^{\ast}\u3009\le 0. It follows from Lemma 2.5 and (3.3) that {D}_{f}({x}^{\ast},{x}_{n})\to 0 as n\to \mathrm{\infty}. Consequently, {x}_{n}\to {x}^{\ast}.

Case 2. Suppose that there exists a subsequence \{{n}_{j}\} of \{n\} such that

{D}_{f}({x}^{\ast},{x}_{{n}_{j}})<{D}_{f}({x}^{\ast},{x}_{{n}_{j}+1})

for all j\in \mathbb{N}. Then, by Lemma 2.6, there exists a nondecreasing sequence \{{m}_{k}\}\subset \mathbb{N} such that {m}_{k}\to \mathrm{\infty}, {D}_{f}({x}^{\ast},{x}_{{m}_{k}})\le {D}_{f}({x}^{\ast},{x}_{{m}_{k}+1}) and {D}_{f}({x}^{\ast},{x}_{k})\le {D}_{f}({x}^{\ast},{x}_{{m}_{k}+1}) for all k\in \mathbb{N}. From (3.3) and {\alpha}_{n}\to 0, we have

\begin{array}{r}(1-{\alpha}_{{m}_{k}})({D}_{f}({u}_{{m}_{k},1},{x}_{{m}_{k}})+\cdots +{D}_{f}({w}_{{m}_{k}},{u}_{{m}_{k},N-1}))\\ \phantom{\rule{1em}{0ex}}\le ({D}_{f}({x}^{\ast},{x}_{{m}_{k}})-{D}_{f}({x}^{\ast},{x}_{{m}_{k}+1}))+{\alpha}_{{m}_{k}}{D}_{f}({x}^{\ast},{x}_{{m}_{k}})\\ \phantom{\rule{2em}{0ex}}+{\alpha}_{{m}_{k}}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right),{z}_{{m}_{k}}-{x}^{\ast}\u3009\to 0\phantom{\rule{1em}{0ex}}\text{as}k\to \mathrm{\infty},\end{array}

which implies that {D}_{f}({u}_{{m}_{k},1},{x}_{{m}_{k}}),\dots ,{D}_{f}({w}_{{m}_{k}},{u}_{{m}_{k},N-1})\to 0, and hence {u}_{{m}_{k},1}-{x}_{{m}_{k}}\to 0,\dots ,{w}_{{m}_{k}}-{u}_{{m}_{k},N-1}\to 0 as k\to \mathrm{\infty}. Thus, as in Case 1, we obtain that

\underset{k\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right),{z}_{{m}_{k}}-{x}^{\ast}\u3009\le 0.

(3.7)

Furthermore, from (3.3) we have that

{D}_{f}({x}^{\ast},{x}_{{m}_{k}+1})\le (1-{\alpha}_{{m}_{k}}){D}_{f}({x}^{\ast},{x}_{{m}_{k}})+{\alpha}_{{m}_{k}}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right),{z}_{{m}_{k}}-{x}^{\ast}\u3009.

(3.8)

Thus, since {D}_{f}({x}^{\ast},{x}_{{m}_{k}})\le {D}_{f}({x}^{\ast},{x}_{{m}_{k}+1}), we get that

\begin{array}{rcl}{\alpha}_{{m}_{k}}{D}_{f}({x}^{\ast},{x}_{{m}_{k}})& \le & {D}_{f}({x}^{\ast},{x}_{{m}_{k}})-{D}_{f}({x}^{\ast},{x}_{{m}_{k}+1})\\ +{\alpha}_{{m}_{k}}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right),{z}_{{m}_{k}}-{x}^{\ast}\u3009\\ \le & {\alpha}_{{m}_{k}}\u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right),{z}_{{m}_{k}}-{x}^{\ast}\u3009.\end{array}

Moreover, since {\alpha}_{{m}_{k}}>0, we obtain that

{D}_{f}({x}^{\ast},{x}_{{m}_{k}})\le \u3008\mathrm{\nabla}f(u)-\mathrm{\nabla}f\left({x}^{\ast}\right),{z}_{{m}_{k}}-{x}^{\ast}\u3009.

