Uniform continuity of heat equation with L 1 dataI came across the following remark in my reading.Remark. If the initial data is in L 1 , then heat equation solutions approach 0 within the sup-norm. That is, if K(x,t) is the heat kernel and f ∈ L 1 ( R n ), then lim t → ∞ ∫ R n K ( x − y , t ) f ( y ) d y = 0 ,uniformly.I can picture this working if f is uniformly continuous in R n and lim x → ∞ f ( x ) = 0. Does f ∈ L 1 ( R n ) imply this somehow, and I'm not seeing this? I'm relatively new to L p -spaces.If so, then for ϵ &gt; 0, we can find some radius r ϵ &gt; 0 such that | | f ( y ) | &lt; ϵ / 2.Then we can bound the heat kernel, and use the fact that, ∫ K ( x − y , t ) d y = 1 to conclude, more or less.

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Uniform continuity of heat equation with       L          1        dataI came across the following remark in my reading.Remark. If the initial data is in       L          1      , then heat equation solutions approach 0 within the sup-norm. That is, if K(x,t) is the heat kernel and   f  ∈      L    1    (            R        n    ), then      lim          t      →      ∞            ∫                            R                n              K  (  x  −  y  ,  t  )  f  (  y  )  d  y  =  0  ,uniformly.I can picture this working if f is uniformly continuous in             R        n   and       lim          x      →      ∞        f  (  x  )  =  0. Does   f  ∈      L    1    (            R        n    ) imply this somehow, and I&#039;m not seeing this? I&#039;m relatively new to       L          p      -spaces.If so, then for   ϵ  &amp;gt;  0, we can find some radius       r          ϵ        &amp;gt;  0 such that |      |    f  (  y  )      |    &amp;lt;  ϵ      /    2.Then we can bound the heat kernel, and use the fact that,   ∫  K  (  x  −  y  ,  t  )  d  y  =  1 to conclude, more or less.

Alisa Jacobs · Accepted Answer

This is quite straightforward from the definition. In n dimensions, the heat kernel is given by  K  (  x  −  y  ,  t  )  =      1          (      4      π      t              )                  n                      /                    2                          e          −              |            x      −      y                        |                2                    /            4      t        .As       e          −      x        ≤  1 for   x  ≥  0 this implies that       |    K  (  x  −  y  ,  t  )      |    ≤      (    4    π    t          )              −        n                  /                2             and therefore      |          ∫                                    R                    n                      K    (    x    −    y    ,    t    )    f    (    y    )    d    y    |    ≤      (    4    π    t          )              −        n                  /                2                  ∫                            R                n                  |    f  (  y  )      |    d  y  ,which converges to 0 as   t  →  ∞ uniformly in x, as soon as f is in       L    1    (            R        n    ).

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