Proving this integral as a function of $t$ is continuous?












0












$begingroup$


So let $h$ be a function bounded by $M$, so that $|h|< M$.



Also let $g$ be a continuous and non-negative functions.



Define:



$$f(t) = int_0^t g(u-t)h(u) du$$



How can I show $f(t)$ is continuous in $t$? I know that normal integrals are continuous (when the integrated function does not depend on $t$). And since $g$ is continuous it seems like it would be true. But I am struggling to show it.



I have been trying to show it using the normal $epsilon, delta$ definition by trying to find a bound above for



$$|f(t)-f(s)|$$



So that $|t-s| < delta implies |f(t)-f(s)| leq epsilon$, but when I try to do this, the upper bound I get might be negative, which means this is not necessarily true. This is not a homework question, I am working through some textbook proofs, and want to fill in the blanks that are omitted in the proofs as I really struggle with analysis.



I found a previous question here but I do not think it answers my question. As the bounds are not part of the function there.






real-analysis functional-analysis analysis







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    So let $h$ be a function bounded by $M$, so that $|h|< M$.



    Also let $g$ be a continuous and non-negative functions.



    Define:



    $$f(t) = int_0^t g(u-t)h(u) du$$



    How can I show $f(t)$ is continuous in $t$? I know that normal integrals are continuous (when the integrated function does not depend on $t$). And since $g$ is continuous it seems like it would be true. But I am struggling to show it.



    I have been trying to show it using the normal $epsilon, delta$ definition by trying to find a bound above for



    $$|f(t)-f(s)|$$



    So that $|t-s| < delta implies |f(t)-f(s)| leq epsilon$, but when I try to do this, the upper bound I get might be negative, which means this is not necessarily true. This is not a homework question, I am working through some textbook proofs, and want to fill in the blanks that are omitted in the proofs as I really struggle with analysis.



    I found a previous question here but I do not think it answers my question. As the bounds are not part of the function there.






    real-analysis functional-analysis analysis







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      So let $h$ be a function bounded by $M$, so that $|h|< M$.



      Also let $g$ be a continuous and non-negative functions.



      Define:



      $$f(t) = int_0^t g(u-t)h(u) du$$



      How can I show $f(t)$ is continuous in $t$? I know that normal integrals are continuous (when the integrated function does not depend on $t$). And since $g$ is continuous it seems like it would be true. But I am struggling to show it.



      I have been trying to show it using the normal $epsilon, delta$ definition by trying to find a bound above for



      $$|f(t)-f(s)|$$



      So that $|t-s| < delta implies |f(t)-f(s)| leq epsilon$, but when I try to do this, the upper bound I get might be negative, which means this is not necessarily true. This is not a homework question, I am working through some textbook proofs, and want to fill in the blanks that are omitted in the proofs as I really struggle with analysis.



      I found a previous question here but I do not think it answers my question. As the bounds are not part of the function there.






      real-analysis functional-analysis analysis







      share|cite|improve this question









      $endgroup$




      So let $h$ be a function bounded by $M$, so that $|h|< M$.



      Also let $g$ be a continuous and non-negative functions.



      Define:



      $$f(t) = int_0^t g(u-t)h(u) du$$



      How can I show $f(t)$ is continuous in $t$? I know that normal integrals are continuous (when the integrated function does not depend on $t$). And since $g$ is continuous it seems like it would be true. But I am struggling to show it.



      I have been trying to show it using the normal $epsilon, delta$ definition by trying to find a bound above for



      $$|f(t)-f(s)|$$



      So that $|t-s| < delta implies |f(t)-f(s)| leq epsilon$, but when I try to do this, the upper bound I get might be negative, which means this is not necessarily true. This is not a homework question, I am working through some textbook proofs, and want to fill in the blanks that are omitted in the proofs as I really struggle with analysis.



