This is related to these posts and here.

Let $L^1([n,n+1])$ denote the subspace of $L^p$-functions on $[0,\infty)$ essentially supported on $[-n,n]$. Denote the accelerated $\ell^1$-direct sum Banach space $$ \bigoplus_{n \in \mathbb{N}}^{\ell^1} L^{1/2}([0,n]) := \left\{ f \in L^{1}([0,\infty)):\, \sum_{n=1}^{\infty} \left( 2^n\int_{x \in [n,n+1] } \sqrt{|f(x)|} dx \right) <\infty \right\}. $$

Alternatively, equip $L^1_{\mathrm{comp}}$ be the space of essentially compactly-supported $L^p$ functions, with their usual inductive limit topology obtained by the inductive system $$ \left\{f \in L^1_{\mathrm{loc}}: \, \operatorname{esssupp}(f)\subseteq [0,n] \right\} \to \left\{f \in L^1_{\mathrm{loc}}: \, \operatorname{esssupp}(f)\subseteq [0,m] \right\}\qquad n\leq m;\, n,m \in \mathbb{N} . $$ Is the topology of $\oplus^{\ell^1}_{n}L^1([0,n])$ coarser than that of $L^p_{\mathrm{comp}}$ on their intersection?

It seems to me to be the case since $\oplus^{\ell^1}_{n}L^1([0,n])$ is much smaller than $L^p_{\mathrm{comp}}$.

This is related to these posts and here.

Let $L^1([n,n+1])$ denote the subspace of $L^p$-functions on $[0,\infty)$ essentially supported on $[-n,n]$. Denote the accelerated $\ell^1$-direct sum Banach space $$ \bigoplus_{n \in \mathbb{N}}^{\ell^1} L^{1/2}([0,n]) := \left\{ f \in L^{1}([0,\infty)):\, \sum_{n=1}^{\infty} \left( 2^n\int_{x \in [n,n+1] } \sqrt{|f(x)|} dx \right) <\infty \right\}. $$

Alternatively, equip $L^1_{\mathrm{comp}}$ be the space of essentially compactly-supported $L^p$ functions, with their usual inductive limit topology obtained by the inductive system $$ \left\{f \in L^1_{\mathrm{loc}}: \, \operatorname{esssupp}(f)\subseteq [0,n] \right\} \to \left\{f \in L^1_{\mathrm{loc}}: \, \operatorname{esssupp}(f)\subseteq [0,m] \right\}\qquad n\leq m;\, n,m \in \mathbb{N} , $$ in the category of topological vector spaces. Is the topology of $\oplus^{\ell^1}_{n}L^1([0,n])$ coarser than that of $L^p_{\mathrm{comp}}$ on their intersection?

It seems to me to be the case since $\oplus^{\ell^1}_{n}L^1([0,n])$ is much smaller than $L^p_{\mathrm{comp}}$.

This is related to these posts and here.

Let $L^1([n,n+1])$ denote the subspace of $L^p$-functions on $[0,\infty)$ essentially supported on $[-n,n]$. Denote the accelerated $\ell^1$-direct sum Banach space $$ \bigoplus_{n \in \mathbb{N}}^{\ell^1} L^1([0,n]) := \left\{ f \in L^1([0,\infty)):\, \sum_{n=1}^{\infty} \left( 2^n\int_{x \in [n,n+1] } |f(x)| dx \right) <\infty \right\}. $$

Alternatively, equip $L^1_{\mathrm{comp}}$ be the space of essentially compactly-supported $L^p$ functions, with their usual inductive limit topology obtained by the inductive system $$ \left\{f \in L^1_{\mathrm{loc}}: \, \operatorname{esssupp}(f)\subseteq [0,n] \right\} \to \left\{f \in L^1_{\mathrm{loc}}: \, \operatorname{esssupp}(f)\subseteq [0,m] \right\}\qquad n\leq m;\, n,m \in \mathbb{N} . $$ Is the topology of $\oplus^{\ell^1}_{n}L^1([0,n])$ coarser than that of $L^p_{\mathrm{comp}}$ on their intersection?

