Uniform limit of continuously differentiable function

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Consider $$f_n(x)=sqrtx^2+frac1n$$ on $[-1,1]$




I know $f_n(x) rightarrow vert x vert$ pointwise on $[-1,1]$



How to prove this convergence is also uniform? That is, how to make $$Bigvert ;sqrtx^2+frac1n -vert x vert ;; Bigvert $$ as small as possible? Any hint?




real-analysis uniform-convergence




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    Consider $$f_n(x)=sqrtx^2+frac1n$$ on $[-1,1]$




    I know $f_n(x) rightarrow vert x vert$ pointwise on $[-1,1]$



    How to prove this convergence is also uniform? That is, how to make $$Bigvert ;sqrtx^2+frac1n -vert x vert ;; Bigvert $$ as small as possible? Any hint?




    real-analysis uniform-convergence




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      Consider $$f_n(x)=sqrtx^2+frac1n$$ on $[-1,1]$




      I know $f_n(x) rightarrow vert x vert$ pointwise on $[-1,1]$



      How to prove this convergence is also uniform? That is, how to make $$Bigvert ;sqrtx^2+frac1n -vert x vert ;; Bigvert $$ as small as possible? Any hint?




      real-analysis uniform-convergence




      share|cite|improve this question












      Consider $$f_n(x)=sqrtx^2+frac1n$$ on $[-1,1]$




      I know $f_n(x) rightarrow vert x vert$ pointwise on $[-1,1]$



      How to prove this convergence is also uniform? That is, how to make $$Bigvert ;sqrtx^2+frac1n -vert x vert ;; Bigvert $$ as small as possible? Any hint?





      real-analysis uniform-convergence





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 26 at 3:16









      Learning Mathematics

      469213




      469213




















          3 Answers
          3






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          For $a,b ge 0$, we see $$(a+b)^2 ge a^2 + b^2 ,,, implies ,,, a + b ge sqrta^2 + b^2.$$ This shows that $$sqrtx^2 + frac 1 n le lvert x rvert + frac 1 sqrt n.$$ It's also clear that $lvert x rvert le sqrtx^2 + frac 1 n$ and thus $$0 le sqrtx^2 + frac 1 n - lvert x rvertle frac1sqrt n$$ shows that $sqrtx^2 + frac 1 n to lvert x rvert$ uniformly for $x in mathbb R$.






          share|cite|improve this answer





















          • you mean $M_n=sup ; vert f_n - f vert = frac1sqrtn rightarrow 0$, so the convergence is uniform ?
            – Learning Mathematics
            Jul 26 at 3:29






          • 1




            Exactly. Rather, I only showed that $$sup lvert f_n - f rvert le frac1sqrt n,$$ but of course, the bound is achieved at $x = 0$.
            – User8128
            Jul 26 at 3:41










          • ok Thank you sir!
            – Learning Mathematics
            Jul 26 at 3:46

















          up vote
          2
          down vote













          Note that



          $$sqrtx^2+frac1n -vert x vert =left(sqrtx^2+frac1n -vert x vertright) fracsqrtx^2+frac1n +vert x vertsqrtx^2+frac1n +vert x vert = fracx^2 + frac1n- x^2sqrtx^2+frac1n +vert x vert leqslant frac1sqrtn
          $$






          share|cite|improve this answer





















          • Thank you for sharing your thought sir!
            – Learning Mathematics
            Jul 26 at 3:34


















          up vote
          1
          down vote













          Note that
          $$
          lvert xrvert leq f_n(x) leq lvert xrvert + frac1sqrtnqquad forall xin[-1,1]tag$dagger$
          $$
          the second inequality e.g. by observing that, for any $x$,
          $$
          left(lvert xrvert + frac1sqrtn right)^2 = f_n(x)^2 + 2fraclvert xrvertsqrtn geq f_n(x)^2,.
          $$






          share|cite|improve this answer





















          • This one helps too! thanks!
            – Learning Mathematics
            Jul 26 at 3:32










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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          For $a,b ge 0$, we see $$(a+b)^2 ge a^2 + b^2 ,,, implies ,,, a + b ge sqrta^2 + b^2.$$ This shows that $$sqrtx^2 + frac 1 n le lvert x rvert + frac 1 sqrt n.$$ It's also clear that $lvert x rvert le sqrtx^2 + frac 1 n$ and thus $$0 le sqrtx^2 + frac 1 n - lvert x rvertle frac1sqrt n$$ shows that $sqrtx^2 + frac 1 n to lvert x rvert$ uniformly for $x in mathbb R$.






          share|cite|improve this answer





















          • you mean $M_n=sup ; vert f_n - f vert = frac1sqrtn rightarrow 0$, so the convergence is uniform ?
            – Learning Mathematics
            Jul 26 at 3:29