      I found a previous question here but I do not think it answers my question. As the bounds are not part of the function there.






      real-analysis functional-analysis analysis




      real-analysis functional-analysis analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 13 '18 at 0:23









      XiaomiXiaomi

      1,066115




      1,066115






















          1 Answer
          1






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          1












          $begingroup$

          Hint:
          $$eqalign{|f(t) - f(s)| &= left|int_0^t g(u-t) h(u) ; du - int_0^s g(u-s) h(u); duright|cr
          &le left|int_0^s (g(u-t) - g(u-s))h(u); duright| +left| int_s^t g(u-t) h(u); duright|}$$
          Now use the fact that on a bounded interval, $g$ is bounded and uniformly continuous.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thats what I got originally. But I don't understand how I can guarantee that the RHS won't make $epsilon - RHS$ negative using the uniform continuity of $g$. Isn't it possible that $epsilon - |int_0^s dots du|$ will be negative for certain $s$?
            $endgroup$
            – Xiaomi
            Dec 13 '18 at 1:03












          • $begingroup$
            If you choose $delta > 0$ appropriately, you can make each term on the right less than $epsilon/2$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 2:49










          • $begingroup$
            Thus, uniform continuity says for any $eta > 0$ there is $delta > 0$ such that $|x-y| < delta$ implies $|g(x)-g(y)|<eta$. Then $int_0^s (g(u-t) - g(u-s)) h(u); du < eta M s$. So you want $eta < epsilon/(2Ms)$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 2:53










          • $begingroup$
            Oh I see! Thank you!
            $endgroup$
            – Xiaomi
            Dec 13 '18 at 4:27











          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

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          1












          $begingroup$

          Hint:
          $$eqalign{|f(t) - f(s)| &= left|int_0^t g(u-t) h(u) ; du - int_0^s g(u-s) h(u); duright|cr
          &le left|int_0^s (g(u-t) - g(u-s))h(u); duright| +left| int_s^t g(u-t) h(u); duright|}$$
          Now use the fact that on a bounded interval, $g$ is bounded and uniformly continuous.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thats what I got originally. But I don't understand how I can guarantee that the RHS won't make $epsilon - RHS$ negative using the uniform continuity of $g$. Isn't it possible that $epsilon - |int_0^s dots du|$ will be negative for certain $s$?
            $endgroup$
            – Xiaomi
            Dec 13 '18 at 1:03












          • $begingroup$
            If you choose $delta > 0$ appropriately, you can make each term on the right less than $epsilon/2$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 2:49










          • $begingroup$
            Thus, uniform continuity says for any $eta > 0$ there is $delta > 0$ such that $|x-y| < delta$ implies $|g(x)-g(y)|<eta$. Then $int_0^s (g(u-t) - g(u-s)) h(u); du < eta M s$. So you want $eta < epsilon/(2Ms)$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 2:53










          • $begingroup$
            Oh I see! Thank you!
            $endgroup$
            – Xiaomi
            Dec 13 '18 at 4:27
















          1












          $begingroup$

          Hint:
          $$eqalign{|f(t) - f(s)| &= left|int_0^t g(u-t) h(u) ; du - int_0^s g(u-s) h(u); duright|cr
          &le left|int_0^s (g(u-t) - g(u-s))h(u); duright| +left| int_s^t g(u-t) h(u); duright|}$$
          Now use the fact that on a bounded interval, $g$ is bounded and uniformly continuous.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thats what I got originally. But I don't understand how I can guarantee that the RHS won't make $epsilon - RHS$ negative using the uniform continuity of $g$. Isn't it possible that $epsilon - |int_0^s dots du|$ will be negative for certain $s$?
            $endgroup$
            – Xiaomi
            Dec 13 '18 at 1:03












          • $begingroup$
            If you choose $delta > 0$ appropriately, you can make each term on the right less than $epsilon/2$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 2:49










          • $begingroup$
            Thus, uniform continuity says for any $eta > 0$ there is $delta > 0$ such that $|x-y| < delta$ implies $|g(x)-g(y)|<eta$. Then $int_0^s (g(u-t) - g(u-s)) h(u); du < eta M s$. So you want $eta < epsilon/(2Ms)$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 2:53










          • $begingroup$
            Oh I see! Thank you!
            $endgroup$
            – Xiaomi
            Dec 13 '18 at 4:27














          1












          1








          1





          $begingroup$

          Hint:
          $$eqalign{|f(t) - f(s)| &= left|int_0^t g(u-t) h(u) ; du - int_0^s g(u-s) h(u); duright|cr
          &le left|int_0^s (g(u-t) - g(u-s))h(u); duright| +left| int_s^t g(u-t) h(u); duright|}$$
          Now use the fact that on a bounded interval, $g$ is bounded and uniformly continuous.






          share|cite|improve this answer









          $endgroup$



          Hint:
          $$eqalign{|f(t) - f(s)| &= left|int_0^t g(u-t) h(u) ; du - int_0^s g(u-s) h(u); duright|cr
          &le left|int_0^s (g(u-t) - g(u-s))h(u); duright| +left| int_s^t g(u-t) h(u); duright|}$$
          Now use the fact that on a bounded interval, $g$ is bounded and uniformly continuous.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 0:50









