I am preparing for my exam and need help with the following task:Let f : R → R be a continuous function with the estimation x 2 ≤ f ( x ) ∀ x ∈ R . Show that f takes on its absolute minima.If a function is continuous, then lim x → a f ( x ) = f ( a ).Our function has an absolute minima in x 0 ∈ R , if f ( x ) ≥ f ( x 0 ) for x ∈ R .At first I thought the task is pretty easy. We learned how to prove that if f : [ a , b ] → R is continuous, then f has an absolute maxima and an absolute Minima in [a,b]. The Problem here is, that the domain of our function here is unbounded. Thats why I don't have any idea what I could and should use for the proof. We should probably use the estimation x 2 ≤ f ( x ) ∀ x ∈ R . This gives us the information, that f is bounded from below with f ( x ) ≥ 0 ∀ x ∈ R . But how does this help me? And what else do we have?

Question

I am preparing for my exam and need help with the following task:Let    f  :      R    →      R   be a continuous function with the estimation       x    2    ≤  f  (  x  )  ∀  x  ∈      R  . Show that f takes on its absolute minima.If a function is continuous, then       lim          x      →      a            f    (    x    )  =  f  (  a  ).Our function has an absolute minima in       x    0    ∈      R  , if   f  (  x  )  ≥  f  (      x    0    ) for   x  ∈      R  .At first I thought the task is pretty easy. We learned how to prove that if   f  :  [  a  ,  b  ]  →      R   is continuous, then f has an absolute maxima and an absolute Minima in [a,b]. The Problem here is, that the domain of our function here is unbounded. Thats why I don&#039;t have any idea what I could and should use for the proof. We should probably use the estimation       x    2    ≤  f  (  x  )   ∀  x  ∈      R  . This gives us the information, that f is bounded from below with   f  (  x  )  ≥  0   ∀  x  ∈      R  . But how does this help me? And what else do we have?

dasse9 · Accepted Answer

Let   a  =  f  (  0  )  ≥  0. Then, if       |    x      |    ≥      a  , we have:                    (1)                    f        (        x        )        ≥                  x          2                ≥        a        =        f        (        0        )            As   [  −      a    ,      a    ] is compact and   f is continuous on it, there is some       x    0    ∈  [  −      a    ,      a    ], such that:  f  (      x    0    )  =      min          [      −              a            ,              a            ]        fClearly,   f  (      x    0    )  ≤  f  (  0  ). Therefore by (1), we have:  f  (      x    0    )  =      min                  R              f

jlo2ni5x · Accepted Answer

It helps, thank you