          • 1




            Exactly. Rather, I only showed that $$sup lvert f_n - f rvert le frac1sqrt n,$$ but of course, the bound is achieved at $x = 0$.
            – User8128
            Jul 26 at 3:41










          • ok Thank you sir!
            – Learning Mathematics
            Jul 26 at 3:46














          up vote
          1
          down vote



          accepted










          For $a,b ge 0$, we see $$(a+b)^2 ge a^2 + b^2 ,,, implies ,,, a + b ge sqrta^2 + b^2.$$ This shows that $$sqrtx^2 + frac 1 n le lvert x rvert + frac 1 sqrt n.$$ It's also clear that $lvert x rvert le sqrtx^2 + frac 1 n$ and thus $$0 le sqrtx^2 + frac 1 n - lvert x rvertle frac1sqrt n$$ shows that $sqrtx^2 + frac 1 n to lvert x rvert$ uniformly for $x in mathbb R$.






          share|cite|improve this answer





















          • you mean $M_n=sup ; vert f_n - f vert = frac1sqrtn rightarrow 0$, so the convergence is uniform ?
            – Learning Mathematics
            Jul 26 at 3:29






          • 1




            Exactly. Rather, I only showed that $$sup lvert f_n - f rvert le frac1sqrt n,$$ but of course, the bound is achieved at $x = 0$.
            – User8128
            Jul 26 at 3:41










          • ok Thank you sir!
            – Learning Mathematics
            Jul 26 at 3:46












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          For $a,b ge 0$, we see $$(a+b)^2 ge a^2 + b^2 ,,, implies ,,, a + b ge sqrta^2 + b^2.$$ This shows that $$sqrtx^2 + frac 1 n le lvert x rvert + frac 1 sqrt n.$$ It's also clear that $lvert x rvert le sqrtx^2 + frac 1 n$ and thus $$0 le sqrtx^2 + frac 1 n - lvert x rvertle frac1sqrt n$$ shows that $sqrtx^2 + frac 1 n to lvert x rvert$ uniformly for $x in mathbb R$.






          share|cite|improve this answer













          For $a,b ge 0$, we see $$(a+b)^2 ge a^2 + b^2 ,,, implies ,,, a + b ge sqrta^2 + b^2.$$ This shows that $$sqrtx^2 + frac 1 n le lvert x rvert + frac 1 sqrt n.$$ It's also clear that $lvert x rvert le sqrtx^2 + frac 1 n$ and thus $$0 le sqrtx^2 + frac 1 n - lvert x rvertle frac1sqrt n$$ shows that $sqrtx^2 + frac 1 n to lvert x rvert$ uniformly for $x in mathbb R$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 26 at 3:22









          User8128

          10.2k1522




          10.2k1522











          • you mean $M_n=sup ; vert f_n - f vert = frac1sqrtn rightarrow 0$, so the convergence is uniform ?
            – Learning Mathematics
            Jul 26 at 3:29






          • 1




            Exactly. Rather, I only showed that $$sup lvert f_n - f rvert le frac1sqrt n,$$ but of course, the bound is achieved at $x = 0$.
            – User8128
            Jul 26 at 3:41










          • ok Thank you sir!
            – Learning Mathematics
            Jul 26 at 3:46
















          • you mean $M_n=sup ; vert f_n - f vert = frac1sqrtn rightarrow 0$, so the convergence is uniform ?
            – Learning Mathematics
            Jul 26 at 3:29






          • 1




            Exactly. Rather, I only showed that $$sup lvert f_n - f rvert le frac1sqrt n,$$ but of course, the bound is achieved at $x = 0$.
            – User8128
            Jul 26 at 3:41










          • ok Thank you sir!
            – Learning Mathematics
            Jul 26 at 3:46















          you mean $M_n=sup ; vert f_n - f vert = frac1sqrtn rightarrow 0$, so the convergence is uniform ?
          – Learning Mathematics
          Jul 26 at 3:29




          you mean $M_n=sup ; vert f_n - f vert = frac1sqrtn rightarrow 0$, so the convergence is uniform ?
          – Learning Mathematics
          Jul 26 at 3:29




          1




          1




          Exactly. Rather, I only showed that $$sup lvert f_n - f rvert le frac1sqrt n,$$ but of course, the bound is achieved at $x = 0$.
          – User8128
          Jul 26 at 3:41