          Robert IsraelRobert Israel

          325k23214468




          325k23214468












          • $begingroup$
            Thats what I got originally. But I don't understand how I can guarantee that the RHS won't make $epsilon - RHS$ negative using the uniform continuity of $g$. Isn't it possible that $epsilon - |int_0^s dots du|$ will be negative for certain $s$?
            $endgroup$
            – Xiaomi
            Dec 13 '18 at 1:03












          • $begingroup$
            If you choose $delta > 0$ appropriately, you can make each term on the right less than $epsilon/2$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 2:49










          • $begingroup$
            Thus, uniform continuity says for any $eta > 0$ there is $delta > 0$ such that $|x-y| < delta$ implies $|g(x)-g(y)|<eta$. Then $int_0^s (g(u-t) - g(u-s)) h(u); du < eta M s$. So you want $eta < epsilon/(2Ms)$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 2:53










          • $begingroup$
            Oh I see! Thank you!
            $endgroup$
            – Xiaomi
            Dec 13 '18 at 4:27


















          • $begingroup$
            Thats what I got originally. But I don't understand how I can guarantee that the RHS won't make $epsilon - RHS$ negative using the uniform continuity of $g$. Isn't it possible that $epsilon - |int_0^s dots du|$ will be negative for certain $s$?
            $endgroup$
            – Xiaomi
            Dec 13 '18 at 1:03












          • $begingroup$
            If you choose $delta > 0$ appropriately, you can make each term on the right less than $epsilon/2$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 2:49










          • $begingroup$
            Thus, uniform continuity says for any $eta > 0$ there is $delta > 0$ such that $|x-y| < delta$ implies $|g(x)-g(y)|<eta$. Then $int_0^s (g(u-t) - g(u-s)) h(u); du < eta M s$. So you want $eta < epsilon/(2Ms)$.
            $endgroup$
            – Robert Israel
            Dec 13 '18 at 2:53










          • $begingroup$
            Oh I see! Thank you!
            $endgroup$
            – Xiaomi
            Dec 13 '18 at 4:27
















          $begingroup$
          Thats what I got originally. But I don't understand how I can guarantee that the RHS won't make $epsilon - RHS$ negative using the uniform continuity of $g$. Isn't it possible that $epsilon - |int_0^s dots du|$ will be negative for certain $s$?
          $endgroup$
          – Xiaomi
          Dec 13 '18 at 1:03






          $begingroup$
          Thats what I got originally. But I don't understand how I can guarantee that the RHS won't make $epsilon - RHS$ negative using the uniform continuity of $g$. Isn't it possible that $epsilon - |int_0^s dots du|$ will be negative for certain $s$?
          $endgroup$
          – Xiaomi
          Dec 13 '18 at 1:03














          $begingroup$
          If you choose $delta > 0$ appropriately, you can make each term on the right less than $epsilon/2$.
          $endgroup$
          – Robert Israel
          Dec 13 '18 at 2:49




          $begingroup$
          If you choose $delta > 0$ appropriately, you can make each term on the right less than $epsilon/2$.
          $endgroup$
          – Robert Israel
          Dec 13 '18 at 2:49












          $begingroup$
          Thus, uniform continuity says for any $eta > 0$ there is $delta > 0$ such that $|x-y| < delta$ implies $|g(x)-g(y)|<eta$. Then $int_0^s (g(u-t) - g(u-s)) h(u); du < eta M s$. So you want $eta < epsilon/(2Ms)$.
          $endgroup$
          – Robert Israel
          Dec 13 '18 at 2:53




          $begingroup$
          Thus, uniform continuity says for any $eta > 0$ there is $delta > 0$ such that $|x-y| < delta$ implies $|g(x)-g(y)|<eta$. Then $int_0^s (g(u-t) - g(u-s)) h(u); du < eta M s$. So you want $eta < epsilon/(2Ms)$.
          $endgroup$
          – Robert Israel
          Dec 13 '18 at 2:53












          $begingroup$
          Oh I see! Thank you!
          $endgroup$
          – Xiaomi
          Dec 13 '18 at 4:27




          $begingroup$
          Oh I see! Thank you!
          $endgroup$
          – Xiaomi
          Dec 13 '18 at 4:27


















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