It follows from (3.7) that {D}_{f}({x}^{\ast},{x}_{{m}_{k}})\to 0 as k\to \mathrm{\infty}. This together with (3.8) implies that {D}_{f}({x}^{\ast},{x}_{{m}_{k}+1})\to 0. Therefore, since {D}_{f}({x}^{\ast},{x}_{k})\le {D}_{f}({x}^{\ast},{x}_{{m}_{k}+1}) for all k\in \mathbb{N}, we conclude that {x}_{k}\to {x}^{\ast} as k\to \mathrm{\infty}. Hence, both cases imply that \{{x}_{n}\} converges strongly to {x}^{\ast}={P}_{F}^{f}u and the proof is complete. □

If in Theorem 3.1 N=1, then we get the following corollary.

**Corollary 3.2** *Let* f:E\to \mathbb{R} *be a strongly coercive Legendre function which is bounded*, *uniformly Fréchet differentiable and totally convex on bounded subsets of* *E*. *Let* *C* *be a nonempty*, *closed and convex subset of* int(domf), *and let* A:C\to {E}^{\ast} *be a continuous monotone mapping with* \mathit{VI}(C,A)\ne \mathrm{\varnothing}. *Let* {\{{x}_{n}\}}_{n\ge 0} *be a sequence defined by* (3.1),

\{\begin{array}{l}{x}_{0}=u\in C\phantom{\rule{1em}{0ex}}\mathit{\text{chosen arbitrarily}},\\ {w}_{n}={T}_{{r}_{n}}{x}_{n},\\ {x}_{n+1}={P}_{C}^{f}\mathrm{\nabla}{f}^{\ast}({\alpha}_{n}\mathrm{\nabla}f(u)+(1-{\alpha}_{n})\mathrm{\nabla}f({w}_{n})),\end{array}

(3.9)

*where* {T}_{\gamma}x:=\{z\in C:\u3008Az,y-z\u3009+\frac{1}{\gamma}\u3008\mathrm{\nabla}f(z)-\mathrm{\nabla}f(x),y-z\u3009\ge 0,\mathrm{\forall}y\in C\} *for all* x\in E; {\alpha}_{n}\in (0,1) *satisfies* {lim}_{n\to \mathrm{\infty}}{\alpha}_{n}=0 *and* {\sum}_{n=1}^{\mathrm{\infty}}{\alpha}_{n}=\mathrm{\infty} *and* \{{r}_{n}\}\subset [{c}_{1},\mathrm{\infty}) *for some* {c}_{1}>0. *Then the sequence* {\{{x}_{n}\}}_{n\ge 0} *converges strongly to a point* {x}^{\ast}={P}_{\mathit{VI}(C,A)}(u).

If C=E, then \mathit{VI}(C,A)={A}^{-1}(0) and hence the following corollary holds.

**Corollary 3.3** *Let* f:E\to \mathbb{R} *be a strongly coercive Legendre function which is bounded*, *uniformly Fréchet differentiable and totally convex on bounded subsets of* *E*. *Let* {A}_{i}:E\to {E}^{\ast}, *for* i=1,2,\dots ,N, *be a finite family of continuous monotone mappings*. *Let* \mathcal{F}:={\bigcap}_{i=1}^{N}\mathit{VI}(C,{A}_{i})={\bigcap}_{i=1}^{N}{A}^{-1}(0)\ne \mathrm{\varnothing}. *Let* {\{{x}_{n}\}}_{n\ge 0} *be a sequence defined by* (3.1). *Then* \{{x}_{n}\} *converges strongly to* {x}^{\ast}={P}_{\mathcal{F}}^{f}(u).

If in Theorem 3.1 we assume u=0, then the scheme converges strongly to the common minimum-norm zero of a finite family of continuous monotone mappings. In fact, we have the following corollary.

**Corollary 3.4** *Let* f:E\to \mathbb{R} *be a strongly coercive Legendre function which is bounded*, *uniformly Fréchet differentiable and totally convex on bounded subsets of* *E*. *Let* *C* *be a nonempty*, *closed and convex subset of* int(domf), *and let* {A}_{i}:C\to {E}^{\ast}, *for* i=1,2,\dots ,N, *be a finite family of continuous monotone mappings with* \mathcal{F}:={\bigcap}_{i=1}^{N}\mathit{VI}(C,{A}_{i})\ne \mathrm{\varnothing}. *Let* {\{{x}_{n}\}}_{n\ge 0} *be a sequence defined by* (3.1) *with* u=0. *Then* \{{x}_{n}\} *converges strongly to* {x}^{\ast}={P}_{\mathcal{F}}^{f}(0), *which is the common minimum*-*norm* (*with respect to the Bregman distance*) *solution of the variational inequalities*.