It seems to me to be the case since $\oplus^{\ell^1}_{n}L^1([0,n])$ is much smaller than $L^p_{\mathrm{comp}}$.

This is related to these posts and here.

Let $L^1([n,n+1])$ denote the subspace of $L^p$-functions on $[0,\infty)$ essentially supported on $[-n,n]$. Denote the accelerated $\ell^1$-direct sum Banach space $$ \bigoplus_{n \in \mathbb{N}}^{\ell^1} L^{1/2}([0,n]) := \left\{ f \in L^{1}([0,\infty)):\, \sum_{n=1}^{\infty} \left( 2^n\int_{x \in [n,n+1] } \sqrt{|f(x)|} dx \right) <\infty \right\}. $$

Alternatively, equip $L^1_{\mathrm{comp}}$ be the space of essentially compactly-supported $L^p$ functions, with their usual inductive limit topology obtained by the inductive system $$ \left\{f \in L^1_{\mathrm{loc}}: \, \operatorname{esssupp}(f)\subseteq [0,n] \right\} \to \left\{f \in L^1_{\mathrm{loc}}: \, \operatorname{esssupp}(f)\subseteq [0,m] \right\}\qquad n\leq m;\, n,m \in \mathbb{N} . $$ Is the topology of $\oplus^{\ell^1}_{n}L^1([0,n])$ coarser than that of $L^p_{\mathrm{comp}}$ on their intersection?

It seems to me to be the case since $\oplus^{\ell^1}_{n}L^1([0,n])$ is much smaller than $L^p_{\mathrm{comp}}$.

This is related to these posts and here.

Let $L^1([n,n+1])$ denote the subspace of $L^p$-function on $[0,\infty)$ essentialy supported on $[-n,n]$. Denote the accelerated $\ell^1$-direct sum Banach space $$ \bigoplus_{n \in \mathbb{N}}^{\ell^1} L^1([0,n]) := \left\{ f \in L^1([0,\infty)):\, \sum_{n=1}^{\infty} \left( 2^n\int_{x \in [n,n+1] } |f(x)| dx \right) <\infty \right\}. $$

Alternatively, equip $L^1_{comp}$ be the space of essentially compactly-supported $L^p$ functions, with their usual inductive limit topology obtained by the inductive system $$ \left\{f \in L^1_{loc}: \, esssupp(f)\subseteq [0,n] \right\} \to \left\{f \in L^1_{loc}: \, esssupp(f)\subseteq [0,m] \right\}\qquad n\leq m;\, n,m \in \mathbb{N} . $$ Is the topology of $\oplus^{\ell^1}_{n}L^1([0,n])$ coarser than that of $L^p_{comp}$ on their intersection?

It seems to me to be the case since $\oplus^{\ell^1}_{n}L^1([0,n])$ is much smaller than $L^p_{comp}$.

This is related to these posts and here.

Let $L^1([n,n+1])$ denote the subspace of $L^p$-functions on $[0,\infty)$ essentially supported on $[-n,n]$. Denote the accelerated $\ell^1$-direct sum Banach space $$ \bigoplus_{n \in \mathbb{N}}^{\ell^1} L^1([0,n]) := \left\{ f \in L^1([0,\infty)):\, \sum_{n=1}^{\infty} \left( 2^n\int_{x \in [n,n+1] } |f(x)| dx \right) <\infty \right\}. $$

Alternatively, equip $L^1_{\mathrm{comp}}$ be the space of essentially compactly-supported $L^p$ functions, with their usual inductive limit topology obtained by the inductive system $$ \left\{f \in L^1_{\mathrm{loc}}: \, \operatorname{esssupp}(f)\subseteq [0,n] \right\} \to \left\{f \in L^1_{\mathrm{loc}}: \, \operatorname{esssupp}(f)\subseteq [0,m] \right\}\qquad n\leq m;\, n,m \in \mathbb{N} . $$ Is the topology of $\oplus^{\ell^1}_{n}L^1([0,n])$ coarser than that of $L^p_{\mathrm{comp}}$ on their intersection?

It seems to me to be the case since $\oplus^{\ell^1}_{n}L^1([0,n])$ is much smaller than $L^p_{\mathrm{comp}}$.