Please help me to prove that $|f(x)| \le M \Vert x\Vert$ around $0$ when $|f(x)| \le \Vert x \Vert^\alpha$ around $0$ Question:
Suppose that $0<r<1$ and that $f\colon B_1(0) \to \Bbb R$ is continuously differentiable. 
If there is an $\alpha>0$ such that $|f(x)| \le \Vert x \Vert^{\alpha}$ for all $x\in B_r(0)$, then prove that there is an $M>0$ such that $|f(x)| \le M\Vert x \Vert$ for $x\in B_r(0)$ 

Solution trial: let $x_k\in B_r(0)$ be a sequence with $x_k\to 0$ as $k\to \infty$ 
Since $f$ is continuously differentiable, $f$ is continouos as well. 
Since $f$ is continouos and $x_k\to 0$, it must be that $f(x_k)\to f(0)$.
That is, $$\lim_{k\to\infty}|f(x_k)|\le \lim_{k\to\infty}\Vert x_k \Vert^{\alpha}=0.$$

This is just an idea that may not work. Please help me find a valid proof. Thank you. 
 A: Lemma: There exist $\delta > 0$ and $M>0$ such that whenever $\Vert x \Vert < \delta$, $|f(x)| \le M \Vert x \Vert$.
Warning: I have not mucked about with this sort of mathematics in a number of years, so please check my argument carefully.
Since $f$ has continuous first partial derivatives at $0$, it is differentiable at $0$.
That is,
$$\lim_{x \to 0} \frac {f(x) - f(0) - x \cdot \nabla f(0)}{\Vert x \Vert} = 0.$$
Since $|f(0)| \le \Vert 0 \Vert^\alpha = 0$, $f(0)=0$.
Thus $$\lim_{x \to 0} \frac {f(x) - x \cdot \nabla f(0)}{\Vert x \Vert} = 0.$$
That is, for any $\epsilon > 0$ there is a $\delta > 0$ such that whenever $\Vert x \Vert < \delta$, $$\left\vert  \frac {f(x) - x \cdot \nabla f(0)}{\Vert x \Vert} \right\vert < \epsilon.$$
Rearranging the fraction,
$$\left\vert \frac {f(x)}{\Vert x \Vert} - \frac {x \cdot \nabla f(0)}{\Vert x \Vert} \right\vert < \epsilon.$$
Now the quantity $$q(x) := \frac{x\cdot\nabla f(0)}{\Vert x \Vert}$$ does not depend on the magnitude of $x$, so by a continuity and compactness argument, it is bounded.
Thus the quantity $$\frac {f(x)}{\Vert x \Vert}$$ is also bounded.
$\square$
A: The following argument/discussion makes no assumption about the dimension of the domain of
$f(x)$; that is, we take all balls $B_s(0) \subset R^N$ for some positive integer $N$.
That being said, I break this problem into three different cases:
First, consider the situation $\alpha = 1$; then we have $|f(x)| \le ||x||$, and choosing $M= 1$, we are done.
The second case I will consider is $\alpha > 1$;  here we may write $|f(x) | \le ||x||^{\alpha} = |x||^{\alpha - 1}||x||$, and since $||x||^{\alpha - 1} < r^{\alpha - 1}$ for $x \in B_r(0)$, we can set $M = \rho^{\alpha - 1}$ for any $\rho \ge r$ and we are done.
NOTA BENE:  It should be observed that here I am taking $B_r(0)$ to be the open ball of radius $r$ centered at $0$, that is, $B_r(0) = \{x : ||x|| < r\}$.
