Uniform continuity and antiderivative of x ↦ e − x 2 Show that the antiderivatives of x ↦ e − x 2 are uniformly continuous in R .So we know that for a function to be uniformly continuous there has to exists ξ s.t when | x − y | &lt; δ implies | f ( x ) − f ( y ) | &lt; ε.But how do we go about this since we cannot really integrate e − x 2 and we need the antiderivative?

Question

Uniform continuity and antiderivative of   x  ↦      e          −              x        2            Show that the antiderivatives of   x  ↦      e          −              x        2             are uniformly continuous in       R  .So we know that for a function to be uniformly continuous there has to exists   ξ s.t when       |    x  −  y      |    &amp;lt;  δ implies       |    f  (  x  )  −  f  (  y  )      |    &amp;lt;  ε.But how do we go about this since we cannot really integrate       e          −              x        2             and we need the antiderivative?

jmibanezla · Accepted Answer

Step 1Every function f defined and differentiable over an interval, with a bounded derivative, is uniformly continuous. Indeed, by the mean value theorem, if   x  ≠  y, we can say that            f      (      x      )      −      f      (      y      )              x      −      y        =      f    ′    (  c  ) with c between x and y. Therefore, if       |        f    ′    (  x  )      |    ≤  L for every x, we get      |    f  (  x  )  −  f  (  y  )      |    ≤  L      |    x  −  y      |  Step 2Thus f is not only uniformly continuous, but also Lipschitzian, which is a stronger property.Since   0  ≤      e          −              x        2              ≤  1, we&#039;re done.

glycleWogry · Accepted Answer

Step 1Let   F  (  x  )  =      ∫          −      ∞              x            e          −              t        2              d  t be the antiderivative of       e          −              x        2              ..Without loss of generalities (indeed, the problem is symmetric if we exchange x and y), let&#039;s assume   y  &amp;gt;  x and consider the following:      |    F  (  y  )  −  F  (  x  )      |    =      |          ∫              −        ∞                    y                    e              −                  t          2                      d    t    −          ∫              −        ∞                    x                    e              −                  t          2                      d    t    |    =      |          ∫              x                    y                    e              −                  t          2                      d    t    |    =      ∫          x              y            e          −              t        2              d  t  .Step 2Since       e          −              t        2              ≤  1, then:      |    F  (  y  )  −  F  (  x  )      |    =      ∫          x              y            e          −              t        2              d  t  ≤      ∫          x              y        d  t  =  y  −  x  .Now, for any   ε  &amp;gt;  0, there exists a   δ  &amp;gt;  0, such that if we choose       |    y  −  x      |    &amp;lt;  δ  =  ε, then       |    F  (  y  )  −  F  (  x  )      |    &amp;lt;  ε  ..Hence, the antiderivative of       e          −              x        2             is uniformly continuous.

Uniform continuity and antiderivative of x ↦ e <mrow class="MJX-TeX

Answered question

Answer & Explanation

New Questions in Calculus 1