          Exactly. Rather, I only showed that $$sup lvert f_n - f rvert le frac1sqrt n,$$ but of course, the bound is achieved at $x = 0$.
          – User8128
          Jul 26 at 3:41












          ok Thank you sir!
          – Learning Mathematics
          Jul 26 at 3:46




          ok Thank you sir!
          – Learning Mathematics
          Jul 26 at 3:46










          up vote
          2
          down vote













          Note that



          $$sqrtx^2+frac1n -vert x vert =left(sqrtx^2+frac1n -vert x vertright) fracsqrtx^2+frac1n +vert x vertsqrtx^2+frac1n +vert x vert = fracx^2 + frac1n- x^2sqrtx^2+frac1n +vert x vert leqslant frac1sqrtn
          $$






          share|cite|improve this answer





















          • Thank you for sharing your thought sir!
            – Learning Mathematics
            Jul 26 at 3:34















          up vote
          2
          down vote













          Note that



          $$sqrtx^2+frac1n -vert x vert =left(sqrtx^2+frac1n -vert x vertright) fracsqrtx^2+frac1n +vert x vertsqrtx^2+frac1n +vert x vert = fracx^2 + frac1n- x^2sqrtx^2+frac1n +vert x vert leqslant frac1sqrtn
          $$






          share|cite|improve this answer





















          • Thank you for sharing your thought sir!
            – Learning Mathematics
            Jul 26 at 3:34













          up vote
          2
          down vote










          up vote
          2
          down vote









          Note that



          $$sqrtx^2+frac1n -vert x vert =left(sqrtx^2+frac1n -vert x vertright) fracsqrtx^2+frac1n +vert x vertsqrtx^2+frac1n +vert x vert = fracx^2 + frac1n- x^2sqrtx^2+frac1n +vert x vert leqslant frac1sqrtn
          $$






          share|cite|improve this answer













          Note that



          $$sqrtx^2+frac1n -vert x vert =left(sqrtx^2+frac1n -vert x vertright) fracsqrtx^2+frac1n +vert x vertsqrtx^2+frac1n +vert x vert = fracx^2 + frac1n- x^2sqrtx^2+frac1n +vert x vert leqslant frac1sqrtn
          $$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 26 at 3:25









          RRL

          43.5k42260




          43.5k42260











          • Thank you for sharing your thought sir!
            – Learning Mathematics
            Jul 26 at 3:34

















          • Thank you for sharing your thought sir!
            – Learning Mathematics
            Jul 26 at 3:34
















          Thank you for sharing your thought sir!
          – Learning Mathematics
          Jul 26 at 3:34





          Thank you for sharing your thought sir!
          – Learning Mathematics
          Jul 26 at 3:34











          up vote
          1
          down vote













          Note that
          $$
          lvert xrvert leq f_n(x) leq lvert xrvert + frac1sqrtnqquad forall xin[-1,1]tag$dagger$
          $$
          the second inequality e.g. by observing that, for any $x$,
          $$
          left(lvert xrvert + frac1sqrtn right)^2 = f_n(x)^2 + 2fraclvert xrvertsqrtn geq f_n(x)^2,.
          $$






          share|cite|improve this answer





















          • This one helps too! thanks!
            – Learning Mathematics
            Jul 26 at 3:32














          up vote
          1
          down vote













          Note that
          $$
          lvert xrvert leq f_n(x) leq lvert xrvert + frac1sqrtnqquad forall xin[-1,1]tag$dagger$
          $$
          the second inequality e.g. by observing that, for any $x$,
          $$
          left(lvert xrvert + frac1sqrtn right)^2 = f_n(x)^2 + 2fraclvert xrvertsqrtn geq f_n(x)^2,.
          $$






          share|cite|improve this answer





















          • This one helps too! thanks!
            – Learning Mathematics
            Jul 26 at 3:32












          up vote
          1
          down vote










          up vote
          1
          down vote









          Note that
          $$
          lvert xrvert leq f_n(x) leq lvert xrvert + frac1sqrtnqquad forall xin[-1,1]tag$dagger$
          $$
          the second inequality e.g. by observing that, for any $x$,
          $$
          left(lvert xrvert + frac1sqrtn right)^2 = f_n(x)^2 + 2fraclvert xrvertsqrtn geq f_n(x)^2,.
          $$






          share|cite|improve this answer













          Note that
          $$
          lvert xrvert leq f_n(x) leq lvert xrvert + frac1sqrtnqquad forall xin[-1,1]tag$dagger$
          $$
          the second inequality e.g. by observing that, for any $x$,
          $$
          left(lvert xrvert + frac1sqrtn right)^2 = f_n(x)^2 + 2fraclvert xrvertsqrtn geq f_n(x)^2,.
          $$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 26 at 3:22









          Clement C.

          47k33682




          47k33682











          • This one helps too! thanks!
            – Learning Mathematics
            Jul 26 at 3:32
















          • This one helps too! thanks!
            – Learning Mathematics
            Jul 26 at 3:32















          This one helps too! thanks!
          – Learning Mathematics
          Jul 26 at 3:32




          This one helps too! thanks!
          – Learning Mathematics
          Jul 26 at 3:32












           

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