we can prove $b_{1}=b_{3}=0$? for any reanl numbers $t\in[-\sqrt{2},\sqrt{2}]$,then have
$$-\dfrac{1}{2}\le t^4+b_{3}t^3+b_{2}t^2+b_{1}t+b_{0}\le\dfrac{1}{2}$$
prove or disprove
$$b_{1}=b_{3}=0$$

my idea: I can only prove that
$$b_{1}+b_{3}=0$$
can't prove $b_{1}=b_{3}=0$
my prove $b_{1}+b_{3}=0$ methods:

let $$f(t)=t^4+b_{3}t^3+b_{2}t^2+b_{1}t+b_{0}$$
and let $$g(t)=\dfrac{f(t)+f(-t)}{2}=t^4+b_{2}t^2+b_{0}$$

so we easy to have
$$g(\sqrt{2})+g(0)-2g(1)=2$$
and $$|g(t)|\le\dfrac{1}{2}$$
then we must have
$$g(\sqrt{2})=\dfrac{1}{2},g(0)=\dfrac{1}{2},g(1)=-\dfrac{1}{2}$$
then
$$4+2b_{2}+b_{0}=\dfrac{1}{2},b_{0}=\dfrac{1}{2},1+b_{2}+b_{0}=-\dfrac{1}{2}$$
then 
$$b_{0}=\dfrac{1}{2},b_{2}=-2$$
and 
$$|f(1)|\le\dfrac{1}{2}\Longrightarrow |1+b_{3}+b_{2}+b_{1}+b_{0}|\le\dfrac{1}{2}\Longrightarrow 0\le b_{1}+b_{3}\le 1$$
$$|f(-1)|\le\dfrac{1}{2}\Longrightarrow |1-b_{3}+b_{2}-b_{1}+b_{0}|\le\dfrac{1}{2}\Longrightarrow -1 \le b_{1}+b_{3}\le 0$$
so $$b_{1}+b_{3}=0$$

But I Guess we following can prove   $$b_{1}=b_{3}=0?$$
if ture? How prove it

if not true,why not?

Thank you everyone
As you have shown, $f(t)=t^4+at^3-2t^2-at+\frac{1}{2}$ where $a=b_3$.  Then

$f(\sqrt{2})=4+2\sqrt{2}a-4-\sqrt{2}a+\frac{1}{2}=\sqrt{2}a+\frac{1}{2}$, so $-\frac{1}{2}\le f(\sqrt{2})\le \frac{1}{2}$ gives $-1\le \sqrt{2}a\le0$.

Similarly, $f(-\sqrt{2})=4-2\sqrt{2}a-4+\sqrt{2}a+\frac{1}{2}=-\sqrt{2}a+\frac{1}{2}$, so $-\frac{1}{2}\le f(-\sqrt{2})\le \frac{1}{2}$ gives $-1\le -\sqrt{2}a\le0$ and so $0\le\sqrt{2}a\le1$.  Therefore $\sqrt{2}a=0$, so $b_3=a=0$.
your proof seems questionable.
Since $g^{\prime}(t)=t^3+2b_2t$ the monotonicity of $g(t)$ will depend on the value of $b_2$.

Why do you assert that

$$g(\sqrt{2})=\dfrac{1}{2}\qquad g(0)=\dfrac{1}{2} g(1)=-\dfrac{1}{2}$$