Let $p<q$ be positive integers (with the allowance that $q$ may be $\infty$). How can we show that the sum of $L^p$ and $L^q$ is a Banach space under the norm $\|f\|=\inf\{\|g\|_p+\|h\|_q: g+h=f\}$?

Let $\{f_n\}$ be a sequence in $L^p+L^q$, such that $$\sum_{n=1}^\infty \|f_n\|<\infty.$$

We would like to conclude that this implies that $\sum_{n=1}^\infty f_n$ converges to something in $L^p+L^q$, at which point it follows that $L^p+L^q$ is complete by a basic theorem. We could easily do this if for each $f_i$ it were possible to express $f_i=g_i+h_i$, where $g_i\in L^p, h_i\in L^q$ and $\|f_i\|=\|g_i\|_p+\|h_i\|_q$. Though we know Cauchy sequences in $L^p$ converge, it is not clear that for a given $f$, all sequences (or some sequence) $\{g_n+h_n\}$ such that the $(\|g_n\|_p+\|h_n\|_q)\to \|f_n\|$ have the property that $\{g_n\},\{h_n\}$ converge in their respective spaces. It seems possible to imagine the sum converging without the summands converging in their respective spaces.

Any help would be much appreciated.


2 Answers 2


Hint: for each $n$, choose $g_n \in L^p$, $h_n \in L^q$ such that $f_n = g_n + h_n$ and $||g_n||_p + ||h_n||_q \le ||f_n|| + 2^{-n}$.


You want in fact to show the following result:

Let $(X_1,||\cdot||_1)$ and $(X_2,||\cdot||_2)$ two Banach spaces such that $X_i\subset V$ where $V$ is a vector space. We define $X=\{x_1+x_2,x_1\in X_1,x_2\in X_2\}$ endowed with the norm $||x||_X:=\inf\{||x_1||_1+||x_2||_2,x=x_1+x_2\}$. Then $(X,||\cdot||_X)$ is a Banach space.

To see that, take $\{x^{(n)}\}$ a Cauchy sequence in $X$. We can extract a subsequence, denoted $\{y^{(k)}\}$ such that $||y^{(k+1)}-y^{(k)}||_X\leq 2^{-k}$ for all $k$. Let $(y_1^{(k)}, y_2^{(k)})\in X_1\times X_2$ such that $||y^{(k+1)}-y^{(k)}||_X+2^{-k}\geq ||y_1^{(k)}||_{X_1}+||y_2^{(k)}||_{X_2}$ and $y^{(k+1)}-y^{(k)}=y_1^{(k)}+y_2^{(k)}$. Since $X_1$ and $X_2$ are Banach spaces we can define $y_1:=\sum_{k=0}^{+\infty}y_1^{(k)}$ and $y_2:=\sum_{k=0}^{+\infty}y_2^{(k)}$. We have $$y^{(n+1)}=y^{(0)}+\sum_{k=0}^ny^{(k+1)}-y^{(k)}+y^{(0)}=\sum_{k=1}^ny_1^{(k)}+\sum_{k=1}^ny_2^{(k)}+y^{(0)},$$ which shows that $y^{(n)}$ converges to $\sum_{k=0}^{+\infty}y_1^{(k)}+\sum_{k=0}^{+\infty}y_2^{(k)}+y^{(0)}$.

  • $\begingroup$ Could you please explain why $\|\cdot\|_{X}$ is a norm on $X$? $\endgroup$
    – Feng
    Oct 10, 2019 at 7:57

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