Show that $f(x)=x^4$ is convex for $x\in (0,\infty)$ show $f(x)=x^4$ is convex. 
I know it is convex since $f''(x)>0$ 
. How can we show by using definition?
do we have to use 
Let L be linear space. $t\in[0,1],y\in L,f(xt+y(1-t))=(xt)^4+4(xt)^3((1-t)y)^1+6(xt)^2((1-t)y)^2+4(xt)(((1-t)y)^3+((1-t)y)^4$
edit:
$(xt)^4+4(xt)^3((1-t)y)^1+6(xt)^2((1-t)y)^2+4(xt)(((1-t)y)^3+((1-t)y)^4\le tf(x)+4tf(x)+10tf(x)(1-t)f(y)+(1-t)f(y)$

It is easy to show that $f(x) = x^2$ is convex and increasing on $\mathbb{R}_+$. 
Hence $\forall x, y \in \mathbb{R}_+, t \in [0, 1]$ we have:
$$(tx + (1-t)y)^4 = ((tx + (1-t)y)^2)^2 \stackrel{(1)}\leqslant (tx^2 + (1-t)y^2)^2
\stackrel{(2)}\leqslant \\ t(x^2)^2 + (1-t)(y^2)^2 = tx^4 + (1-t)y^4.$$
$(1)$: using that $x^2$ is convex and increasing.
$(2)$: again using that $x^2$ is convex.
Note also that in your question $L = \mathbb{R}_+$. This is not a linear space and it should not be. But it must be convex because we can speak about convexity of a function only on a convex subset of its domain. 