The remaining, and of course the most difficult case, is $0 < \alpha < 1$; here we may still affirm that $||x||^\alpha = ||x||^{\alpha  -1}||x||$, but only for $||x|| > 0$; since $\alpha - 1 < 0$, 
$||x||^{\alpha -1} = \frac{1}{||x||^{1 - \alpha}}$
is in fact undefined for $x= 0$, and becomes arbitrarily large as $x \to 0$; therefore it can't serve as the starting point to derive the kind if estimate we need.  These difficulties may be circumvented, however, by  exploiting the fact that $f(x)$ is differentiable in the vicinity of $0$.  Let's do it like this:  pick $r'$ such that $0 < r' < r$, and consider the open annular region $A(r', r) = B_r(0) - \bar{B}_{r'}(0)$, where $\bar{B}_{r'}(0)$ denotes the closure of $B_{r'}(0)$;  $A(r', r)$ looks like the orthogonal projection of a donut (or "torus" for the mathematically inclined ;) ) into a plane in the two dimensional case.  Now for $x \in A(r', r)$ we may legitimately assert that
$|f(x)| \le ||x||^\alpha = ||x||^{\alpha -1}||x|| = \frac{1}{||x||^{1 - \alpha}}||x|| < \frac{1}{(r')^{1 - \alpha}}||x||$
since $||x|| > r'$ , hence $||x||^{1 - \alpha} > (r')^{1 - \alpha}$for such $x$.  So we can take
$M' = \frac{1}{(r')^{1 - \alpha}}$,
on $A(r', r)$, and we obtain $|f(x)| \le M'||x||$, again on $A(r', r)$  But it remains to deal with $x \in \bar{B}_{r'}(0)$; this is where the derivative of $f(x)$ enters the picture.  For $x \in \bar{B}_{r'}(0)$, consider the ray
$tx$ where $0 \le t \le 1$; we compute $\frac{df(xt)}{dt}$ along this path; we have, by the chain rule,
$\frac{df(xt)}{dt} = \nabla f(xt) \cdot \frac{d(xt)}{dt} = \nabla f(xt) \cdot x$;
furthermore,
$f(x) = f(x) - f(0) = \int_0^1 \frac{df(xt)}{dt}dt =  \int_0^1 (\nabla f(xt) \cdot x)dt$,
and since by hypothesis $f(0) = 0$ (by virtue of the assumption that $|f(x)| \le ||x||^\alpha$, as dfeuer pointed out in his answer), we may write
$f(x) = \int_0^1 (\nabla f(xt) \cdot x)dt$,
whence 
$|f(x)| = |\int_0^1 (\nabla f(xt) \cdot x)dt| \le \int_0^1 |(\nabla f(xt) \cdot x)|dt$
$\le \int_0^1 (||(\nabla f(xt)|| \: ||x||)dt = ||x||\int_0^1 (||\nabla f(xt))||dt$,
this last equality holding since $||x||$ is constant with respect to $t$.  We now invoke the hypothesis that $f$ is continuously differentiable on $B_1(0)$, hence on $\bar B_{r'}(0) \subset B_1(0)$; thus $\nabla f(x)$ is continuous on $\bar B_{r'}(0)$, and thus (finally!) $||\nabla f(x)||$ is continuous on $\bar B_{r'}(0)$; thus, since $\bar B_{r'}(0)$ is compact, $||\nabla f(x)||$ is bounded on $\bar B_{r'}(0)$ (it actually obtains both its maximum and minimum values over $\bar B_{r'}(0)$ at points in $\bar B_{r'}(0)$), thus for some positive real $K$,
$||\nabla f(x)|| < K$
on $\bar B_{r'}(0)$.  Using this fact in the last inequality yields
$|f(x)| \le ||x||\int_0^1 (||\nabla f(xt))||dt \le ||x|| \int_0^1 Kdt = K ||x||$,
holding on $\bar B_{r'}(0)$.  We thus have $|f(x)| \le K ||x||$ holding on $\bar B_{r'}(0)$, and  $|f(x)| \le M' ||x||$ holding on $A(r', r)$; setting $M = \max \{K, M'\}$ we finally arrive at $|f(x)| \le M ||x||$, valid on $B_r(0) = \bar B_{r'}(0) \cup A(r', r)$.  And we are done, and this time I mean, "DONE!"
Whew!  Lots of details in that one!
The problem being resolved, a few (hopefully germane) comments may be made:
In looking at the derivations given in the answers, one might ask just why bother with the arguments based directly on $||x||^\alpha$, when the differentiability argument
used both by dfeuer and myself, though in slightly different forms, seems like it might be capable of resolving the question.  One reason is that $f(x)$, though differentiable on 
$B_1(0)$, may in fact not be differentiable on $\bar B_1(0)$; for instance, $||\nabla f||$ may become unbounded as $||x|| \to 1$.  An example of such an $f(x)$ is provided by the function $y(x) = 1 - \sqrt{(1 - x^2)}$ (which describes the "bottom half" of the circle
$x^2 + (y - 1)^2 = 1)$, defined on the open interval $(-1, 1) \subset R$; $y(x)$ can in fact be defined on the closed interval $[-1, 1]$, but $y'(x)$ only exists on its interior, $(-1, 1)$.  Since the derivative $y'(x)$ becomes arbitrarily large on $(-1, 1)$,
an estimate based on a simple bound of the derivative fails in a case such as this
And I see no reason, off the top of my head, why we can't allow $r = 1$!
